24 Oct
2012
24 Oct
'12
3:50 p.m.
I have a data frame called t:
dput(t)
structure(list(Volume = c(2625941L, 4685483L, 3160694L, 2627816L,
2430273L, 2498011L), SLA = c(28L, 44L, 12L, 28L, 4L, 28L), Duration = c(21L,
25L, 15L, 13L, 15L, 20L)), .Names = c("Volume", "SLA",
"Duration"
), row.names = 2:7, class = "data.frame")
I would like to create a linear model and also prection using zelig library. When I do
this:
Ø z.out1 <- zelig(Duration ~ Volume, model = "ls", data = t)
I get this error. I also tried gamma, logit, the same error:
Error in `rownames<-`(`*tmp*`, value = c(1L, 0L)) :
attempt to set rownames on object with no dimensions
Any ideas what I am doing wrong?
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24 Oct
24 Oct
3:53 p.m.
New subject: zelig to perfom linear model
Try renaming your data frame. There is an R function called t, so R is
probably confused.
If you still have problems and are on Zelig version 4x, try Zelig version
3.5.x.
Best, Olivia
On Oct 24, 2012 12:50 PM, "Ceylan, Hasan"
<hasan.ceylan(a)credit-suisse.com>
wrote:
I have a data frame called t:****
** **
dput(t)****
structure(list(Volume = c(2625941L, 4685483L, 3160694L, 2627816L, ****
2430273L, 2498011L), SLA = c(28L, 44L, 12L, 28L, 4L, 28L), Duration =
c(21L, ****
25L, 15L, 13L, 15L, 20L)), .Names = c("Volume", "SLA",
"Duration"****
), row.names = 2:7, class = "data.frame")****
** **
I would like to create a linear model and also prection using zelig
library. When I do this:****
** **
**Ø ** z.out1 <- zelig(Duration ~ Volume, model = "ls", data = t)****
** **
I get this error. I also tried gamma, logit, the same error:****
** **
Error in `rownames<-`(`*tmp*`, value = c(1L, 0L)) : ****
attempt to set rownames on object with no dimensions****
** **
Any ideas what I am doing wrong?****
**
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4:03 p.m.
New subject: zelig to perfom linear model
Olivia,
That was it. I changed t to df. It work. I also did this:
out1 <- zelig(Duration ~ Volume, model = "probit", data = df)
x.out1 <- setx(out1, Volume = 20000000)
s.out1 <- sim(z.out1, x = x.out1)
I have this graph. How do you read this graph?
Hasan Ceylan
Tools & Engineering NY
From: Olivia Lau [mailto:monkeykupo@gmail.com]
Sent: Wednesday, October 24, 2012 3:54 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
Try renaming your data frame. There is an R function called t, so R is probably confused.
If you still have problems and are on Zelig version 4x, try Zelig version 3.5.x.
Best, Olivia
On Oct 24, 2012 12:50 PM, "Ceylan, Hasan" <hasan.ceylan(a)credit-suisse.com>
wrote:
I have a data frame called t:
dput(t)
structure(list(Volume = c(2625941L, 4685483L, 3160694L, 2627816L,
2430273L, 2498011L), SLA = c(28L, 44L, 12L, 28L, 4L, 28L), Duration = c(21L,
25L, 15L, 13L, 15L, 20L)), .Names = c("Volume", "SLA",
"Duration"
), row.names = 2:7, class = "data.frame")
I would like to create a linear model and also prection using zelig library. When I do
this:
Ø z.out1 <- zelig(Duration ~ Volume, model = "ls", data = t)
I get this error. I also tried gamma, logit, the same error:
Error in `rownames<-`(`*tmp*`, value = c(1L, 0L)) :
attempt to set rownames on object with no dimensions
Any ideas what I am doing wrong?
==============================================================================
Please access the attached hyperlink for an important electronic communications
disclaimer:
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4:15 p.m.
New subject: zelig to perfom linear model
In this case, you just need to do summary(s.out1) and look at the
Expected Value summary. This will give you a numeric summary of the
plot that you pasted below (mean, sd, etc). Basically, the plot (and
the EV summary) give you the Duration that you should expect when you
have Volume = 20000000.
Note, if Duration is actually a time interval, the linear model is
probably not what you want. Try something like exponential, weibull,
lognormal, or cox proportional hazards.
On Wed, Oct 24, 2012 at 1:03 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
Olivia,
That was it. I changed t to df. It work. I also did this:
out1 <- zelig(Duration ~ Volume, model = "probit", data = df)
x.out1 <- setx(out1, Volume = 20000000)
s.out1 <- sim(z.out1, x = x.out1)
I have this graph. How do you read this graph?
Hasan Ceylan
Tools & Engineering NY
From: Olivia Lau [mailto:monkeykupo@gmail.com]
Sent: Wednesday, October 24, 2012 3:54 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
Try renaming your data frame. There is an R function called t, so R is probably
confused.
If you still have problems and are on Zelig version 4x, try Zelig version 3.5.x.
Best, Olivia
On Oct 24, 2012 12:50 PM, "Ceylan, Hasan"
<hasan.ceylan(a)credit-suisse.com> wrote:
I have a data frame called t:
dput(t)
structure(list(Volume = c(2625941L, 4685483L, 3160694L, 2627816L,
2430273L, 2498011L), SLA = c(28L, 44L, 12L, 28L, 4L, 28L), Duration = c(21L,
25L, 15L, 13L, 15L, 20L)), .Names = c("Volume", "SLA",
"Duration"
), row.names = 2:7, class = "data.frame")
I would like to create a linear model and also prection using zelig library. When I do
this:
Ø z.out1 <- zelig(Duration ~ Volume, model = "ls", data = t)
I get this error. I also tried gamma, logit, the same error:
Error in `rownames<-`(`*tmp*`, value = c(1L, 0L)) :
attempt to set rownames on object with no dimensions
Any ideas what I am doing wrong?
==============================================================================
Please access the attached hyperlink for an important electronic communications
disclaimer:
http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html
==============================================================================
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4:29 p.m.
New subject: zelig to perfom linear model
Olivia:
Two questions:
1. http://127.0.0.1:30545/doc/html/Search?package=Zelig&agrep=FALSE&ty…
Only shows a few model names, where can I get all the models that zelig supports. Duration
is in seconds.
2. this is the ouput of summary(s.out1)
Expected Values: E(Y|X)
mean sd 50% 2.5% 97.5%
22.92 1.14 22.845 20.847 25.437
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%
22.906 8.788 21.514 9.662 42.936
Under predicted Values, 42 is interpreted as 97% probability that duration will be
42.936?
-----Original Message-----
From: monkeykupo(a)gmail.com [mailto:monkeykupo@gmail.com] On Behalf Of Olivia Lau
Sent: Wednesday, October 24, 2012 4:16 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
In this case, you just need to do summary(s.out1) and look at the
Expected Value summary. This will give you a numeric summary of the
plot that you pasted below (mean, sd, etc). Basically, the plot (and
the EV summary) give you the Duration that you should expect when you
have Volume = 20000000.
Note, if Duration is actually a time interval, the linear model is
probably not what you want. Try something like exponential, weibull,
lognormal, or cox proportional hazards.
On Wed, Oct 24, 2012 at 1:03 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
Olivia,
That was it. I changed t to df. It work. I also did this:
out1 <- zelig(Duration ~ Volume, model = "probit", data = df)
x.out1 <- setx(out1, Volume = 20000000)
s.out1 <- sim(z.out1, x = x.out1)
I have this graph. How do you read this graph?
Hasan Ceylan
Tools & Engineering NY
From: Olivia Lau [mailto:monkeykupo@gmail.com]
Sent: Wednesday, October 24, 2012 3:54 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
Try renaming your data frame. There is an R function called t, so R is probably
confused.
If you still have problems and are on Zelig version 4x, try Zelig version 3.5.x.
Best, Olivia
On Oct 24, 2012 12:50 PM, "Ceylan, Hasan"
<hasan.ceylan(a)credit-suisse.com> wrote:
I have a data frame called t:
dput(t)
structure(list(Volume = c(2625941L, 4685483L, 3160694L, 2627816L,
2430273L, 2498011L), SLA = c(28L, 44L, 12L, 28L, 4L, 28L), Duration = c(21L,
25L, 15L, 13L, 15L, 20L)), .Names = c("Volume", "SLA",
"Duration"
), row.names = 2:7, class = "data.frame")
I would like to create a linear model and also prection using zelig library. When I do
this:
Ø z.out1 <- zelig(Duration ~ Volume, model = "ls", data = t)
I get this error. I also tried gamma, logit, the same error:
Error in `rownames<-`(`*tmp*`, value = c(1L, 0L)) :
attempt to set rownames on object with no dimensions
Any ideas what I am doing wrong?
==============================================================================
Please access the attached hyperlink for an important electronic communications
disclaimer:
http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html
==============================================================================
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==============================================================================
===============================================================================
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disclaimer:
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===============================================================================
4:31 p.m.
New subject: zelig to perfom linear model
See the full manual online: http://r.iq.harvard.edu/docs/zelig.pdf
On Wed, Oct 24, 2012 at 1:29 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
Olivia:
Two questions:
1. http://127.0.0.1:30545/doc/html/Search?package=Zelig&agrep=FALSE&ty…
Only shows a few model names, where can I get all the models that zelig supports.
Duration is in seconds.
2. this is the ouput of summary(s.out1)
Expected Values: E(Y|X)
mean sd 50% 2.5% 97.5%
22.92 1.14 22.845 20.847 25.437
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%
22.906 8.788 21.514 9.662 42.936
Under predicted Values, 42 is interpreted as 97% probability that duration will be
42.936?
-----Original Message-----
From: monkeykupo(a)gmail.com [mailto:monkeykupo@gmail.com] On Behalf Of Olivia Lau
Sent: Wednesday, October 24, 2012 4:16 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
In this case, you just need to do summary(s.out1) and look at the
Expected Value summary. This will give you a numeric summary of the
plot that you pasted below (mean, sd, etc). Basically, the plot (and
the EV summary) give you the Duration that you should expect when you
have Volume = 20000000.
Note, if Duration is actually a time interval, the linear model is
probably not what you want. Try something like exponential, weibull,
lognormal, or cox proportional hazards.
On Wed, Oct 24, 2012 at 1:03 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
Olivia,
That was it. I changed t to df. It work. I also did this:
out1 <- zelig(Duration ~ Volume, model = "probit", data = df)
x.out1 <- setx(out1, Volume = 20000000)
s.out1 <- sim(z.out1, x = x.out1)
I have this graph. How do you read this graph?
Hasan Ceylan
Tools & Engineering NY
From: Olivia Lau [mailto:monkeykupo@gmail.com]
Sent: Wednesday, October 24, 2012 3:54 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
Try renaming your data frame. There is an R function called t, so R is probably
confused.
If you still have problems and are on Zelig version 4x, try Zelig version 3.5.x.
Best, Olivia
On Oct 24, 2012 12:50 PM, "Ceylan, Hasan"
<hasan.ceylan(a)credit-suisse.com> wrote:
I have a data frame called t:
dput(t)
structure(list(Volume = c(2625941L, 4685483L, 3160694L, 2627816L,
2430273L, 2498011L), SLA = c(28L, 44L, 12L, 28L, 4L, 28L), Duration = c(21L,
25L, 15L, 13L, 15L, 20L)), .Names = c("Volume", "SLA",
"Duration"
), row.names = 2:7, class = "data.frame")
I would like to create a linear model and also prection using zelig library. When I do
this:
Ø z.out1 <- zelig(Duration ~ Volume, model = "ls", data = t)
I get this error. I also tried gamma, logit, the same error:
Error in `rownames<-`(`*tmp*`, value = c(1L, 0L)) :
attempt to set rownames on object with no dimensions
Any ideas what I am doing wrong?
==============================================================================
Please access the attached hyperlink for an important electronic communications
disclaimer:
http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html
==============================================================================
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==============================================================================
===============================================================================
Please access the attached hyperlink for an important electronic communications
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===============================================================================
25 Oct
25 Oct
10:12 a.m.
New subject: zelig to perfom linear model
Thank you Olivia. I just have one quick question. Results from summary(s.out) states
this:
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%
22.77 8.266 21.826 9.764 41.305
Does this say, 97.5% confident that the result will be 41.305?
Thanks,
hasan
-----Original Message-----
From: monkeykupo(a)gmail.com [mailto:monkeykupo@gmail.com] On Behalf Of Olivia Lau
Sent: Wednesday, October 24, 2012 4:32 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
See the full manual online: http://r.iq.harvard.edu/docs/zelig.pdf
On Wed, Oct 24, 2012 at 1:29 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
Olivia:
Two questions:
1. http://127.0.0.1:30545/doc/html/Search?package=Zelig&agrep=FALSE&ty…
Only shows a few model names, where can I get all the models that zelig supports.
Duration is in seconds.
2. this is the ouput of summary(s.out1)
Expected Values: E(Y|X)
mean sd 50% 2.5% 97.5%
22.92 1.14 22.845 20.847 25.437
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%
22.906 8.788 21.514 9.662 42.936
Under predicted Values, 42 is interpreted as 97% probability that duration will be
42.936?
-----Original Message-----
From: monkeykupo(a)gmail.com [mailto:monkeykupo@gmail.com] On Behalf Of Olivia Lau
Sent: Wednesday, October 24, 2012 4:16 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
In this case, you just need to do summary(s.out1) and look at the
Expected Value summary. This will give you a numeric summary of the
plot that you pasted below (mean, sd, etc). Basically, the plot (and
the EV summary) give you the Duration that you should expect when you
have Volume = 20000000.
Note, if Duration is actually a time interval, the linear model is
probably not what you want. Try something like exponential, weibull,
lognormal, or cox proportional hazards.
On Wed, Oct 24, 2012 at 1:03 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
===============================================================================
Please access the attached hyperlink for an important electronic communications
disclaimer:
http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html
===============================================================================
Olivia,
That was it. I changed t to df. It work. I also did this:
out1 <- zelig(Duration ~ Volume, model = "probit", data = df)
x.out1 <- setx(out1, Volume = 20000000)
s.out1 <- sim(z.out1, x = x.out1)
I have this graph. How do you read this graph?
Hasan Ceylan
Tools & Engineering NY
From: Olivia Lau [mailto:monkeykupo@gmail.com]
Sent: Wednesday, October 24, 2012 3:54 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
Try renaming your data frame. There is an R function called t, so R is probably
confused.
If you still have problems and are on Zelig version 4x, try Zelig version 3.5.x.
Best, Olivia
On Oct 24, 2012 12:50 PM, "Ceylan, Hasan"
<hasan.ceylan(a)credit-suisse.com> wrote:
I have a data frame called t:
dput(t)
structure(list(Volume = c(2625941L, 4685483L, 3160694L, 2627816L,
2430273L, 2498011L), SLA = c(28L, 44L, 12L, 28L, 4L, 28L), Duration = c(21L,
25L, 15L, 13L, 15L, 20L)), .Names = c("Volume", "SLA",
"Duration"
), row.names = 2:7, class = "data.frame")
I would like to create a linear model and also prection using zelig library. When I do
this:
Ø z.out1 <- zelig(Duration ~ Volume, model = "ls", data = t)
I get this error. I also tried gamma, logit, the same error:
Error in `rownames<-`(`*tmp*`, value = c(1L, 0L)) :
attempt to set rownames on object with no dimensions
Any ideas what I am doing wrong?
==============================================================================
Please access the attached hyperlink for an important electronic communications
disclaimer:
http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html
==============================================================================
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===============================================================================
12:12 p.m.
New subject: zelig to perfom linear model
Nope, the output means that the 95% confidence interval is (9.8, 41.3). We
are 95% "confident" that the result will be in that interval.
On Thu, Oct 25, 2012 at 10:12 AM, Ceylan, Hasan <
hasan.ceylan(a)credit-suisse.com> wrote:
Thank you Olivia. I just have one quick question.
Results from
summary(s.out) states this:
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%
22.77 8.266 21.826 9.764 41.305
Does this say, 97.5% confident that the result will be 41.305?
Thanks,
hasan
-----Original Message-----
From: monkeykupo(a)gmail.com [mailto:monkeykupo@gmail.com] On Behalf Of
Olivia Lau
Sent: Wednesday, October 24, 2012 4:32 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
See the full manual online: http://r.iq.harvard.edu/docs/zelig.pdf
On Wed, Oct 24, 2012 at 1:29 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
>
If you still have problems and are on Zelig version 4x, try Zelig version 3.5.x.
>
Best, Olivia
>
> On Oct 24, 2012 12:50 PM, "Ceylan, Hasan" <
hasan.ceylan(a)credit-suisse.com> wrote:
>
I have a data frame called t:
>
>
>
> dput(t)
>
> structure(list(Volume = c(2625941L, 4685483L, 3160694L, 2627816L,
>
> 2430273L, 2498011L), SLA = c(28L, 44L, 12L, 28L, 4L, 28L), Duration = c(21L,
>
25L, 15L, 13L, 15L, 20L)), .Names = c("Volume", "SLA",
"Duration"
>
> ), row.names = 2:7, class = "data.frame")
>
>
>
> I would like to create a linear model and also prection using zelig library.
When I do this:
>
>
> Ø z.out1 <- zelig(Duration ~ Volume, model = "ls", data = t)
>
>
>
> I get this error. I also tried gamma, logit, the same error:
>
>
>
> Error in `rownames<-`(`*tmp*`, value = c(1L, 0L)) :
>
> attempt to set rownames on object with no dimensions
>
>
>
> Any ideas what I am doing wrong?
>
>
>
>
==============================================================================
>
>
>
> -
> --
> Zelig Mailing List, served by HUIT
> Send messages: zelig(a)lists.gking.harvard.edu
> [un]subscribe Options: http://lists.gking.harvard.edu/mailman/listinfo/zelig
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===============================================================================
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communications disclaimer:
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===============================================================================
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--
Patrick Lam
Department of Government and Institute for Quantitative Social Science,
Harvard University
http://www.people.fas.harvard.edu/~plam
Olivia:
Two questions:
1.
http://127.0.0.1:30545/doc/html/Search?package=Zelig&agrep=FALSE&ty…
Only shows a few model names, where can I get all
the models that zelig
supports. Duration is in seconds.
2. this is the ouput of summary(s.out1)
Expected Values: E(Y|X)
mean sd 50% 2.5% 97.5%
22.92 1.14 22.845 20.847 25.437
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%
22.906 8.788 21.514 9.662 42.936
Under predicted Values, 42 is interpreted as 97% probability that
duration will be
42.936?
-----Original Message-----
From: monkeykupo(a)gmail.com [mailto:monkeykupo@gmail.com] On Behalf Of
Olivia Lau
Sent: Wednesday, October 24, 2012 4:16 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
In this case, you just need to do summary(s.out1) and look at the
Expected Value summary. This will give you a numeric summary of the
plot that you pasted below (mean, sd, etc). Basically, the plot (and
the EV summary) give you the Duration that you should expect when you
have Volume = 20000000.
Note, if Duration is actually a time interval, the linear model is
probably not what you want. Try something like exponential, weibull,
lognormal, or cox proportional hazards.
On Wed, Oct 24, 2012 at 1:03 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
>
> Olivia,
>
>
>
> That was it. I changed t to df. It work. I also did this:
>
>
>
> out1 <- zelig(Duration ~ Volume, model = "probit", data = df)
>
> x.out1 <- setx(out1, Volume = 20000000)
>
> s.out1 <- sim(z.out1, x = x.out1)
>
>
>
> I have this graph. How do you read this graph?
>
>
>
>
>
> Hasan Ceylan
>
> Tools & Engineering NY
>
>
>
> From: Olivia Lau [mailto:monkeykupo@gmail.com]
> Sent: Wednesday, October 24, 2012 3:54 PM
> To: Ceylan, Hasan
> Cc: zelig(a)lists.gking.harvard.edu
> Subject: Re: [zelig] zelig to perfom linear model
>
>
>
> Try renaming your data frame. There is an R function called t, so R is
probably confused.
> Please access the attached hyperlink for an
important electronic
communications disclaimer:
==============================================================================
> Zelig program information:
http://gking.harvard.edu/zelig/
> Zelig mailing list
> Zelig(a)lists.gking.harvard.edu
>
> To unsubscribe from this list or get other information:
>
> https://lists.gking.harvard.edu/mailman/listinfo/zelig
>
>
>
==============================================================================
> Please access the attached hyperlink for an
important electronic
communications disclaimer:
==============================================================================
Please access the attached hyperlink for an
important electronic
communications disclaimer:
===============================================================================
12:22 p.m.
New subject: zelig to perfom linear model
Are you sure? As I understand, this sounds like out of 1000 simulation, 97.5% of the time,
it was 41.305?
Hasan Ceylan
Tools & Engineering NY
From: pkphlam(a)gmail.com [mailto:pkphlam@gmail.com] On Behalf Of Patrick Lam
Sent: Thursday, October 25, 2012 12:13 PM
To: Ceylan, Hasan
Cc: Olivia Lau; zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
Nope, the output means that the 95% confidence interval is (9.8, 41.3). We are 95%
"confident" that the result will be in that interval.
On Thu, Oct 25, 2012 at 10:12 AM, Ceylan, Hasan <hasan.ceylan(a)credit-suisse.com>
wrote:
Thank you Olivia. I just have one quick question. Results from summary(s.out) states
this:
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%
22.77 8.266 21.826 9.764 41.305
Does this say, 97.5% confident that the result will be 41.305?
Thanks,
hasan
-----Original Message-----
From: monkeykupo(a)gmail.com [mailto:monkeykupo@gmail.com] On Behalf Of Olivia Lau
Sent: Wednesday, October 24, 2012 4:32 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
See the full manual online: http://r.iq.harvard.edu/docs/zelig.pdf
On Wed, Oct 24, 2012 at 1:29 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
Olivia:
Two questions:
1. http://127.0.0.1:30545/doc/html/Search?package=Zelig&agrep=FALSE&ty…
Only shows a few model names, where can I get all the models that zelig supports.
Duration is in seconds.
2. this is the ouput of summary(s.out1)
Expected Values: E(Y|X)
mean sd 50% 2.5% 97.5%
22.92 1.14 22.845 20.847 25.437
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%
22.906 8.788 21.514 9.662 42.936
Under predicted Values, 42 is interpreted as 97% probability that duration will be
42.936?
-----Original Message-----
From: monkeykupo(a)gmail.com [mailto:monkeykupo@gmail.com] On Behalf Of Olivia Lau
Sent: Wednesday, October 24, 2012 4:16 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
In this case, you just need to do summary(s.out1) and look at the
Expected Value summary. This will give you a numeric summary of the
plot that you pasted below (mean, sd, etc). Basically, the plot (and
the EV summary) give you the Duration that you should expect when you
have Volume = 20000000.
Note, if Duration is actually a time interval, the linear model is
probably not what you want. Try something like exponential, weibull,
lognormal, or cox proportional hazards.
On Wed, Oct 24, 2012 at 1:03 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
===============================================================================
Please access the attached hyperlink for an important electronic communications
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Patrick Lam
Department of Government and Institute for Quantitative Social Science, Harvard
University
http://www.people.fas.harvard.edu/~plam
===============================================================================
Please access the attached hyperlink for an important electronic communications
disclaimer:
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===============================================================================
Olivia,
That was it. I changed t to df. It work. I also did this:
out1 <- zelig(Duration ~ Volume, model = "probit", data = df)
x.out1 <- setx(out1, Volume = 20000000)
s.out1 <- sim(z.out1, x = x.out1)
I have this graph. How do you read this graph?
Hasan Ceylan
Tools & Engineering NY
From: Olivia Lau [mailto:monkeykupo@gmail.com]
Sent: Wednesday, October 24, 2012 3:54 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
Try renaming your data frame. There is an R function called t, so R is probably
confused.
If you still have problems and are on Zelig version 4x, try Zelig version 3.5.x.
Best, Olivia
On Oct 24, 2012 12:50 PM, "Ceylan, Hasan"
<hasan.ceylan(a)credit-suisse.com> wrote:
I have a data frame called t:
dput(t)
structure(list(Volume = c(2625941L, 4685483L, 3160694L, 2627816L,
2430273L, 2498011L), SLA = c(28L, 44L, 12L, 28L, 4L, 28L), Duration = c(21L,
25L, 15L, 13L, 15L, 20L)), .Names = c("Volume", "SLA",
"Duration"
), row.names = 2:7, class = "data.frame")
I would like to create a linear model and also prection using zelig library. When I do
this:
Ø z.out1 <- zelig(Duration ~ Volume, model = "ls", data = t)
I get this error. I also tried gamma, logit, the same error:
Error in `rownames<-`(`*tmp*`, value = c(1L, 0L)) :
attempt to set rownames on object with no dimensions
Any ideas what I am doing wrong?
==============================================================================
Please access the attached hyperlink for an important electronic communications
disclaimer:
http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html
==============================================================================
-
--
Zelig Mailing List, served by HUIT
Send messages: zelig(a)lists.gking.harvard.edu
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==============================================================================
===============================================================================
Please access the attached hyperlink for an important electronic communications
disclaimer:
http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html
===============================================================================
12:25 p.m.
New subject: zelig to perfom linear model
No that is incorrect. The 97.5% and 2.5% numbers are the 97.5 and 2.5
quantiles of the distribution of simulations.
On Thu, Oct 25, 2012 at 12:22 PM, Ceylan, Hasan <
hasan.ceylan(a)credit-suisse.com> wrote:
Are you sure? As I understand, this sounds like out of
1000 simulation,
97.5% of the time, it was 41.305?****
** **
Hasan Ceylan****
Tools & Engineering NY****
** **
*From:* pkphlam(a)gmail.com [mailto:pkphlam@gmail.com] *On Behalf Of *Patrick
Lam
*Sent:* Thursday, October 25, 2012 12:13 PM
*To:* Ceylan, Hasan
*Cc:* Olivia Lau; zelig(a)lists.gking.harvard.edu
*Subject:* Re: [zelig] zelig to perfom linear model****
** **
Nope, the output means that the 95% confidence interval is (9.8, 41.3).
We are 95% "confident" that the result will be in that interval.****
On Thu, Oct 25, 2012 at 10:12 AM, Ceylan, Hasan <
hasan.ceylan(a)credit-suisse.com> wrote:****
Thank you Olivia. I just have one quick question. Results from
summary(s.out) states this:****
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%****
22.77 8.266 21.826 9.764 41.305
Does this say, 97.5% confident that the result will be 41.305?
Thanks,
hasan****
-----Original Message-----
From: monkeykupo(a)gmail.com [mailto:monkeykupo@gmail.com] On Behalf Of
Olivia Lau****
Sent: Wednesday, October 24, 2012 4:32 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
See the full manual online: http://r.iq.harvard.edu/docs/zelig.pdf
On Wed, Oct 24, 2012 at 1:29 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
>
If you still have problems and are on Zelig version 4x, try Zelig version 3.5.x.
>
Best, Olivia
>
> On Oct 24, 2012 12:50 PM, "Ceylan, Hasan" <
hasan.ceylan(a)credit-suisse.com> wrote:
>
I have a data frame called t:
>
>
>
> dput(t)
>
> structure(list(Volume = c(2625941L, 4685483L, 3160694L, 2627816L,
>
> 2430273L, 2498011L), SLA = c(28L, 44L, 12L, 28L, 4L, 28L), Duration = c(21L,
>
25L, 15L, 13L, 15L, 20L)), .Names = c("Volume", "SLA",
"Duration"
>
> ), row.names = 2:7, class = "data.frame")
>
>
>
> I would like to create a linear model and also prection using zelig library.
When I do this:
>
>
> Ø z.out1 <- zelig(Duration ~ Volume, model = "ls", data = t)
>
>
>
> I get this error. I also tried gamma, logit, the same error:
>
>
>
> Error in `rownames<-`(`*tmp*`, value = c(1L, 0L)) :
>
> attempt to set rownames on object with no dimensions
>
>
>
> Any ideas what I am doing wrong?
>
>
>
>
==============================================================================
>
>
>
> -
> --
> Zelig Mailing List, served by HUIT
> Send messages: zelig(a)lists.gking.harvard.edu
> [un]subscribe Options: http://lists.gking.harvard.edu/mailman/listinfo/zelig
===============================================================================
===============================================================================
Please access the attached hyperlink for an important electronic
communications disclaimer:
http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html
===============================================================================
-
--
Zelig Mailing List, served by HUIT
Send messages: zelig(a)lists.gking.harvard.edu
[un]subscribe Options:
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Zelig mailing list
Zelig(a)lists.gking.harvard.edu
To unsubscribe from this list or get other information:
https://lists.gking.harvard.edu/mailman/listinfo/zelig****
****
** **
--
Patrick Lam
Department of Government and Institute for Quantitative Social Science,
Harvard University
http://www.people.fas.harvard.edu/~plam****
**
==============================================================================
Please access the attached hyperlink for an important electronic
communications disclaimer:
http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html
==============================================================================
****
--
Patrick Lam
Department of Government and Institute for Quantitative Social Science,
Harvard University
http://www.people.fas.harvard.edu/~plam
Olivia:
Two questions:
1.
http://127.0.0.1:30545/doc/html/Search?package=Zelig&agrep=FALSE&ty…
Only shows a few model names, where can I get all
the models that zelig
supports. Duration is in seconds.
2. this is the ouput of summary(s.out1)
Expected Values: E(Y|X)
mean sd 50% 2.5% 97.5%
22.92 1.14 22.845 20.847 25.437
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%
22.906 8.788 21.514 9.662 42.936
Under predicted Values, 42 is interpreted as 97% probability that
duration will be
42.936?
-----Original Message-----
From: monkeykupo(a)gmail.com [mailto:monkeykupo@gmail.com] On Behalf Of
Olivia Lau
Sent: Wednesday, October 24, 2012 4:16 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
In this case, you just need to do summary(s.out1) and look at the
Expected Value summary. This will give you a numeric summary of the
plot that you pasted below (mean, sd, etc). Basically, the plot (and
the EV summary) give you the Duration that you should expect when you
have Volume = 20000000.
Note, if Duration is actually a time interval, the linear model is
probably not what you want. Try something like exponential, weibull,
lognormal, or cox proportional hazards.
On Wed, Oct 24, 2012 at 1:03 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
>
> Olivia,
>
>
>
> That was it. I changed t to df. It work. I also did this:
>
>
>
> out1 <- zelig(Duration ~ Volume, model = "probit", data = df)
>
> x.out1 <- setx(out1, Volume = 20000000)
>
> s.out1 <- sim(z.out1, x = x.out1)
>
>
>
> I have this graph. How do you read this graph?
>
>
>
>
>
> Hasan Ceylan
>
> Tools & Engineering NY
>
>
>
> From: Olivia Lau [mailto:monkeykupo@gmail.com]
> Sent: Wednesday, October 24, 2012 3:54 PM
> To: Ceylan, Hasan
> Cc: zelig(a)lists.gking.harvard.edu
> Subject: Re: [zelig] zelig to perfom linear model
>
>
>
> Try renaming your data frame. There is an R function called t, so R is
probably confused.
> Please access the attached hyperlink for an
important electronic
communications disclaimer:
==============================================================================
> Zelig program information:
http://gking.harvard.edu/zelig/
> Zelig mailing list
> Zelig(a)lists.gking.harvard.edu
>
> To unsubscribe from this list or get other information:
>
> https://lists.gking.harvard.edu/mailman/listinfo/zelig
>
>
>
==============================================================================
> Please access the attached hyperlink for an
important electronic
communications disclaimer:
==============================================================================
Please access the attached hyperlink for an
important electronic
communications disclaimer:
===============================================================================
12:38 p.m.
New subject: zelig to perfom linear model
Lam:
Thank you for your explanation. I have one other question. Using different models,
sometimes I don't get "Predicted Values". Does this mean, this model is not
able to perform the predictive analysis? How can we easily tell which models are the best
one given our data?
Regards,
hasan
From: pkphlam(a)gmail.com [mailto:pkphlam@gmail.com] On Behalf Of Patrick Lam
Sent: Thursday, October 25, 2012 12:26 PM
To: Ceylan, Hasan
Cc: Olivia Lau; zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
No that is incorrect. The 97.5% and 2.5% numbers are the 97.5 and 2.5 quantiles of the
distribution of simulations.
On Thu, Oct 25, 2012 at 12:22 PM, Ceylan, Hasan <hasan.ceylan(a)credit-suisse.com>
wrote:
Are you sure? As I understand, this sounds like out of 1000 simulation, 97.5% of the time,
it was 41.305?
Hasan Ceylan
Tools & Engineering NY
From: pkphlam(a)gmail.com [mailto:pkphlam@gmail.com] On Behalf Of Patrick Lam
Sent: Thursday, October 25, 2012 12:13 PM
To: Ceylan, Hasan
Cc: Olivia Lau; zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
Nope, the output means that the 95% confidence interval is (9.8, 41.3). We are 95%
"confident" that the result will be in that interval.
On Thu, Oct 25, 2012 at 10:12 AM, Ceylan, Hasan <hasan.ceylan(a)credit-suisse.com>
wrote:
Thank you Olivia. I just have one quick question. Results from summary(s.out) states
this:
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%
22.77 8.266 21.826 9.764 41.305
Does this say, 97.5% confident that the result will be 41.305?
Thanks,
hasan
-----Original Message-----
From: monkeykupo(a)gmail.com [mailto:monkeykupo@gmail.com] On Behalf Of Olivia Lau
Sent: Wednesday, October 24, 2012 4:32 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
See the full manual online: http://r.iq.harvard.edu/docs/zelig.pdf
On Wed, Oct 24, 2012 at 1:29 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
Olivia:
Two questions:
1. http://127.0.0.1:30545/doc/html/Search?package=Zelig&agrep=FALSE&ty…
Only shows a few model names, where can I get all the models that zelig supports.
Duration is in seconds.
2. this is the ouput of summary(s.out1)
Expected Values: E(Y|X)
mean sd 50% 2.5% 97.5%
22.92 1.14 22.845 20.847 25.437
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%
22.906 8.788 21.514 9.662 42.936
Under predicted Values, 42 is interpreted as 97% probability that duration will be
42.936?
-----Original Message-----
From: monkeykupo(a)gmail.com [mailto:monkeykupo@gmail.com] On Behalf Of Olivia Lau
Sent: Wednesday, October 24, 2012 4:16 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
In this case, you just need to do summary(s.out1) and look at the
Expected Value summary. This will give you a numeric summary of the
plot that you pasted below (mean, sd, etc). Basically, the plot (and
the EV summary) give you the Duration that you should expect when you
have Volume = 20000000.
Note, if Duration is actually a time interval, the linear model is
probably not what you want. Try something like exponential, weibull,
lognormal, or cox proportional hazards.
On Wed, Oct 24, 2012 at 1:03 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
===============================================================================
Please access the attached hyperlink for an important electronic communications
disclaimer:
http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html
===============================================================================
-
--
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Send messages: zelig(a)lists.gking.harvard.edu
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Zelig mailing list
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--
Patrick Lam
Department of Government and Institute for Quantitative Social Science, Harvard
University
http://www.people.fas.harvard.edu/~plam
==============================================================================
Please access the attached hyperlink for an important electronic communications
disclaimer:
http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html
==============================================================================
--
Patrick Lam
Department of Government and Institute for Quantitative Social Science, Harvard
University
http://www.people.fas.harvard.edu/~plam
===============================================================================
Please access the attached hyperlink for an important electronic communications
disclaimer:
http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html
===============================================================================
Olivia,
That was it. I changed t to df. It work. I also did this:
out1 <- zelig(Duration ~ Volume, model = "probit", data = df)
x.out1 <- setx(out1, Volume = 20000000)
s.out1 <- sim(z.out1, x = x.out1)
I have this graph. How do you read this graph?
Hasan Ceylan
Tools & Engineering NY
From: Olivia Lau [mailto:monkeykupo@gmail.com]
Sent: Wednesday, October 24, 2012 3:54 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
Try renaming your data frame. There is an R function called t, so R is probably
confused.
If you still have problems and are on Zelig version 4x, try Zelig version 3.5.x.
Best, Olivia
On Oct 24, 2012 12:50 PM, "Ceylan, Hasan"
<hasan.ceylan(a)credit-suisse.com> wrote:
I have a data frame called t:
dput(t)
structure(list(Volume = c(2625941L, 4685483L, 3160694L, 2627816L,
2430273L, 2498011L), SLA = c(28L, 44L, 12L, 28L, 4L, 28L), Duration = c(21L,
25L, 15L, 13L, 15L, 20L)), .Names = c("Volume", "SLA",
"Duration"
), row.names = 2:7, class = "data.frame")
I would like to create a linear model and also prection using zelig library. When I do
this:
Ø z.out1 <- zelig(Duration ~ Volume, model = "ls", data = t)
I get this error. I also tried gamma, logit, the same error:
Error in `rownames<-`(`*tmp*`, value = c(1L, 0L)) :
attempt to set rownames on object with no dimensions
Any ideas what I am doing wrong?
==============================================================================
Please access the attached hyperlink for an important electronic communications
disclaimer:
http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html
==============================================================================
-
--
Zelig Mailing List, served by HUIT
Send messages: zelig(a)lists.gking.harvard.edu
[un]subscribe Options: http://lists.gking.harvard.edu/mailman/listinfo/zelig
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Zelig mailing list
Zelig(a)lists.gking.harvard.edu
To unsubscribe from this list or get other information:
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==============================================================================
Please access the attached hyperlink for an important electronic communications
disclaimer:
http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html
==============================================================================
===============================================================================
Please access the attached hyperlink for an important electronic communications
disclaimer:
http://www.credit-suisse.com/legal/en/disclaimer_email_ib.html
===============================================================================
12:40 p.m.
New subject: zelig to perfom linear model
Hasan,
Please start by reading the link to the manual that I sent you
yesterday. Most Zelig users start there. It explains everything.
Thanks,
Olivia
On Thu, Oct 25, 2012 at 9:38 AM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
Lam:
Thank you for your explanation. I have one other question. Using different
models, sometimes I don’t get “Predicted Values”. Does this mean, this model
is not able to perform the predictive analysis? How can we easily tell which
models are the best one given our data?
Regards,
hasan
From: pkphlam(a)gmail.com [mailto:pkphlam@gmail.com] On Behalf Of Patrick Lam
Sent: Thursday, October 25, 2012 12:26 PM
To: Ceylan, Hasan
Cc: Olivia Lau; zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
No that is incorrect. The 97.5% and 2.5% numbers are the 97.5 and 2.5
quantiles of the distribution of simulations.
On Thu, Oct 25, 2012 at 12:22 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
Are you sure? As I understand, this sounds like out of 1000 simulation,
97.5% of the time, it was 41.305?
Hasan Ceylan
Tools & Engineering NY
From: pkphlam(a)gmail.com [mailto:pkphlam@gmail.com] On Behalf Of Patrick Lam
Sent: Thursday, October 25, 2012 12:13 PM
To: Ceylan, Hasan
Cc: Olivia Lau; zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
Nope, the output means that the 95% confidence interval is (9.8, 41.3). We
are 95% "confident" that the result will be in that interval.
On Thu, Oct 25, 2012 at 10:12 AM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
Thank you Olivia. I just have one quick question. Results from
summary(s.out) states this:
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%
22.77 8.266 21.826 9.764 41.305
Does this say, 97.5% confident that the result will be 41.305?
Thanks,
hasan
-----Original Message-----
From: monkeykupo(a)gmail.com [mailto:monkeykupo@gmail.com] On Behalf Of Olivia
Lau
Sent: Wednesday, October 24, 2012 4:32 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
See the full manual online: http://r.iq.harvard.edu/docs/zelig.pdf
On Wed, Oct 24, 2012 at 1:29 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
Olivia:
Two questions:
1.
http://127.0.0.1:30545/doc/html/Search?package=Zelig&agrep=FALSE&ty…
Only shows a few model names, where can I get all the models that zelig
supports. Duration is in seconds.
2. this is the ouput of summary(s.out1)
Expected Values: E(Y|X)
mean sd 50% 2.5% 97.5%
22.92 1.14 22.845 20.847 25.437
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%
22.906 8.788 21.514 9.662 42.936
Under predicted Values, 42 is interpreted as 97% probability that duration
will be 42.936?
-----Original Message-----
From: monkeykupo(a)gmail.com [mailto:monkeykupo@gmail.com] On Behalf Of
Olivia Lau
Sent: Wednesday, October 24, 2012 4:16 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
In this case, you just need to do summary(s.out1) and look at the
Expected Value summary. This will give you a numeric summary of the
plot that you pasted below (mean, sd, etc). Basically, the plot (and
the EV summary) give you the Duration that you should expect when you
have Volume = 20000000.
Note, if Duration is actually a time interval, the linear model is
probably not what you want. Try something like exponential, weibull,
lognormal, or cox proportional hazards.
On Wed, Oct 24, 2012 at 1:03 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com> wrote:
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Olivia,
That was it. I changed t to df. It work. I also did this:
out1 <- zelig(Duration ~ Volume, model = "probit", data = df)
x.out1 <- setx(out1, Volume = 20000000)
s.out1 <- sim(z.out1, x = x.out1)
I have this graph. How do you read this graph?
Hasan Ceylan
Tools & Engineering NY
From: Olivia Lau [mailto:monkeykupo@gmail.com]
Sent: Wednesday, October 24, 2012 3:54 PM
To: Ceylan, Hasan
Cc: zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
Try renaming your data frame. There is an R function called t, so R is
probably confused.
If you still have problems and are on Zelig version 4x, try Zelig version
3.5.x.
Best, Olivia
On Oct 24, 2012 12:50 PM, "Ceylan, Hasan"
<hasan.ceylan(a)credit-suisse.com> wrote:
I have a data frame called t:
dput(t)
structure(list(Volume = c(2625941L, 4685483L, 3160694L, 2627816L,
2430273L, 2498011L), SLA = c(28L, 44L, 12L, 28L, 4L, 28L), Duration =
c(21L,
25L, 15L, 13L, 15L, 20L)), .Names = c("Volume", "SLA",
"Duration"
), row.names = 2:7, class = "data.frame")
I would like to create a linear model and also prection using zelig
library. When I do this:
Ø z.out1 <- zelig(Duration ~ Volume, model = "ls", data = t)
I get this error. I also tried gamma, logit, the same error:
Error in `rownames<-`(`*tmp*`, value = c(1L, 0L)) :
attempt to set rownames on object with no dimensions
Any ideas what I am doing wrong?
==============================================================================
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2:34 p.m.
New subject: zelig to perfom linear model
I've read the pdf file. It does not answer my questions. It is hard to read the
plots:
For example, given summary(s.out1):
> See the full manual online: http://r.iq.harvard.edu/docs/zelig.pdf
<http://r.iq.harvard.edu/docs/zelig.pdf
>
>
>
>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>> From: Olivia Lau [mailto:monkeykupo@gmail.com]
<mailto:[mailto:monkeykupo@gmail.com]
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>> Zelig program information: http://gking.harvard.edu/zelig/
<http://gking.harvard.edu/zelig/
>>
>>> https://lists.gking.harvard.edu/mailman/listinfo/zelig
<https://lists.gking.harvard.edu/mailman/listinfo/zelig
>>
>>
>>
>>
>>
>
>
>
> [un]subscribe Options: http://lists.gking.harvard.edu/mailman/listinfo/zelig
<http://lists.gking.harvard.edu/mailman/listinfo/zelig
> Zelig program information: http://gking.harvard.edu/zelig/
<http://gking.harvard.edu/zelig/
> https://lists.gking.harvard.edu/mailman/listinfo/zelig
<https://lists.gking.harvard.edu/mailman/listinfo/zelig
===============================================================================
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===============================================================================
summary(s.out1)
Model: gamma
Number of simulations: 1000
Values of X
(Intercept) Volume
2 1 2625941
attr(,"assign")
[1] 0 1
Expected Values: E(Y|X)
mean sd 50% 2.5% 97.5%
22.888 1.255 22.785 20.678 25.617
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%
22.763 7.846 22.258 10.305 39.864
What does those graph mean? What is in xaxis and what is is yaxis?
-----Original Message-----
From: monkeykupo(a)gmail.com [mailto:monkeykupo@gmail.com] On Behalf Of Olivia Lau
Sent: Thursday, October 25, 2012 12:40 PM
To: Ceylan, Hasan
Cc: Patrick Lam; zelig(a)lists.gking.harvard.edu
Subject: Re: [zelig] zelig to perfom linear model
Hasan,
Please start by reading the link to the manual that I sent you
yesterday. Most Zelig users start there. It explains everything.
Thanks,
Olivia
On Thu, Oct 25, 2012 at 9:38 AM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com <mailto:hasan.ceylan@credit-suisse.com> >
wrote:
Lam:
Thank you for your explanation. I have one other
question. Using different
models, sometimes I don't get "Predicted
Values". Does this mean, this model
is not able to perform the predictive analysis? How
can we easily tell which
models are the best one given our data?
Regards,
hasan
From: pkphlam(a)gmail.com
<mailto:pkphlam@gmail.com> [mailto:pkphlam@gmail.com]
<mailto:[mailto:pkphlam@gmail.com]> On Behalf Of Patrick Lam
Sent: Thursday, October 25, 2012 12:26 PM
To: Ceylan, Hasan
> Cc: Olivia Lau; zelig(a)lists.gking.harvard.edu
<mailto:zelig@lists.gking.harvard.edu Subject: Re: [zelig] zelig to perfom linear model
No that is incorrect. The 97.5% and 2.5% numbers are
the 97.5 and 2.5
quantiles of the distribution of simulations.
On Thu, Oct 25, 2012 at 12:22 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com
<mailto:hasan.ceylan@credit-suisse.com> > wrote:
Are you sure? As I understand, this sounds like out of
1000 simulation,
97.5% of the time, it was 41.305?
Hasan Ceylan
Tools & Engineering NY
From: pkphlam(a)gmail.com
<mailto:pkphlam@gmail.com> [mailto:pkphlam@gmail.com]
<mailto:[mailto:pkphlam@gmail.com]> On Behalf Of Patrick Lam
Sent: Thursday, October 25, 2012 12:13 PM
To: Ceylan, Hasan
> Cc: Olivia Lau; zelig(a)lists.gking.harvard.edu
<mailto:zelig@lists.gking.harvard.edu Subject: Re: [zelig] zelig to perfom linear model
Nope, the output means that the 95% confidence
interval is (9.8, 41.3). We
are 95% "confident" that the result will be
in that interval.
On Thu, Oct 25, 2012 at 10:12 AM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com
<mailto:hasan.ceylan@credit-suisse.com> > wrote:
Thank you Olivia. I just have one quick question.
Results from
summary(s.out) states this:
Predicted Values: Y|X
mean sd 50% 2.5% 97.5%
22.77 8.266 21.826 9.764 41.305
Does this say, 97.5% confident that the result will be
41.305?
Thanks,
hasan
-----Original Message-----
From: monkeykupo(a)gmail.com
<mailto:monkeykupo@gmail.com> [mailto:monkeykupo@gmail.com]
<mailto:[mailto:monkeykupo@gmail.com]> On Behalf Of Olivia
Lau
Sent: Wednesday, October 24, 2012 4:32 PM
To: Ceylan, Hasan
> Cc: zelig(a)lists.gking.harvard.edu <mailto:zelig@lists.gking.harvard.edu Subject: Re: [zelig] zelig to perfom linear model
On Wed, Oct 24, 2012 at 1:29 PM, Ceylan, Hasan
<hasan.ceylan(a)credit-suisse.com
<mailto:hasan.ceylan@credit-suisse.com> > wrote:
> Olivia:
> > Two questions:
> > 1.
>>
http://127.0.0.1:30545/doc/html/Search?package=Zelig&agrep=FALSE&ty…
<http://127.0.0.1:30545/doc/html/Search?package=Zelig&agrep=FALSE&types=demo > Only shows a few model names, where can I get all
the models that zelig
> supports. Duration is in seconds.
> > 2. this is the ouput of summary(s.out1)
> > Expected Values: E(Y|X)
> mean sd 50% 2.5% 97.5%
> 22.92 1.14 22.845 20.847 25.437
> Predicted Values: Y|X
> mean sd 50% 2.5% 97.5%
> 22.906 8.788 21.514 9.662 42.936
> > Under predicted Values, 42 is interpreted as 97%
probability that duration
> will be 42.936?
> -----Original Message-----
> From: monkeykupo(a)gmail.com
<mailto:monkeykupo@gmail.com> [mailto:monkeykupo@gmail.com]
<mailto:[mailto:monkeykupo@gmail.com]> On Behalf Of
> Olivia Lau
> Sent: Wednesday, October 24, 2012 4:16 PM
> To: Ceylan, Hasan
>> Cc: zelig(a)lists.gking.harvard.edu <mailto:zelig@lists.gking.harvard.edu Subject: Re: [zelig] zelig to perfom linear model
> > In this case, you just need to do summary(s.out1)
and look at the
> Expected Value summary. This will give you a
numeric summary of the
> plot that you pasted below (mean, sd, etc).
Basically, the plot (and
> the EV summary) give you the Duration that you
should expect when you
> have Volume = 20000000.
> > Note, if Duration is actually a time interval, the
linear model is
> probably not what you want. Try something like
exponential, weibull,
> lognormal, or cox proportional hazards.
> > On Wed, Oct 24, 2012 at 1:03 PM, Ceylan, Hasan
> <hasan.ceylan(a)credit-suisse.com
<mailto:hasan.ceylan@credit-suisse.com> > wrote:
>> >> Olivia,
>> >> That was it. I changed t to df. It work. I
also did this:
>> >> out1 <- zelig(Duration ~ Volume, model =
"probit", data = df)
>> >> x.out1 <- setx(out1, Volume = 20000000)
>> >> s.out1 <- sim(z.out1, x = x.out1)
>> >> I have this graph. How do you read this graph?
>> Hasan Ceylan
>> Tools & Engineering NY
>> >> Sent: Wednesday, October 24, 2012 3:54 PM
>> To: Ceylan, Hasan
>>> Cc: zelig(a)lists.gking.harvard.edu
<mailto:zelig@lists.gking.harvard.edu Subject: Re: [zelig] zelig
to perfom linear model
>> >> Try renaming your data frame. There is an R
function called t, so R is
>> probably confused.
>> >> If you still have problems and are on Zelig
version 4x, try Zelig version
>> 3.5.x.
>> >> Best, Olivia
>> >> On Oct 24, 2012 12:50 PM, "Ceylan,
Hasan"
>>
<hasan.ceylan(a)credit-suisse.com <mailto:hasan.ceylan@credit-suisse.com> >
wrote:
>> >> I have a data frame called t:
>> >> dput(t)
>> >> structure(list(Volume = c(2625941L, 4685483L,
3160694L, 2627816L,
>> >> 2430273L, 2498011L), SLA = c(28L, 44L, 12L,
28L, 4L, 28L), Duration =
>> c(21L,
>> >> 25L, 15L, 13L, 15L, 20L)), .Names =
c("Volume", "SLA", "Duration"
>> >> ), row.names = 2:7, class =
"data.frame")
>> >> I would like to create a linear model and also
prection using zelig
>> library. When I do this:
>> >> Ø z.out1 <- zelig(Duration ~ Volume,
model = "ls", data = t)
>> >> I get this error. I also tried gamma, logit,
the same error:
>> >> Error in `rownames<-`(`*tmp*`, value =
c(1L, 0L)) :
>> >> attempt to set rownames on object with no
dimensions
>> >> Any ideas what I am doing wrong?
>> >>
==============================================================================
>> Please access the attached hyperlink for an
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>> communications disclaimer:
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Department of Government and Institute for
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Harvard University
> http://www.people.fas.harvard.edu/~plam
<http://www.people.fas.harvard.edu/~plam
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4224
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12 comments
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Patrick Lam