Hi,
I am curious if the status of RWinEdt and WinEdt 6.0 has changed since this
thread
http://r.789695.n4.nabble.com/RWinEdt-in-WinEdt-6-td2174285.html
Thanks for the help (especially Uwe Ligges for writing RWinEdt).
Chris
[[alternative HTML version deleted]]
Hi,
I am wondering if there is a way to call Gnuplot from R and/or if anyone can
recommend a package on CRAN capable of doing this?
Thanks,
Chris
PS - Please cc me on the response.
[[alternative HTML version deleted]]
__
R-help@r-project.org
Hi,
I have 45 models that I have named: 1, 2, 3, ... , 45 and I am trying to
plot them in order of ascending BIC values. I am however unclear as to
how I can get the models to line up on the x-axis by BIC and not by
numeric order. For example, if model 5 has a lower BIC than 1, I want it
to
Hi Hadley,
Thanks for the reply and the great graphing package. That code is giving
me the following error:
qplot(reorder(model,delta),delta,data=growthm.bic)
Error in UseMethod(reorder) : no applicable method for reorder
Cheers,
Chris
On 6/28/09 8:21 PM, hadley wickham wrote:
Hi Chris,
Thanks Hadley that worked.
Chris
On 6/29/09 11:05 AM, hadley wickham wrote:
In that case, try:
qplot(reorder(factor(model),delta),delta,data=growthm.bic)
Deepayan: do you think there should also be a numeric method for reorder?
Hadley
On Mon, Jun 29, 2009 at 10:39 AM, Christopher
Hi,
I am trying to combine two data frames by ID. The first data frame is
the whole data set and the second data frame is a subset of the first.
What I would like to do is take the values from variable, p1, from the
second data frame and merge them back into that variable in the first
data
Hi,
I am trying to do run the following model saved in C:/bugs/sus.bug
model {
for (i in 1:n){
y[i] ~ dpois(lamdba[i])
log(lambda[i]) - mu+bmale[male[i]]+bschn[schn[i]]+epsilon[i] #
epsilon[i] ~ dnorm(0,tau.epsilon)
}
mu ~ dnorm(0,.0001)
bmale ~ dnorm(0,.0001)
tau.epsilon -
Hi,
How can I perform the following
y1y2y3y4
1 0 3 2
0 1 2 1
3 2 0 1
0 5 1 0
into ...
y
1 0 3 2 0 1 2 1 3 2 0 1 0 5 1 0
Please cc me on reply as I subscribe to the digest.
Thanks!
Chris
Thanks. That worked.
Chris
On 7/19/09 6:41 PM, jim holtman wrote:
Try this:
x
y1 y2 y3 y4
1 1 0 3 2
2 0 1 2 1
3 3 2 0 1
4 0 5 1 0
as.vector(t(x))
[1] 1 0 3 2 0 1 2 1 3 2 0 1 0 5 1 0
On Sun, Jul 19, 2009 at 7:25 PM, Christopher
Hi,
When I zoom into a graph created in ggplot2 with the
coord_cartesian(ylim=c(0,5)) option, I have no values labelled on my y-axis.
For this graph ggplot2 only puts labels the y-axis at intervals of 10 (i.e.
0, 10, 20, ...). However, the major portion of the graph I am interested in
is located
Hi,
I have the following data:
est
sch190 sch107 sch290 sch256 sch287 sch130
sch139
4.16656026 2.64306071 4.22579866 6.12024789 4.49624748 11.12799127
1.17353917
sch140 sch282 sch161 sch193 sch156 sch288
sch352
') +
geom_pointrange(aes(x=as.factor(sch), y=est, ymin=lower.95ci,
ymax=upper.95ci))+
xlab('School') + ylab(Value-added)+theme_bw()
On 07/07/2011 05:55 PM, Christopher Desjardins wrote:
Hi,
I have the following data:
est
sch190 sch107 sch290 sch256 sch287
Hi,
I have the follow ggplot2 code I am running:
ggplot(data=bb.res.math,aes(x=factor(id.bb),y=bb.math.comb,fill=BB)) +
geom_bar() + facet_grid(BB~.) + scale_fill_brewer(pal=Set1) + ylab(Average
Student Residual (Math)) + xlab(Student ID)
The number of unique id.bb is 2207 and so my X-axis has
Hi,
I am using ggplot2 to with the following code:
gmathk2 -
qplot(time,math,colour=Kids,data=kids.ach.lm.k5,geom=smooth,method=lm,formula=y~ns(x,1))
+ opts(title=Smoother Plot: Math K-5) + xlab(Time) + ylab(Math) +
scale_colour_brewer(pal=Set1); gmathk2
This plots all the smoother for all the x
Hi,
I am trying to get TextWrangler to work with LaTeX and Sweave. Ideally
I would call a script from TextWrangler that would run Sweave on a
document, then LaTeX (using SyncTeX), and finally
open the corresponding pdf in Skim. Of course I don't always need to
run Sweave and would be looking for
I have 3 vectors: p1, p2, and p3. I would like each vector to be any
possible value between 0 and 1 and p1 + p2 + p3 = 1. I want to graph these
and I've thought about using scatterplot3d(). Here's what I have so far.
library(scatterplot3d)
p1 -
- Original Message -
From: Greg Snow greg.s...@imail.org
Date: Tuesday, March 29, 2011 1:02 pm
Subject: Re: [R] Creating 3 vectors that sum to 1
To: Christopher Desjardins cddesjard...@gmail.com,
r-help@r-project.org r-help@r-project.org
Do a search
Hi,
I am trying to write a loop to recode my data from -999 to NA in R. What's
the most efficient way to do this? Below is what I'm presently doing, which
is inefficient. Thanks,
Chris
dat0 - read.table(time1.dat)
colnames(dat0) - c(e1dq, e1arcp, e1dev, s1prcp, s1nrcp, s1ints,
a1gpar,
question or
script writing.
Thanks,
Chris
On Wed, Mar 30, 2011 at 10:15 AM, Christopher Desjardins
cddesjard...@gmail.com wrote:
Hi,
I am trying to write a loop to recode my data from -999 to NA in R.
What's
the most efficient way to do this? Below is what I'm presently doing,
which
On Wed, Mar 30, 2011 at 10:51 AM, peter dalgaard pda...@gmail.com wrote:
On Mar 30, 2011, at 16:05 , Christopher Desjardins wrote:
dat0 - read.table('tim1.dat', na = -999)
Ah ... yes. I knew that but clearly didn't at the time of my question or
script writing.
Thanks,
Chris
Hi,
I have longitudinal school suspension data on students. I would like to
figure out how many students (id_r) have no suspensions (sus), i.e. have a
code of '0'. My data is in long format and the first 20 records look like
the following:
suslm[1:20,c(1,7)]
id_r sus
11 0
15 10
16
On Wed, Apr 6, 2011 at 4:03 PM, Douglas Bates ba...@stat.wisc.edu wrote:
On Wed, Apr 6, 2011 at 3:44 PM, Christopher Desjardins
cddesjard...@gmail.com wrote:
Hi,
I have longitudinal school suspension data on students. I would like to
figure out how many students (id_r) have no suspensions
2224252630
3132
FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE
FALSE FALSE
33
FALSE
names(r[ r == TRUE ])
[1] 15 25
Regards,
Jorge
On Wed, Apr 6, 2011 at 5:03 PM, Christopher Desjardins wrote:
Thanks. And how many could I find
On Thu, Apr 7, 2011 at 8:07 AM, Christopher Desjardins
cddesjard...@gmail.com wrote:
Hi Jorge,
I want to make sure this does what I want.
So I want to get a count of students that never get a suspension. Once a
student has a non-zero I don't want to count that student. Each id_r is may
Hi,
I am wondering if anyone knows how to perform an F-test on the change in R
square between hierarchical models in R? SPSS provides this information and
a researcher that I am working with is interested in getting this
information. Alternatively, if someone knows how I can calculate the test
Hi,
I have a pretty simple problem. Here is the code:
dat1=complete(dat.mice,1)
dat2=complete(dat.mice,2)
dat3=complete(dat.mice,3)
dat4=complete(dat.mice,4)
dat5=complete(dat.mice,5)
dat6=complete(dat.mice,6)
dat7=complete(dat.mice,7)
dat8=complete(dat.mice,8)
dat9=complete(dat.mice,9)
Hi,
I am wondering if there is a way to change the pattern of the fill in
histogram in ggplot2? By default the fill is solid and I'd like to add some
sort of pattern to make it more visible that these are different levels of a
factor.
Thanks!
Chris
[[alternative HTML version deleted]]
Hi I am running the following code:
sym - c(sym1,sym2,sym4)
lifedxm - c(O-BD,O-WELL,O-UNI)
life - c(lifedxm,lifedxm,lifedxm)
tp - c(TP-ANY,TP-ANY, TP-ANY, TP-SUB, TP-SUB, TP-SUB, TP-CLIN
, TP-CLIN, TP-CLIN)
data - data.frame(sym,life,tp)
Hi,
I have the following data management issue. I am trying to combine multiple
years of ethnicity data into one variable called ethnic. The data looks
similar to the following
idethnic07ethnic08 ethnic09ethnic10
1 1 1 11
2
/chris/Dropbox/phd/analysis/Simulation/zinbmodels/summaries/zinbsumstat-,j,k,l,m,.rdata,
sep=)
load(fsumstatZINB)
print(zinbMeans[,1])
result[[paste(j,k,l,m)]] - zinbMeans
}
}
}
}
On Mon, Feb 4, 2013 at 12:02 PM, Christopher Desjardins
cddesjard
Hi,
Let's say I have the following data:
a=matrix(c(1,2,4,4,2,1,1,2,4),nrow=3,byrow=T)
a
[,1] [,2] [,3]
[1,]124
[2,]421
[3,]124
What syntax should I use to get R to tell me the column that corresponds to
the maximum value for each row?
For my
Hi,
I want to submit a package to CRAN and I am getting the following Note:
* checking CRAN incoming feasibility ... NOTE
New submission
How can I take care of this? And/or is it a big deal?
Thanks and sorry if this is something that I easily overlooked I have
googled this topic for a while
Hi,
I have the following data:
data[1:20,c(1,2,20)]
idr schyear year
1 80
1 91
1 10 NA
2 4 NA
2 5 -1
2 60
2 71
2 82
2 93
2 104
2 11 NA
2 126
3 4 NA
3 5 -2
3 6 -1
3
6
3.13 3 4 -3
3.14 3 5 -2
3.15 3 6 -1
3.16 3 70
3.17 3 81
3.18 3 92
3.19 3 103
3.20 3 114
On Sat, Nov 3, 2012 at 1:14 PM, Christopher Desjardins
cddesjard...@gmail.com wrote:
Hi,
I have
.
Thanks,
Chris
On Sat, Nov 3, 2012 at 5:50 PM, Christopher Desjardins
cddesjard...@gmail.com wrote:
Hi Jim,
Thank you so much. That does exactly what I want.
Chris
On Sat, Nov 3, 2012 at 1:30 PM, jim holtman jholt...@gmail.com wrote:
x - read.table(text = idr schyear year
+ 1 8
1120
A.K.
- Original Message -
From: William Dunlap wdun...@tibco.com
To: arun smartpink...@yahoo.com; Christopher Desjardins
cddesjard...@gmail.com
Cc: R help r-help@r-project.org
Sent: Saturday, November 3, 2012 9:21 PM
Subject: RE: [R] Replacing NAs in long
Hi,
I have my data the following way:
y A B C
0 11 2
0 12 1
1 11 2
0 11 2
1 11 2
1 12 1
0 12 2
.
.
.
And so on. How can I make my data look like the following:
y A B C
2 1 1 2
1 1 2 1
0 1 2 2
.
.
.
of Anthropology
Texas AM University
College Station, TX 77843-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Christopher Desjardins
Sent: Thursday, July 19, 2012 7:35 PM
To: R help
Subject: [R] Changing ungrouped cases
12
0 1 2 2 1
.
.
.
So I would know there were 2 successes out of 4.
Thanks!
Chris
On Fri, Jul 20, 2012 at 10:41 AM, Christopher Desjardins
cddesjard...@gmail.com wrote:
Thanks the aggregate() command is what I was looking for.
Chris
On Thu, Jul 19, 2012 at 9:03 PM, David L
Hi,
I am wondering if there is an easy way to access the hat matrix for
zeroinfl and hurdle objects in the pscl library?
Thanks,
Chris
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Hi,
I have the following data.
set.seed(6245)
data - data.frame(x1=rnorm(5),x2=rnorm(5),x3=rnorm(5),x4=rnorm(5))
round(data,digits=3)
x1 x2 x3 x4
1 0.482 1.320 -0.859 -0.142
2 -0.753 -0.041 -0.063 0.886
3 0.028 -0.256 -0.069 0.354
4 -0.086 0.475 0.244 0.781
5 0.690
)]-NA;x})
data2
# x1 x2 x3 x4
#1 0.482 1.320 -0.859 -0.142
#2 -0.753 -0.041 NA NA
#3 0.028 -0.256 -0.069 0.354
#4 -0.086 0.475 0.244 0.781
#5 0.690 -0.181 1.274 1.633
A.K.
- Original Message -
From: Christopher Desjardins cddesjard
Does this do what you want?
library(Amelia)
library(Zelig)
library(stargazer)
library(xtable)
data(africa)
m = 10
imp1 - amelia(x = africa,cs=country,m=m)
imp2 - amelia(x = africa,cs=country,m=m)
lm.imputed1 - zelig(infl ~ trade + civlib, model=ls,data = imp1)
lm.imputed2 - zelig(infl ~ trade +
)
but then I get my error:
Error: Unrecognized object type.
Even though your example is insightful, I can't figure out how to solve
my problem.
Any advice is very welcome.
Regards,
f.
On 17 August 2013 17:02, Christopher Desjardins cddesjard...@gmail.comwrote:
Does this do what you want
Oh and are you looking for just the summarized results over all the imputed
runs? i thought you wanted them from each iteration.
On Sat, Aug 17, 2013 at 11:38 AM, Christopher Desjardins
cddesjard...@gmail.com wrote:
What do you mean by results? Do you want just the estimated parameters
Seems you're after the pooled results. Would the following work?
library(Amelia)
library(Zelig)
library(xtable)
data(africa)
m = 10
imp1 - amelia(x = africa,cs=country,m=m)
imp2 - amelia(x = africa,cs=country,m=m)
lm.imputed1 - zelig(gdp_pc ~ trade + civlib, model=ls,data = imp1)
lm.imputed2
.
Is it possible that there is no way to get nicely latex formatted tables
concerning multiply imputed data-set?
But maybe I should open a separate thread on this.
Thanks a lot for your kind and patient help.
Best regards,
f.
On 18 August 2013 15:56, Christopher Desjardins
cddesjard
Hi,
I am running the following code based on the cpm vignette's code. I believe
the code is syntactically correct but it just seems to hang R. I can get
this to run if I set the sims to 100 but with 2000 it just hangs. Any ideas
why?
Thanks,
Chris
library(cpm)
cpmTypes -
Hi,
I am having trouble with syntax for a for loop. Here is what I am trying to
do.
class=c(rep(1,3),rep(2,3),rep(3,3))
out1=rnorm(length(class))
out2=rnorm(length(class))
out3=rnorm(length(class))
data=data.frame(class,out1,out2,out3)
dat.split=split(data,data$class)
for(i in 1:3){
Hi,
I am trying to replicate Lambert (1992)'s simulation with zero-inflated
Poisson models. The citation is here:
@article{lambert1992zero,
Author = {Lambert, D.},
Journal = {Technometrics},
Pages = {1--14},
Publisher = {JSTOR},
Title = {Zero-inflated {P}oisson regression, with an application to
On Fri, Apr 27, 2012 at 1:53 AM, Achim Zeileis achim.zeil...@uibk.ac.atwrote:
On Thu, 26 Apr 2012, Christopher Desjardins wrote:
Hi,
I am trying to replicate Lambert (1992)'s simulation with zero-inflated
Poisson models. The citation is here:
@article{lambert1992zero,
Author = {Lambert
Hi,
I am a little confused at the output from predict() for a zeroinfl object.
Here's my confusion:
## From zeroinfl package
fm_zinb2 - zeroinfl(art ~ . | ., data = bioChemists, dist = negbin)
## The raw zero-inflated overdispersed data
table(bioChemists$art)
0 1 2 3 4 5 6 7
This might be useful: http://adv-r.had.co.nz/
On Tue, Apr 5, 2016 at 10:31 AM, Francisco Banha wrote:
> Dear All,
>
> I'm currently working on a project with
> the purpose of remotely executing R code, which requires me to have to
> work with the code of R itself.
Hi Jose,
You're referring to your response variable when you're saying it's missing
some of the choices, right? Are your response choices ever known or do they
just occur with extremely low frequency? Either way, I think the mlogit
package would be inappropriate for you. I imagine you would have
54 matches
Mail list logo