Suppose that we have the following dataframe:
set.seed(1)
(tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace =
TRUE), R2 = sample(LETTERS[1:5], 10, replace = TRUE)))
x R1 R2
1 1 B B
2 2 B A
3 3 C D
4 4 E B
5 5 B D
6 6 E C
7 7 E D
8 8 D E
9 9
swapme = (as.numeric(tmp$R1) - as.numeric(tmp$R2)) %% 2 !=
> 0),
> {
> tmp[swapme, "R1"] <- r2[swapme]
> tmp[swapme, "R2"] <- r1[swapme]
> tmp
> })
>
> Best,
> Ista
>
> On Thu, Jul 6, 2017 at 4:06 PM, Gang Chen wrote:
>>
With the following data in data.frame:
subject QMemotion yi
s1 75.1017 neutral -75.928276
s2 -47.3512 neutral -178.295990
s3 -68.9016 neutral -134.753906
s1 17.2099 negative -104.168312
s2 -53.1114 negative -182.373474
s3 -33.0322 negative -137.420410
I can
s3 -50.9669 -136.08716
>
>
>
> Jim Holtman
> Data Munger Guru
>
> What is the problem that you are trying to solve?
> Tell me what you want to do, not how you want to do it.
>
> On Thu, Jul 28, 2016 at 4:40 PM, Gang Chen wrote:
>>
>> With the follo
This is a simple question: With a dataframe like the following
myData <- data.frame(X=c(1, 2, 3, 4), Y=c(4, 3, 2, 1), Z=c('A', 'A', 'B', 'B'))
how can I get the cross product between X and Y for each level of
factor Z? My difficulty is that I don't know how to deal with the fact
that crossprod()
ge Station, TX 77840-4352
>
>
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Jim Lemon
> Sent: Tuesday, August 23, 2016 6:02 PM
> To: Gang Chen; r-help mailing list
> Subject: Re: [R] aggregate
>
> Hi Gang Chen,
> I
], x[, 2])))
> Z CP
> A A 10
> B B 10
>
> David C
>
>
> -Original Message-
> From: Gang Chen [mailto:gangch...@gmail.com]
> Sent: Wednesday, August 24, 2016 10:17 AM
> To: David L Carlson
> Cc: Jim Lemon; r-help mailing list
> Subject: Re: [R] aggregate
cture(c(1L,
> 1L, 2L, 2L, 1L, 1L, 2L, 2L), .Label = c("A", "B"), class = "factor")), .Names
> = c("X",
> "Y", "S", "Z"), row.names = c(NA, -8L), class = "data.frame")
>
> Combining two labels just requires the
38 3.2 S2 A
> S2B 22 3.2 S2 B
>
> David C
>
> -Original Message-
> From: Gang Chen [mailto:gangch...@gmail.com]
> Sent: Wednesday, August 24, 2016 2:51 PM
> To: David L Carlson
> Cc: r-help mailing list
> Subject: Re: [R] aggregate
>
> Thanks again for patiently o
I want to do the following: if a string does not contain a colon (:),
no change is needed; if it contains one or more colons, break the
string into multiple strings using the colon as a separator. For
example, "happy:" becomes
"happy" ":"
":sad" turns to
":" "sad"
and "happy:sad" changes to
"h
I'm having some trouble with Anova() in package "car". When the model
formula is explicitly expressed:
library('nlme')
library('car')
fm <- lme(distance ~ age + Sex, data = Orthodont, random = ~ 1)
Anova() works fine:
Anova(fm)
However, if the model formula is scanned from an external source:
fter the
> package is next built on R-Forge, usually in a day or so.
>
> Best,
> John
>
>> -Original Message-
>> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Gang Chen
>> Sent: January-13-15 1:48 PM
>> To: r-help
>> Subject: [R] Pro
A random effect formulation for R package nlme is read in as a string
of characters from an input file:
ranEff <- "pdCompSymm(~1+Age)"
I need to convert 'ranEff' to a formula class. However, as shown below:
> as.formula(ranEff)
~1 + Age
the "pdCompSymm" is lost in the conversion. Any solutions?
Thanks for the help! However, I just need to get
pdCompSymm(~1 + Age)
without a tilde (~) at the beginning.
On Wed, Aug 27, 2014 at 3:34 PM, David Winsemius wrote:
>
> On Aug 27, 2014, at 11:19 AM, Gang Chen wrote:
>
>> A random effect formulation for R package nlme is read
rocess automatic.
On Wed, Aug 27, 2014 at 3:49 PM, David Winsemius wrote:
>
> On Aug 27, 2014, at 12:44 PM, Gang Chen wrote:
>
>> Thanks for the help! However, I just need to get
>>
>> pdCompSymm(~1 + Age)
>
> That's not a formula in the R sense of th
Sorry for the misspelling! And more importantly, thanks a lot for the
nice solution and for the quick help!
On Wed, Aug 27, 2014 at 4:22 PM, David Winsemius wrote:
>
> On Aug 27, 2014, at 1:11 PM, Gang Chen wrote:
>
>> Good point!
>>
>> Here is an example:
>>
&g
When R starts in GUI (e.g., /Applications/R.app/Contents/MacOS/R) on
my Mac OS X 10.7.5, the startup configuration in .Rprofile works fine.
However, when R starts on the terminal (e.g.,
/Library/Frameworks/R.framework/Resources/bin/R), it does not work at
all. What could be the reason for the failu
at does R say is the user's home directory? Did you make
> *any* changes to Rprofile.site, or Renviron?
>
> What is the output from Sys.getenv() in gui and cli, and do they differ?
>
>
>> On Sep 18, 2014, at 11:18 AM, Gang Chen wrote:
>>
>> When R starts in GUI
I’m running R 3.2.2 on a Linux server (Redhat 4.4.7-16), and having the
following problem.
It works fine with the following:
require('MASS’)
var(mvrnorm(n = 1000, rep(0, 2), Sigma=matrix(c(10,3,3,2),2,2)))
However, when running the following in a loop with simulated data (Sigma):
# Sigma defin
>
> Doing anything in 190 dimensions is bound to be fraught with numeric
> peril.
>
> cheers,
>
> Rolf Turner
>
> --
> Technical Editor ANZJS
> Department of Statistics
> University of Auckland
> Phone: +64-9-373-7599 ext. 88276
>
>
> On 19/11/15 08:28,
Hi,
I have a data set collected from 10 measurements (response variables)
on two groups (healthy and patient) of subjects performing 4 different
tasks. In other words there are two fixed factors (group and task),
and 10 response variables. I could analyze the data with aov() or
lme() in package nl
Hi,
I want to store some number of outputs from running a bunch of
analyses such as lm() into an array. I know how to do this with a
one-dimensional array (vector) by creating
myArray <- vector(mode='list', length=10)
and storing each lm() result into a component of myArray.
My question is, how
MAIL PROTECTED]
> +44 (0)20 8525 0696
> http://www.burns-stat.com
> (home of S Poetry and "A Guide for the Unwilling S User")
>
> Gang Chen wrote:
>>
>> Hi,
>>
>> I want to store some number of outputs from running a bunch of
>> analyses such as lm(
Anybody knows what functions can be used to calculate
variance/covariance with complex numbers? var and cov don't seem to
work:
> a
1
V1 0.00810014+0.00169366i
V2 0.00813054+0.00158251i
V3 0.00805489+0.00163295i
V4 0.00809141+0.00159533i
V5 0.00813976+0.00161850i
> var(a)
e elegant
modification than mine?):
> crossprod(t(apply(xri, 1, '-', colMeans(xri/(nrow(xri)-1)
Do you agree?
Gang
On Sat, Mar 27, 2010 at 7:07 PM, Charles C. Berry wrote:
> On Sat, 27 Mar 2010, Gang Chen wrote:
>
>> Anybody knows what functions can be used to calculate
&
I've written a function, myFunc, that works fine with myFunc(data,
...), but when I use apply() to run it with an array of data
apply(myArray, 1, myFunc, ...)
I get a strange error:
Error in match.fun(FUN) : '1' is not a function, character or symbol
which really puzzles me because '1' is meant
I have some bits stored like the following variable nn
(nn <- c(1, 0, 0, 1, 0, 1,0))
[1] 1 0 0 1 0 1 0
not in the format of
1001010
and I need to convert them to numbers in base 10. What's an easy way to do it?
TIA,
Gang
__
R-help@r-project.org mail
t;,collapse="")
>>
>> See ?paste for more information.
>>
>> HTH,
>>
>> Jorge
>>
>>
>> On Thu, Apr 9, 2009 at 5:23 PM, Gang Chen wrote:
>>
>>> I have some bits stored like the following variable nn
>>>
>&g
In a classical meta analysis model y_i = X_i * beta_i + e_i, data
{y_i} are assumed to be independent effect sizes. However, I'm
encountering the following two scenarios:
(1) Each source has multiple effect sizes, thus {y_i} are not fully
independent with each other.
(2) Each source has multiple e
ngapore
> http://courses.nus.edu.sg/course/psycwlm/internet/
> -
>
> On Mon, Feb 8, 2010 at 6:48 AM, Gang Chen wrote:
>> Dear Mike,
>>
>> Thanks a lot for the kind help!
>>
>> Actually a
This is most likely a silly question.
First I run the following:
require(car)
mod.ok <- lm(cbind(pre.1, pre.2, pre.3, pre.4, pre.5, post.1, post.2,
post.3, post.4, post.5,
fup.1, fup.2, fup.3, fup.4, fup.5) ~
treatment*gender, data=OBrienKaiser)
phase <- factor(rep(c("pr
uld just call that function directly. If not, you can copy that
> section of the code into your own function to call and have it return the
> object rather than printing.
>
>
> On Wed, Jul 17, 2013 at 3:38 PM, Gang Chen wrote:
>>
>> This is most likely a silly questi
Hi, I have two sets of sensitivity, specificity, positive predictive
value, and negative predictive value, and accuracy from two tests on
the same subjects. Is there an R package that does such paired
comparisons?
Thanks,
Gang Chen
__
R-help@r
Suppose that I need to run a multivariate linear model
Y = X B + E
many times with the same model matrix X but each time with different
response matrix Y. Is there a function available in 'car' package
similar to refit() in lme4 package so that the model matrix X would
not be reassembled each tim
I have a matrix 'dd' defined as below:
dd <- t(matrix(c(153.0216306, 1, 7.578366e-35,
13.3696538, 1, 5.114571e-04,
0.8476713, 1, 7.144239e-01,
1.2196050, 1, 5.388764e-01,
2.6349405, 1, 2.090719e-01,
6.0507714, 1, 2.780045e-02), nrow=3, ncol=6))
dimnames(dd)[[2]] <- c('# Chisq', 'DF', 'Pr(>Ch
; 2.6349405 1 2.090719e-01 # ISO
> 6.0507714 1 2.780045e-02 # SEC
>
>
>David Carlson
>
>-----Original Message-
>From: r-help-boun...@r-project.org
>[mailto:r-help-boun...@r-project.org] On Behalf Of Gang Chen
>Sent: Thursday, July 3, 2014 2:56 PM
>To: r-h
8366e-35 # Sex    Â
>#2 13.3696538 1 5.114571e-04 # Volume Â
>#3  0.8476713 1 7.144239e-01 # Weight Â
>#4Â Â 1.2196050Â 1 5.388764e-01 # Intensity
>#5Â Â 2.6349405Â 1 2.090719e-01 # ISOÂ Â Â Â Â
>#6Â Â 6.0507714Â 1 2.780045e-02 # SECÂ Â Â Â Â
>
>A.K.
&g
Suppose I have the following dataframe:
L4 <- LETTERS[1:4]
fac <- sample(L4, 10, replace = TRUE)
(d <- data.frame(x = 1, y = 1:10, fac = fac))
x y fac
1 1 1 B
2 1 2 B
3 1 3 D
4 1 4 A
5 1 5 C
6 1 6 D
7 1 7 C
8 1 8 B
9 1 9 B
10 1 10 B
I'd like to add an
mple data and use dput() to include it in your
> email.
>
> Sarah
>
> On Thu, Jul 17, 2014 at 11:00 AM, Gang Chen wrote:
>> Suppose I have the following dataframe:
>>
>> L4 <- LETTERS[1:4]
>> fac <- sample(L4, 10, replace = TRUE)
>> (d <- data.f
range of values you have to work with then one of the other
> more efficient methods may still be a better choice for this specific task.
>
> Hadley Wickham's "tidy data" [1] principles address this concern more
> thoroughly than I have.
>
> [1] Google this phrase...
I wrote an R program that does heavy computations with hundreds of
lines of code. It's running fine both interactively and in batch mode
on a Mac OS X computer. The program also has no problem running on a
Linux system (Fedora 14) interactively. However, when I try it on the
terminal in batch mode
à 12:54 -0500, Gang Chen a écrit :
>> I wrote an R program that does heavy computations with hundreds of
>> lines of code. It's running fine both interactively and in batch mode
>> on a Mac OS X computer. The program also has no problem running on a
>> Linux system (Fedo
Suppose I have a dataframe defined as
L3 <- LETTERS[1:3]
(d <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace
= TRUE)))
x y fac
1 1 1 C
2 1 2 A
3 1 3 B
4 1 4 C
5 1 5 B
6 1 6 B
7 1 7 A
8 1 8 A
9 1 9 B
10 1 10 A
I want to extract
Perfect! Thanks a lot, A.K!
On Fri, Dec 13, 2013 at 4:21 PM, arun wrote:
>
>
> Hi,
> Try:
> d[match(unique(d$fac),d$fac),]
> A.K.
>
>
> On Friday, December 13, 2013 4:17 PM, Gang Chen
> wrote:
> Suppose I have a dataframe defined as
>
> L3 <- LE
en everyone who tries it will get the same data
> frame d.
>
> Sarah
>
>
> On Fri, Dec 13, 2013 at 4:15 PM, Gang Chen wrote:
> > Suppose I have a dataframe defined as
> >
> > L3 <- LETTERS[1:3]
> > (d <- data.frame(cbind(x = 1, y = 1:10), fac
Suppose I have a dataframe 'd' defined as
L3 <- LETTERS[1:3]
d0 <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace
= TRUE))
(d <- d0[d0$fac %in% c('A', 'B'),])
x y fac
2 1 2 B
3 1 3 A
4 1 4 A
5 1 5 A
6 1 6 B
8 1 8 A
Even though factor 'fac' in 'd' onl
gt;
>
> best
> daniel
>
> Feladó: r-help-boun...@r-project.org [r-help-boun...@r-project.org] ;
> meghatalmazó: Gang Chen [gangch...@gmail.com]
> Küldve: 2013. december 14. 21:09
> To: r-help
> Tárgy: [R] Change factor levels
>
> Suppose I have
Hi,
I'm trying to install the gsl wrapper source code
(http://cran.r-project.org/src/contrib/gsl_1.9-8.tar.gz) on a Linux
system (OpenSuse 11.1), but encountering the following problem. I've
already installed 'gsl' version 1.14
(ftp://ftp.gnu.org/gnu/gsl/gsl-1.14.tar.gz) on the system. What's
miss
You nailed it, Prof. Ripley! Thanks a lot...
Gang
On Sat, Oct 30, 2010 at 2:58 PM, Prof Brian Ripley
wrote:
> On Sat, 30 Oct 2010, Gang Chen wrote:
>
>> Hi,
>>
>> I'm trying to install the gsl wrapper source code
>> (http://cran.r-project.org/src/contrib/gsl
I want to create some 3D scatter plot with a diagonal line. In addition, I'd
like to have those points plus the diagonal line projected to those three
planes (xy, yz and xz). Which package can I use to achieve this,
scatterplot3d or something else?
Thanks,
Gang
[[alternative HTML version
Thanks a lot for the quick help! How to project the scatter plot with the
diagonal line to the three planes with scatterplot3d? I could not find such
an example demonstrating that in the vignette.
Thanks,
Gang
2011/1/8 Uwe Ligges
>
>
> On 08.01.2011 16:38, Gang Chen wrote:
>
Yes, too bad I didn't realize that it's so simple like that! Thanks...
On Sat, Jan 8, 2011 at 12:45 PM, David Winsemius wrote:
>
> On Jan 8, 2011, at 11:21 AM, Gang Chen wrote:
>
> Thanks a lot for the quick help! How to project the scatter plot with the
>> diago
I know how to convert a simple dataframe from wide to long format with one
varying factor. However, for a dataset with two factors like the following,
Subj T1_Cond1 T1_Cond2 T2_Cond1 T2_Cond2
1 0.125869 4.108232 1.099392 5.556614
2 1.427940 2.170026 0.120748 1.176353
How to eleg
, 4,8) )
>
> ubj substr(variable, 1, 2)Cond1Cond2
> 1 1 T1 0.125869 4.108232
> 2 1 T2 1.099392 5.556614
> 3 2 T1 1.427940 2.170026
> 4 2 T2 0.120748 1.176353
>
> The modifications to
A very simple question. With a data frame like this:
> n = c(2, 3, 5)
> s = c("aa", "bb", "cc")
> df = data.frame(n, s)
I want df$s[1] or df[1,2], but how can I get rid of the extra line in
the output about the factor levels:
> df$s[1]
[1] aa
Levels: aa bb cc
Thanks,
Gang
_
or.
> --
> Robert Tirrell | r...@stanford.edu | (607) 437-6532
> Program in Biomedical Informatics | Butte Lab | Stanford University
>
>
> On Thu, Mar 10, 2011 at 13:35, Gang Chen wrote:
>>
>> n = c(2, 3, 5)
>> > s = c("aa", "bb", "cc
Statistical Computing Facility
> Department of Statistics
> UC Berkeley
> spec...@stat.berkeley.edu
>
>
>
> On Thu, 10 Mar 2011, Gang Chen w
I read somewhere that vector graphics such as eps or dpf are more favorable
than alternatives (jpeg, bmp or png) for publication because vector graphics
scale properly when enlarged. However, my problem is that the file generated
from a graph of fixed size is too large (in the order of 10MB) becaus
Hi, I have a question about SVAR modeling with the package vars. How does it
handle the situation where the A (structural) matrix has a non-recursive
structure in the SVAR model? In other words, what kind of algorithm does
vars adopt to deal with the unidentifiable issue in a non-recursive model?
I define the following function to convert a t-value with degrees of freedom
DF to another t-value with different degrees of freedom fullDF:
tConvert <- function(tval, DF, fullDF) ifelse(DF>=1, qt(pt(tval, DF),
fullDF), 0)
It works as expected with the following case:
> tConvert(c(2,3), c(10,12)
c(0,12), 12)
[1] 0 3
However, I feel my solution is a little kludged. Any better idea?
Thanks,
Gang
On Thu, Jul 7, 2011 at 9:04 PM, David Winsemius wrote:
>
> On Jul 7, 2011, at 8:52 PM, David Winsemius wrote:
>
>
>> On Jul 7, 2011, at 8:47 PM, Gang Chen wrote:
>>
>&
I have some data 'myData' in wide form (attached at the end), and
would like to convert it to long form. I wish to have five variables
in the result:
1) Subj: factor
2) Group: between-subjects factor (2 levels: s / w)
3) Reference: within-subject factor (2 levels: Me / She)
4) F: within-subject fa
ndrew Miles wrote:
> Take a look here.
> http://stackoverflow.com/questions/2185252/reshaping-data-frame-from-wide-to-long-format
> Andrew Miles
> Department of Sociology
> Duke University
> On Oct 6, 2011, at 4:28 PM, Gang Chen wrote:
>
> I have some data 'myData'
value_var = 'value')
> head(mData4, 4)
>
> Subj Group Ref Time F J
> 1 S1 s Me 1 4 5
> 2 S1 s Me 2 3 6
> 3 S1 s She 1 6 10
> 4 S1 s She 2 6 9
>
> mData5 <- cast(mData3, Subj + Group + Ref + Var ~ Time, value_var
S1 s 6 She F1
6S1 s 6 She F2
7S1 s10 She J1
8S1 s 9 She J2
On Fri, Oct 7, 2011 at 7:16 AM, Jim Lemon wrote:
> On 10/07/2011 07:28 AM, Gang Chen wrote:
>>
>> I have some data 'myData' in wide form (
David, thanks a lot for the code! I've learned quite a bit from all
the generous help...
Gang
On Fri, Oct 7, 2011 at 1:37 PM, David Winsemius wrote:
>
> On Oct 7, 2011, at 1:30 PM, David Winsemius wrote:
>
>>
>> On Oct 7, 2011, at 7:40 AM, Gang Chen wrote:
>>
&
Suppose I create an R program called myTest.R with only one line like
the following:
type <- as.integer(readline("input type (1: type1; 2: type2)? "))
Then I'd like to run myTest.R in batch mode by constructing an input
file called answers.R with the following:
source("myTest.R")
1
When I ran t
from a previous run in
interactive mode, s/he may try out the batch mode the next time.
> On Tue, Feb 7, 2012 at 4:05 PM, Gang Chen wrote:
>> Suppose I create an R program called myTest.R with only one line like
>> the following:
>>
>> type <- as.integer(readlin
e "type<- readline(...)" in your script so type
> wouldn't be overwritten in subsequent runs.
>
> If your goal is to batch evaluate multiple answer files from users
> (why else would you ask questions with readline?), then you should
> have enough to go on with my ans
7;Answers.R')
>
> # run again this time with answers available
>> source('myTest.R')
> [1] -1.088665 # skips prompt
> [1] -1.088665 # -1.088^1 (type in Answer.R ==1)
>
> # Now you can also run as batch
> $ R CMD BATCH myTest.R out.R
> $ cat out.R
&g
ve such problem?
Thanks in advance for any information,
Gang Chen
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-proj
Sorry for this dumb question. Suppose I have a named array ww defined as
ww <- 1:5
names(ww) <- c("a", "b", "c", "d", "e")
How can I extract the whole array of numbers without the names?
ww[1:5] does not work while ww[[1]] can only extract one number at a
time.
Thanks,
Gang
Thanks a lot for the suggestions!
Gang
On 4/7/08, Uwe Ligges <[EMAIL PROTECTED]> wrote:
>
>
> Gang Chen wrote:
>
> > Sorry for this dumb question. Suppose I have a named array ww defined as
> >
> > ww <- 1:5
> > names(ww) <- c("a", &qu
I'm trying to analyze a model with two variables, one is Group with
two levels (male and female), and other is Time with four levels (T1,
T2, T3 and T4). And for the convenience of post-hoc testing I wanted
to consider a model with no intercept for factor Time, so I tried
formula
Group*(Time-1)
H
Using the "ergoStool" data cited in Mixed-Effects Models in S and
S-PLUS by Pinheiro and Bates as an example, we have
> library(nlme)
> fm <- lme(effort~Type-1, data=ergoStool, random=~1|Subject)
> summary(fm)
Linear mixed-effects model fit by REML
Data: ergoStool
AIC BIC
-10 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Adjusted p values reported -- single-step method)
On 4/16/08, Simon Blomberg <[EMAIL PROTECTED]> wrote:
> Try glht in package multcomp.
>
> Simon.
>
>
> On
k the 'convenience' is largely lost,
> and packages such as multcomp can post-hoc test any (coherent) set of
> hypotheses you choose, irrespective of the model parametrization.
>
>
>
> >
> > > -Original Message-
> > > From: [EMAIL PROTECTED]
> &g
I want to identify whether a variable is character(0), but get lost.
For example, if I have
> dd<-character(0)
the following doesn't seem to serve as a good identifier:
> dd==character(0)
logical(0)
So how to detect character(0)?
Thanks,
Gang
__
Thanks a lot for all who've provided suggestions!
Gang
On Nov 15, 2007, at 5:09 PM, Duncan Murdoch wrote:
> On 11/15/2007 4:54 PM, Gang Chen wrote:
>> I want to identify whether a variable is character(0), but get
>> lost. For example, if I have
>> > dd&l
Suppose I have a two-way table of nominal category (party
affiliation) X ordinal category (political ideology):
party affiliation X (3 levels) - democratic, independent, and republic
political ideology Y (3 levels) - liberal, moderate, and conservative
The dependent variable is the frequency (o
I have a dataframe DF with 4 columns (variables) A, B, C, and D, and
want to create a new dataframe DF2 by keeping B and C in DF but
counting the frequency of D while collapsing A. I tried
by(DF$D, list(DF$B, DF$C), FUN=summary)
but this is not exactly what I want. What is a good way to do it
Thanks a lot! This is exactly what I wanted.
Gang
On Jan 5, 2008, at 2:20 PM, David Winsemius wrote:
> Gang Chen <[EMAIL PROTECTED]> wrote in
> news:[EMAIL PROTECTED]:
>
>> Suppose I have a two-way table of nominal category (party
>> affiliation) X ordinal c
C1 D243
B2 C2 D123
B2 C2 D243
Thanks,
Gang
On Jan 7, 2008, at 2:06 PM, Duncan Murdoch wrote:
> On 1/7/2008 1:28 PM, Gang Chen wrote:
>> I have a dataframe DF with 4 columns (variables) A, B, C, and D,
>> and want to create a new dataframe DF2 by keeping B and C
Sorry the new column in DF2 should be called FreqD instead of FreqA.
How can I get DF2 with aggregate?
Gang
On Jan 7, 2008, at 2:38 PM, Gang Chen wrote:
> Yes, I misstated it when I said that I would keep B and C. I want to
> collapse column A, but count the frequency of D as a new col
ts
>> x.new <- x[, -1] # delete "A"
>> x.new$FreqD <- counts # add new column
>> # print out unique entries
>> unique(x.new)
>B C D FreqD
> 1 B1 C1 D1 3
> 2 B2 C1 D1 3
> 4 B2 C2 D2 2
> 7 B1 C2 D2 2
>>
>
>
> On
I'm running a categorical data analysis with a two-way design of
nominal by ordinal structure like the Political Ideology Example
(Table 9.5) in Agresti's book Categorical Data Analysis. The nominal
variable is Method while the ordinal variable is Quality (Bad,
Moderate, Good, Excellent). I
With the example you provided, it seems both glht() and contrast()
work fine.
Based on my limited experience with contrast(), if you encounter such
an error message you just mentioned, check
> dat.lme$apVar
You might see something like this
[1] "Non-positive definite approximate variance
For some unknown reason I stopped receiving any messages from the R-
help mailing list. See if this test gets through.
Thanks,
Gang
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http
I'm trying to use the following loop to open a window multiple times
to select files, but only the last window shows up. What am I missing?
library(tcltk)
nWin <- 6
fn <- vector('list', nWin)
for (ii in nWin) {
fn[[ii]] <- tclvalue( tkgetOpenFile( filetypes =
"{{Files} {.1D}} {{All files
--
> John Fox, Professor
> Department of Sociology
> McMaster University
> Hamilton, Ontario, Canada
> web: socserv.mcmaster.ca/jfox
>
>> -Original Message-
>> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
> On
>> Behalf Of Gang Chen
>> Se
Suppose I have a file prog.R stored in a directory under ~/dirname,
and ~/dirname is set in a shell script file (e.g. .cshrc) as one of
the accessible paths on terminal. On a different directory I could run
prog.R interactively by executing
source("~/dirname/prog.R")
It seems that source() does n
> location:
>
> Find(file.exists, file.path(p, "prog.R"))
>
> so you can source that.
>
> On Wed, Oct 1, 2008 at 6:09 PM, Gang Chen <[EMAIL PROTECTED]> wrote:
>> Suppose I have a file prog.R stored in a directory under ~/dirname,
>> and ~/dirname i
I have a list, myList, with each of its 9 components being a 15X15
matrix. I want to run a t-test across the list for each component in
the matrix. For example, the first t-test is on myList[[1]][1, 1],
myList[[2]][1, 1], ..., myList[[9]][1, 1]; and there are totally 15X15
t-tests. How can I run th
I want to run a R program, prog.R, interactively. My question is, is
there a way I can start prog.R on the shell terminal when invoking R,
instead of using source() inside R?
TIA,
Gang
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R-help@r-project.org mailing list
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shouldn't be doing this. I suggest you consult with your local statistician.
>
> -- Bert Gunter
>
> -Original Message-
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
> Behalf Of Gang Chen
> Sent: Thursday, October 02, 2008 11:24 AM
> To: [EMAIL PROTECTED]
>
t; and "R
--interactive < prog.R", and
they all failed.
Any other suggestions?
Thanks,
Gang
On Mon, Oct 6, 2008 at 10:12 PM, Bernardo Rangel Tura
<[EMAIL PROTECTED]> wrote:
> Em Qui, 2008-10-02 às 14:36 -0400, Gang Chen escreveu:
>> I want to run a R program, prog.R, int
.First.sys at the beginning of
> prog.R to get default packages loaded.
>
> luke
>
> On Tue, 7 Oct 2008, Gang Chen wrote:
>
>> Thanks a lot for the suggestion!
>>
>> Unfortunately " R --no-save < prog.R" does not work well with my
>> situation b
EMAIL PROTECTED]>
> wrote:
>> > I am sure you will get helpful answers. I am almost as sure that you
>> > shouldn't be doing this. I suggest you consult with your local
> statistician.
>> >
>> > -- Bert Gunter
>> >
>> > -O
When invoking dev.new() on my Mac OS X 10.4.11, I get an X11 window
instead of quartz which I feel more desirable. So I'd like to set
the default device to quartz. However I'm confused because of the
following:
> Sys.getenv("R_DEFAULT_DEVICE")
R_DEFAULT_DEVICE
"quartz"
> getOption("devi
I've been using parApply() in snow package for parallel computing with
the following lines in R 2.8.1:
library(snow)
nNodes <- 4
cl <- makeCluster(nNodes, type = "SOCK")
fm <- parApply(cl, myData, c(1,2), func1, ...)
Since I have a Mac OS X (version 10.4.11) with two dual-core
processors,
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