On Tue, 22 Jan 2013, Jason Musil wrote:
DeaR all,
I am using mob() for model based partitioning, with a dichotomous
variable (participant's correct/incorrect response to a test item)
regressed onto a continuous predictor related to a given property of the
test item. Although this variable
Since clusterSetupRNG() calls clusterSetupRNGstream() and this calls
.lec.SetPackageSeed(), I could further minimalize the problem:
set.seed(1)
RNGkind(L'Ecuyer-CMRG) # = .Random.seed is of length 7 (first number encodes
the rng kind)
(seed - .Random.seed[2:7]) # should give a valid seed for
Dear List,
I have a set of data which looks like this (small set of sample)
A A 0.431
A A 0.439
A A 0.507
A G 0.508
A A 0.514
I will like to use this data to plot a dot plot, with the X-axis being of type
character, and my y axis of type numeric.
When I try to use the dot
Dear All,
Sorry for asking a newbie question. I want to ask how to import 1000
datasets whose file names are labelled from data1.dat to data1000.dat into
R so that they are named M[1, , ] to M[1000, , ] accordingly. Thank you
very much.
Best Regards,
Ray
[[alternative HTML version
On Tue, 22-Jan-2013 at 08:30AM -0500, Ista Zahn wrote:
| Hi,
|
| ID is not the value column. Your casting call should be
|
| dcast(aa, ... ~ Target, value.var = Eaten)
Thanks for that illumination. That does exactly what I wanted. I
knew there were many ways of achieving the result, but I
At 08:30 23/01/2013, Alma Wilflinger wrote:
Dear Wolfgang and Michael,
thank you very much for your help!
Concerning the Variance: I took the variance I used for CMA (which
is always 1), so I think it should be the right one.
It seems unlikely to me that the variance from each study would
Dear R listers,
I am trying to compute the mean of a dummy variable that is encoded as a
factor. However, even though the levels of my factor are 0 - 1, when I
compute the mean (after coercing the factor to be
numeric), R changes 0 into 1 and 1 into yes, thus altering my expected
result.
Please,
Check R FAQ 7.10: How do I convert factors to numeric?
I hope it helps.
Best,
Dimitris
On 1/23/2013 10:33 AM, Francesco Sarracino wrote:
Dear R listers,
I am trying to compute the mean of a dummy variable that is encoded as a
factor. However, even though the levels of my factor are 0 - 1,
Dear Dimitris,
thanks for your quick reply. I've tried the solutions proposed in 7.10 How
do I convert factors to numeric?
as.numeric(as.character(pp))
and
as.numeric(levels(pp))[as.integer(pp)]
However, whatever I do, I get Warning message: NAs introduced by coercion
and the output is a vector
check also
pp - rep(0:1, 10)
pp - factor(pp, levels=(0:1), labels=c(no,yes))
unclass(pp)
unclass(pp) - 1
Best,
Dimitris
On 1/23/2013 10:48 AM, Francesco Sarracino wrote:
Dear Dimitris,
thanks for your quick reply. I've tried the solutions proposed in 7.10
How do I convert factors to
Thanks,
this works! but I am surprised that R has such a strange behavior and that
there is no way to control it.
BTW, also as.integer(pp)-1 works!
Still, it doesn't look to me as a first best.
At any rate, thanks a lot for your help.
f.
On 23 January 2013 10:53, D. Rizopoulos
Thanks a lot Petr,
for the answer
unfortunately that would convert everything to a matrix
num [1:32002, 1:3] 0 0 0 0 0 0 0 0 0 0 ...
but if you check below you can see that I Want those to form a list.
Regards
Alex
From: PIKAL Petr petr.pi...@precheza.cz
you just need:
mapply(c, Part1$dataset, Part2$dataset, SIMPLIFY = FALSE)
I hope it helps.
Best,
Dimitris
On 1/23/2013 11:01 AM, Alaios wrote:
Thanks a lot Petr,
for the answer
unfortunately that would convert everything to a matrix
num [1:32002, 1:3] 0 0 0 0 0 0 0 0 0 0 ...
but if you
Dear useRs,
I'm pleased to announce that version 1.1.0 of lambda.r is now available on CRAN
(http://cran.r-project.org/web/packages/lambda.r/). This package provides a
complete functional programming environment within R (and is backwards
compatible with S3). Lambda.r introduces many concepts
Dear Wolfgang and Michael,
thank you very much for your help!
Concerning the Variance: I took the variance I used for CMA (which is always
1), so I think it should be the right one.
Thank you for noticing and mentioning though :)
I really appreciate how helpful you both are.
best,
Alma
Dear R-ers,
I've generated data for failure times log(T) from standard normal
distribution. How can I generate data for censoring times from log-normal
distribution which ý can get constatnt hazard rate.
[[alternative HTML version deleted]]
Hi,
I would like to do a meta-analysis, i.e., a mixed-effects regression,
but I don't seem to get what I want using both the nlme or metafor packages.
My question: is there indeed no way to do it?
And if so, is there another package I could use?
Here are the details:
In my meta-analysis I'm
Dear M. Noel:
You may know that there is a list of books on the R web site
(www.r-project.org - books: www.r-project.org/doc/bib/R-books.html).
I'm not sure what you should do to get this book listed there. If no
one else suggests what to do, you might wish to write to Frau Palege in
Thanks a lot.
Unfortunately that did not help either.
num [1:32003, 1:3] 0 0 0 0 0 0 0 0 0 0 ...
- attr(*, dimnames)=List of 2
..$ : chr [1:32003] ...
..$ : NULL
but I want to get
List of 3
$ : num [1:32002] 0 0 0 0 0 0 0 0 0 0 ...
$ : num [1:32002] 0 0 0 0 0 0 0 0 0 0 ...
$
Hi,
thanks. I am indeed interested in the main effects of A and B and their
interaction+ I want to incorporate C (the block or 'repetition' within which
the A and B treatments were applied) as a random variable. So A*B would be
the way, however errors of A and B are different due to
In this example, I get the following:
lis1 - replicate(3, rnorm(5), simplify = FALSE)
lis2 - replicate(3, rnorm(5), simplify = FALSE)
lis1
lis2
mapply(c, lis1, lis2, SIMPLIFY = FALSE)
Best,
Dimitris
On 1/23/2013 11:58 AM, Alaios wrote:
Thanks a lot.
Unfortunately that did not help either.
Hi
Sorry for the rather long message.
I am trying to use the cfa command in the lavaan package to run a CFA however I
am unsure over a couple of issues.
I have @25 dichotomous variables, 300 observations and an EFA on a training
dataset suggests a 3 factor model.
After defining the model I
Hello Christian,
First of all, it's good to see that you are well aware of the fact that lme()
without lmeControl(sigma=1) will lead to the estimation of the residual
variance component, which implies that the sampling variances specified via
varFixed() are only assumed to be known up to a
Or maybe
x-matrix(test,nrow=10)
apply(x,2,mean)
On 23.01.2013, at 00:09, Wim Kreinen wrote:
Hello,
I have vector called test. And now I wish to measure the mean of the first
10 number, the second 10 numbers etc
How does it work?
Thanks Wim
dput (test)
c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
On Wed, Jan 23, 2013 at 9:16 AM, Ray Cheung ray1...@gmail.com wrote:
Dear All,
Sorry for asking a newbie question. I want to ask how to import 1000
datasets whose file names are labelled from data1.dat to data1000.dat into
R so that they are named M[1, , ] to M[1000, , ] accordingly. Thank
As an example:
chars-c(A,A,B)
numbers-as.numeric(as.factor(chars)) #make this numerical
plot(numbers,c(0.4,0.5,0.6),xaxt=n) #xaxt=n says to not plot the x-axis
axis(side=1,at=numbers,labels=chars) #make the axis with labels
On 23.01.2013, at 10:16, Ng Wee Kiat Jeremy wrote:
Dear List,
I
Dear all
I have a data.frame like that :
father mother num_daughterdaughter
291 39060 NULL
275 42190 NULL
273 42361 49410
281 41631 49408
274 42261 49406
295 38692 49403
I don't really understand this data table, but maybe this modification
will give you the idea:
dat - read.table(text=father mother num_daughterdaughter
291 39060 NA
275 42190 NA
273 42361 49410
281 41631 49408
274 42261
Dear all,
I have a matrix with two columns: Names and Values
In names: there are 4 groups they are, CK113234, CK116296, CK116292 and
CK114042
I want to *sort values* (decreasing order) based on each group and
take average of the *top two numbers* in each of the groups.
dput(x)
Dear R Mailinglist,
I want to understand how predictors are associated with a dependent
variable in a regression. I have 3 measurement points. I'm not interested
in understanding the associations of regressors and the predictor at each
measurement separately, instead I would like to use the whole
Hi Wolfgang,
thanks for the instant and comprehensive reply!
On 01/23/2013 12:39 PM, Viechtbauer Wolfgang (STAT) wrote:
[...]
In fact, trying to disentangle that residual variance component from any
random study effects is usually next to impossible. I mention this explicitly
one more
Hi Nico,
You can use tapply:
tapply( x[[Values ]], x[[ Names ]], mean )
CK113234 CK113298 CK114042 CK116292 CK116296
216.5061 190.1725 222.8710 220.4324 204.5741
Alternatively,
tapply( x$Values, x$Names, mean )
CK113234 CK113298 CK114042 CK116292 CK116296
216.5061 190.1725 222.8710
I found a code:
y.ts - ts(data, frequency=12)
aggregate(y.ts, FUN=quantile, probs=0.10)
Seems it works fine even for a big data.frame.
Thanks for your help.
2013/1/22 David Winsemius dwinsem...@comcast.net
On Jan 22, 2013, at 5:58 AM, Simonas Kecorius wrote:
Hey Duncan,
Neither me do
Dear David,
On Wed, 23 Jan 2013 11:19:09 +
David Purves david.pur...@glasgow.ac.uk wrote:
Hi
Sorry for the rather long message.
. . .
I have tried the analysis using John Fox's SEM package / command.
I calculate the correlation matrix with smoothing
Post elsewhere (e.g. stats.stackexchange.com). This is not a
statistical tutorial site.
-- Bert
On Wed, Jan 23, 2013 at 5:28 AM, Torvon tor...@gmail.com wrote:
Dear R Mailinglist,
I want to understand how predictors are associated with a dependent
variable in a regression. I have 3
Gabriela Agostini gabrielaagostini18 at gmail.com writes:
[snip]
I am working with GLMM using the binomial family
I use the following codes
I dropped no significant terms, refitting the model and comparing the
changes with likelihood:
The following 1460 x 1460 matrix can be throught of as 16 distinct 365 x 365
matrices. I'm trying to set off-diaganol terms in the 16 sub-matrices with
indices more than +/- 5 (days) from each other to zero using some for loops.
This works well for some, but not all, of the for loops. The R
I think you need to become more familiar with the factor data type. Reread
the Introduction to R document that comes with R.
---
Jeff NewmillerThe . . Go Live...
You didn't indicate what you want to do with the 101st observation. Arun's
solution creates an 11th group and divides by the number in the group(1),
while Jessica's solution creates an 11th column and divides by 10.
test[101] - 100
unlist(lapply(split(test,((seq_along(test)-1)%/% 10)+1),mean))
Hi,
Try this:
x1-x[rev(order(x$Names,x$Values)),]
do.call(rbind,tapply(x1$Values,list(x1$Names),head,2))
# [,1] [,2]
#CK113234 223.2966 222.6737
#CK113298 192.5964 187.7486
#CK114042 236.3939 232.0223
#CK116292 237.5936 228.0037
#CK116296 223.6372 210.6630
#The average
Hi John
Thanks for your quick reply.
The full warning I got is
' Error in csem(model = model.description, start, opt.flag = 1, typsize =
typsize, :
The matrix is non-invertable.'
The eigenvalues of the tetrachoric correlations are non negative. So it is must
be how I am defining my model.
Hi,
It's not clear regarding those blanks especially, the num_daughter. I guess
the father and mother would be the same as the previous row.
Deleting those rows:
df1 - read.table(text=father mother num_daughter daughter
291 3906 0 NA
275 4219 0 NA
273 4236 1
Hi,
May be this helps:
df1-read.table(text=
father,mother,num_daughter,daughter
291,3906,0,
275,4219,0,
273, 4236,1,49410
281,4163,1,49408
274, 4226,1,49406
295, 3869,2,49403
295,3869,2,49404
287,4113,0,
295, 3871,1,49401
292, 3895,4,49396
292,3895,4, 49397
292,3895,4,49398
292,3895,4,49399
Hello,
How can I judge if a string is in a group of string? For example, I would like
to have
if (subpool in pool){
}else{
}
Where
pool = c(s1,s2)
subpool = c(s1)
How can I write the subpool in pool right in R?
Thanks very much!
Cheers,
Rebecca
Dear David,
It certainly helps to have a reproducible example.
You've left out the error variances (uniquenesses) for the observed
variables. You're also making the specification *much* harder than it needs
to be:
-- snip ---
cfa.mod.1 - cfa()
1: F: a, b, c, d, e, g
2:
Read 1
On 13-01-23 11:14 AM, Yuan, Rebecca wrote:
Hello,
How can I judge if a string is in a group of string? For example, I would like
to have
if (subpool in pool){
}else{
}
if (subpool %in% pool)
Duncan Murdoch
Where
pool = c(s1,s2)
subpool = c(s1)
How can I write the subpool in pool
Not sure what you want this for, but work along the following:
pool = c(s1,s2)
subpool = c(s1)
ifelse(pool==subpool,1,0)
[1] 1 0
Notice:
pool2 = c(s2,s1)
ifelse(pool2==subpool,1,0)
[1] 0 1
Etc.
Hope this helps.
José
José Iparraguirre
Chief Economist
Age UK
-Original
Dear Daniel,
Eventually, FactoMineR (and other Rcmdr plug-ins) really should be updated
to conform to R 3.0.0 requirements, but I've modified the development
version 1.9-4 of the Rcmdr on R-Forge to allow a new option to accommodate
some of these plug-ins. If you set
Hello Duncan,
Thanks for this!
This works!
Best,
Rebecca
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: Wednesday, January 23, 2013 11:23 AM
To: Yuan, Rebecca
Cc: R help
Subject: Re: [R] to check if a character string is in a group of character
On 23-01-2013, at 16:13, emorway emor...@usgs.gov wrote:
The following 1460 x 1460 matrix can be throught of as 16 distinct 365 x 365
matrices. I'm trying to set off-diaganol terms in the 16 sub-matrices with
indices more than +/- 5 (days) from each other to zero using some for loops.
This
Dear Daniel,
Oh, I see I forgot to comment on your second specification in my last reply:
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of David Purves
Sent: Wednesday, January 23, 2013 10:23 AM
To: John Fox
Cc:
Hello Gentlemen,
I mistakenly sent the message twice, because the first time I didn't
receive a notification message so I was unsure if it went through properly.
Your solutions worked great. Thank you! I felt like I was fairly close
just couldn't quite get the final step.
Now, I'm trying to
I think you want %in%
subpool %in% pool
pool %in% subpool
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
+ 32 2 525 02 51
+ 32
Or...
fs-rep(1:ceiling(length(test)/10),each=10)[1:length(test)]
result-by(test,fs,mean)
which will get you the version with the 101st as a single datapoint
so many possibilities..
On 23.01.2013, at 16:43, David L Carlson wrote:
You didn't indicate what you want to do with the 101st
Great! This really helped! One quick follow-up -- is there a trick to
placing a label wherever the line intersects the x-axis (either above or
below the plot)?
On Tue, Jan 22, 2013 at 11:49 PM, PIKAL Petr petr.pi...@precheza.cz wrote:
Hi
This function adds line to each panel
addLine -
Hello,
I'm trying to add a symbol (Delta) to plot legend with text using
expression(paste()) but this disables the text.font that allows to use
bold or italic text.
as follows:
x=c(1:10)
y=c(1:10)
plot(x,y)
legend(1,10,legend=c(A,B,C,expression(paste(Delta, D))),
On Jan 23, 2013, at 1:58 AM, Francesco Sarracino wrote:
Thanks,
this works! but I am surprised that R has such a strange behavior
and that
there is no way to control it.
BTW, also as.integer(pp)-1 works!
Still, it doesn't look to me as a first best.
At any rate, thanks a lot for your help.
I think it is a fair bit of work to interpret the freq=TRUE (prob=FALSE)
version of hist() when the bins have unequal sizes. E.g.,
in the following the bins are sized so that each contains
an equal number of observations. The resulting flat
frequency plot is hard for me to interpret. The
To find the proportion of yess in pp you can use
mean(pp == yes)
and avoid the conversion of a factor to integer (and
subtracting 1). The above works for character and factor
pp.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From:
I need a quick help with the following graph (I'm a lattice newbie):
require(lattice)
npp=1:5
names(npp)=c(A,B,C,D,E)
barchart(npp,origin=0,box.width=1)
# What I want to do, is add a single vertical line
positioned at x = 2
that lays over the bars (say, using a dotted
Dear All,
I have a data frame of vectors of publication names such as 'pub':
pub1 - c('Brown DK, Santos R, Rome DF, Don Juan X')
pub2 - c('Benigni D')
pub3 - c('Arstra SD, Van den Hoops DD, lamarque D')
pub - rbind(pub1, pub2, pub3)
I would like to construct a dataframe with only author's
Thanks!
The comments and the information provided were extremely helpful to me.
According to DRAFT r-sig-mixed-models FAQ, do not compare lmer models
with the corresponding lm fits, or glmer/glm; the log-likelihoods are
not commensurate. The problem is not the link function but rather the
1. Study a regular expression tutorial on the web to learn how to do this.
2. ?regex in R summarizes (tersely! -- but clearly) R's regex's.
3. ?grep tells you about R's regular expression manipulation functions.
-- Bert
On Wed, Jan 23, 2013 at 9:38 AM, Biau David djmb...@yahoo.fr wrote:
Dear
Hello,
Try the following.
fun - function(x, sep = , ){
s - unlist(strsplit(x, sep))
regmatches(s, regexpr([[:alpha:]]*, s))
}
fun(pub)
Hope this helps,
Rui Barradas
Em 23-01-2013 17:38, Biau David escreveu:
Dear All,
I have a data frame of vectors of publication names
plot(1)
legend('topleft',legend=expression(A,italic(A),bolditalic(A),Delta*italic(D)))
On Wed, Jan 23, 2013 at 9:45 AM, raz barvazd...@gmail.com wrote:
Hello,
I'm trying to add a symbol (Delta) to plot legend with text using
expression(paste()) but this disables the text.font that allows to
Hello,
I've just noticed that my first solution would only return the first set
of alphabetic characters, such as Van, not Van den Hoops.
The following will solve that problem.
fun2 - function(x, sep = , ){
x - strsplit(x, sep)
m - lapply(x, function(y) gregexpr(
To Whom It May Concern:
I have noticed that all of the hyphens (-) are changed to periods (.) when
I try to read.table() and the headers contain -
I am using R 2.13 on a RedHat system.
Here is the situation:
I have the following a tab-delimited text file saved as test.txt
File1-a.txt
Hi,
You could try this:
dat1-read.table(text=pub,sep=,,fill=TRUE,stringsAsFactors=F)
dat2- as.data.frame(do.call(cbind,lapply(dat1,function(x) gsub( $,,gsub(^
|\\w+$,,x,stringsAsFactors=F)
dat2
# V1 V2 V3 V4
#1 Brown Santos Rome Don
Thank you for the suggestions.
Just to clarify, my first question was more on what actual coding I
should be using to indicate a nested variable when using the coxph()
function. I asked this after consulting several times with a local
statistician, but unfortunately neither of us are very
Please read ?read.table again, and pay special attention to the
check.names argument.
A - is not allowed in a column name because it would lead to problems like:
mydata$a-b vs mydata$a - b
where
mydata$a.b has no such confusion.
If you must have - instead of ., you can use check.names=FALSE and
Hi,
If the `spaces` in father, mother, num_daughter columns needs to be
replaced by the values in the previous row,
dat1-read.table(text=
father, mother, num_daughter, daughter
291, 3906, 0,
275, 4219, 0,
273, 4236, 1, 49410
281, 4163, 1, 49408
274, 4226, 1, 49406
295, 3869, 2, 49403
, ,,
I need to repeat a function many times, with differing parameters held
constant across iterations. To accomplish this, I would like to create a
list (or vector) of parameters, and then insert that list into the function.
For example:
q-(l,a,b,s)
genericfunction-function(q){
}
##
The
Dear R helpers,
I have following loss data and I need to fit LEFT truncated Log Normal
distribution to this data which is Truncated at 100.
dat =
c(1333834,5710254,9987567,7809469,6940935,3473671,1270209,1102523,1124002,
On Jan 23, 2013, at 12:14 PM, Katherine Gobin wrote:
Dear R helpers,
I have following loss data and I need to fit LEFT truncated Log Normal
distribution to this data which is Truncated at 100.
dat =
c(1333834,5710254,9987567,7809469,6940935,3473671,1270209,1102523,1124002,
Given that your labels are no and yes, what do you expect R to
do? To quote a well-known fortune, R is lacking a mind_read() function!
cheers,
Rolf Turner
On 01/23/2013 10:58 PM, Francesco Sarracino wrote:
Thanks,
this works! but I am surprised that R has such a strange behavior
Thank you all for your replies. Let me try to explain my point: first of
all, let me clarify that I didn't mean to criticize anyone (or anything).
Secondly, what I meant refers to the fact that I've read on an R and
S-plus companion to applied regression about methods to alter the encoding
of
Hi Michael,
The supervisorfor my Master'sThesis told me that my means are the effect size
and cause of this I have to take figure 1 for all standard deviations. So I
hope that was the right information.
From: Michael Dewey i...@aghmed.fsnet.co.uk
Dear Hana,
Thanks for helping.
I am still wondering, why m1 (which should be 2^32-209 [see line 34 in
./src/RngStream.c]) is -767742437 in my case and why the minimal example you
gave was working for you but isn't for me.
Apart from that, ?.Random.seed - L'Ecuyer-CMRG says:
,
| The 6
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Marius Hofert
Sent: Wednesday, January 23, 2013 2:24 PM
To: Hana Sevcikova
Cc: R-help
Subject: Re: [R] How to construct a valid seed for l'Ecuyer's method
withgiven
Marius,
I looked it up in the original L'Ecuyer's paper: The seed must be larger
than 0. Thus, the function defines the seed variable as unsigned long
integer. You're passing a negative number, so I think there is some
overflow going on.
The internal L'Ecuyer RNG is a modification of the
I had the same problem but it now works if you remove some of the #info lines
at the top of the file so that the number of lines is the same as the
example sequence.vcf file before the data starts
--
View this message in context:
Hello R-users
I am getting error messagens when I require some packages or execute some
procedures, like these below:
require(tseries)
Loading required package: tseries
Error in get(Info[i, 1], envir = env) :
cannot allocate memory block of size 2.7 Gb
require (TSA)
Loading required
I'm not following. Printing SEQ to the screen at the intermediate steps
using the following modified R code suggests that 'i' is fine and is not
getting reset to 1 as you suggest? My understanding, or rather my desired
output if someone else is able to weight-in, is that the values in the
second
Hi all,
I am planning to parse some information on a website which includes lots of
Chinese characters. Does someone know how to read/display Chinese in R? Thanks.
url = http://www.teec.org.cn/html/renwujieshao/;
x = readLines(url)
I tried encoding = 'UTF-8' already but it didn't help.
My R
On Jan 23, 2013, at 7:13 AM, emorway wrote:
The following 1460 x 1460 matrix can be throught of as 16 distinct 365 x 365
matrices. I'm trying to set off-diaganol terms in the 16 sub-matrices with
indices more than +/- 5 (days) from each other to zero using some for loops.
This works well
On 13-01-23 8:19 PM, Hui Du wrote:
Hi all,
I am planning to parse some information on a website which includes lots of
Chinese characters. Does someone know how to read/display Chinese in R? Thanks.
url = http://www.teec.org.cn/html/renwujieshao/;
x = readLines(url)
If you look at the
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of emorway
Sent: Wednesday, January 23, 2013 4:59 PM
To: r-help@r-project.org
Subject: Re: [R] setting off-diagonals to zero
I'm not following. Printing SEQ to the screen at the
Hi list,
Could anyone recommend some good website for formatting R code? For
example, when you copy paste R code to gmail and back to R, it loses its
format, the dash symbol causes errors.
I've had someone used it to format my code here on the list, but can't find
it anymore.
Mike
Are you only interested in formatting code from copy and pasting to/from email?
If you are interested in formatting your code in Latex/PDF/HTML take a look at
the knitr package:
http://yihui.name/knitr/
Also, you could check out the formatR package:
Hey Mark,
I am aware of this package. The situation is,
1. I am emailing my code from my machine to a public machine. Installation
is a hassle.
2. Copy pasting for a website.
I know there are websites for formatting Java and C code. So, I am looking
for a website in particular, and I have seen
HI,
Not sure this is what you wanted.
for (i in 731:732) {
SEQ - (i - 5):(i + 5)
print(SEQ)
SEQ - SEQ[SEQ 730 SEQ 1096]
print(SEQ)
vec1-731:741
print(vec1[!vec1%in%SEQ])
}
#[1] 726 727 728 729 730 731 732 733 734 735 736
#[1] 731 732 733 734 735 736
#[1] 737 738 739 740 741
# [1]
Hi Adam,
On Jan 23, 2013, at 11:36 AM, Adam Gabbert wrote:
Hello Gentlemen,
I mistakenly sent the message twice, because the first time I didn't receive
a notification message so I was unsure if it went through properly.
Your solutions worked great. Thank you! I felt like I was
Thanks a lot.
y - iconv(x, gb2312, utf-8) does not work but
y - iconv(x, gb2312, UTF8) works on my machine. Thank you for pointing to
the right direction.
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: Wednesday, January 23, 2013 6:16 PM
To: Hui Du
Here's another approach using the sampling profiler:
prof - function() {
Rprof(memory.profiling=T, interval=0.001)
replicate(100, f())
Rprof(NULL)
summaryRprof(memory = stats)
}
f - function() {
x = seq(1000)
for(i in seq(1000)) {
x[i] - x[i] + 1
}
}
prof()
=
index:
On 23 Jan 2013, at 21:36, Francesco Sarracino f.sarrac...@gmail.com wrote:
what I meant refers to the fact that I've read on an R and
S-plus companion to applied regression about methods to alter the encoding
of factors when using contrasts in regressions. These are options (for
thanks, it works well. I have to work on Arun's previous answer to make it work
too.
David
De : Rui Barradas ruipbarra...@sapo.pt
À : Biau David djmb...@yahoo.fr
Cc : r help list r-help@r-project.org
Envoyé le : Mercredi 23 janvier 2013 19h57
Objet : Re:
On 24-01-2013, at 01:58, emorway emor...@usgs.gov wrote:
I'm not following. Printing SEQ to the screen at the intermediate steps
using the following modified R code suggests that 'i' is fine and is not
getting reset to 1 as you suggest?
You misread. I did not say anything about 'i'.
My
Hi, I'm curious about the ability of these two methods to really wrap (I
mean as in delegate) the target object instead of deep copying and
transforming it to a new structure. First, specifically for vectors.
Second, in general (my hunch is that -say- map-env are transformed copies
but it's just a
HI David,
It could be related to spaces in the data or something else.
Suppose, if the data has some spaces at the end or the beginning.
pub1 - c('Brown DK, Santos R, Rome DF, Don Juan X')
pub2 - c('Benigni D')
pub3 - c('Arstra SD, Van den Hoops DD, lamarque D ')
pubnew-rbind(pub1, pub2,
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