On 03/28/2014 05:27 PM, Luana Marotta wrote:
Hi all,
I'm trying to find out what is the equivalent in R for the following Stata
code:
Let's say I have the variables: a1, a2, a3, a4, a5
forvalues i = 1(1)5 {
summary a`i'
}
That is, I want to know how to loop through variables in R.
Hi L
Hi all,
I'm trying to find out what is the equivalent in R for the following Stata
code:
Let's say I have the variables: a1, a2, a3, a4, a5
forvalues i = 1(1)5 {
summary a`i'
}
That is, I want to know how to loop through variables in R.
Thank you in advance,
Luana
[[alternative H
On 28 Mar 2014, at 00:57, Duncan Murdoch wrote:
> On 27/03/2014, 7:38 PM, Thomas Lumley wrote:
>> You get what you wanted from
>>
>> do.call(plot,list(x=quote(x),y=quote(y)))
>>
>> By the time do.call() gets the arguments it doesn't know how y was
>> originally computed, just what values it has
Dear list members,
I am wondering whether there is any more efficient way to calculate
centered difference on matrix in R? Please see herewith an example:
lon <- matrix(rep(seq(0,2,length.out=1e3), 1e3), 1e3, 1e3)
lat <- matrix(rep(seq(0,2,length.out=1e3), each=1e3), 1e3, 1e3)
x <- matrix(rep(seq
Hello,
Your example leads to error. Anyway:
library(XML)
oneurl="http://www.tutiempo.net/en/Climate/FUKUSHIMA/11-2012/475950.htm";
temp.tables=readHTMLTable(oneurl, header=TRUE)
temp.tables <- temp.tables[[3]]
temp.tables <- temp.tables[1:30,]
HTH,
Pascal
On Fri, Mar 28, 2014 at 2:42 PM, Bill
Hi. I ran some code and I am trying to access the table with the data in
it. I want in particular to delete the 31st row for example
Monthly means and totals:
or remove the names of the columns (or change them). DayT TM Tm
SLPHPP VVV VM VG RA SN TS
Dear All
I have a model of the form Approaches ~ Setup * Minute, where Setup has two
levels and Minute has three levels.
I want to test the difference between the levels of setup (Control vs Test)
within in levels minute e.g. the difference between the two setups for Minute
1, Minute 2 and Mi
Hi,
May be this helps:
set.seed(42)
indx <- with(df,tapply(seq_along(IndividualID), FamilyID,FUN=sample,1))
df[indx,]
# FamilyID IndividualID DadID MomID Sex
#4 1 104 0 0 2
#8 2 204 202 203 2
#or
library(plyr)
ddply(df,.(FamilyID),function(x) x[w
Tena koe Whitney
tapply should work. Try:
tapply(yourData$IndividualID, yourData$FamilyID, sample, size=1) # untested
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Whitney Melroy
Sent: Friday, 28 March
On 28/03/14 12:49, Duncan Murdoch wrote:
On 27/03/2014, 7:17 PM, Rolf Turner wrote:
I was under the impression that
do.call(foo,list(x=x,y=y))
should yield the same result as
foo(x,y).
However if I do
x <- 1:10
y <- (x-5.5)^2
do.call(plot)
I get the expected plot but
Hello,
I have a dataset with family data. For an analysis, I need to select one
subject per family at random.
Here is an example of what my data look like:
FamilyID IndividualID DadIDMomID Sex
1101103 104 1
1
On 27/03/2014, 7:38 PM, Thomas Lumley wrote:
You get what you wanted from
do.call(plot,list(x=quote(x),y=quote(y)))
By the time do.call() gets the arguments it doesn't know how y was
originally computed, just what values it has.
This works, but it doesn't make sense to me. The arguments end
On 27/03/2014, 7:17 PM, Rolf Turner wrote:
I was under the impression that
do.call(foo,list(x=x,y=y))
should yield the same result as
foo(x,y).
However if I do
x <- 1:10
y <- (x-5.5)^2
do.call(plot)
I get the expected plot but with the y-values (sur
You get what you wanted from
do.call(plot,list(x=quote(x),y=quote(y)))
By the time do.call() gets the arguments it doesn't know how y was
originally computed, just what values it has.
-thomas
On Thu, Mar 27, 2014 at 6:17 PM, Rolf Turner wrote:
>
>
> I was under the impression that
>
>
I was under the impression that
do.call(foo,list(x=x,y=y))
should yield the same result as
foo(x,y).
However if I do
x <- 1:10
y <- (x-5.5)^2
do.call(plot,list(x=x,y=y))
I get the expected plot but with the y-values (surrounded by c()) being
printed
On Mar 27, 2014, at 2:34 PM, David Winsemius wrote:
>
> On Mar 27, 2014, at 8:11 AM, el_alisio wrote:
>
>> Hi Kevin,
>>
>> thanks for the advice! The code is now reproducible. Sorry!
>
> No, it is not, since you can only read it on Nabble, yecch. There is some
> sort of weird format stuff th
On Mar 27, 2014, at 8:11 AM, el_alisio wrote:
> Hi Kevin,
>
> thanks for the advice! The code is now reproducible. Sorry!
No, it is not, since you can only read it on Nabble, yecch. There is some sort
of weird format stuff that Nabble does that prevents code from appearing on
Rhelp copies.
?"%%"
PLEASE read an Introduction to R (ships with R) or an online tutorial
of your choice before posting further. You then will not have to waste
nearly as much time (yours and ours) posting and waiting for answers!
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Thanks David,
Neat use of merge there.
I think I can also do this with:
ddply(data,.(id),f)
On Thu, 2014-03-27 at 16:06 -0500, David Carlson wrote:
> It may be possible to do this in a single step, but
>
> > x1 <- aggregate(response~id+age, data, mean)
> > x2 <- data[data$eye=="l", c("id", "res
Hi,
You could try:
library(dplyr)
A %.% group_by(author_id) %.% arrange(desc(prob)) %.%
summarise(paper_id=paste(paper_id,collapse=","))
A.K.
On Thursday, March 27, 2014 11:40 AM, Rohit Gupta wrote:
I have a data A which looks like
author_id paper_id prob
731 24943 1
It may be possible to do this in a single step, but
> x1 <- aggregate(response~id+age, data, mean)
> x2 <- data[data$eye=="l", c("id", "response2")]
> merge(x1, x2)
id age response response2
1 1 2 4.60 High
2 2 9 2.65 High
3 3 5 3.65 High
4 4 2 7.55
I'm no expert on the adonis functions by the error you are getting
usually means the dataset is not quite what you expected. Are you sure
the event dataset is loading as you expect?
On Tue, 2014-03-25 at 16:36 -0400, Maggie Wisniewska wrote:
> Hello,
> I am an R noivce, so excuse my simple questio
Hi all,
I've spent too long in matlab land recently and seem to have forgotten
my R skillz ;-)
I'm sure I'm missing a simple way to do this...
Given a data frame
id<-rep(1:5,2)
eye<-c(rep('l',5),rep('r',5))
age<-rep(round(runif(5,0,12)),2)
response<-round(runif(10,1,10)*10)/10
response2<-sample(c(
Or tweaking Don's solution but using the original mu and sig
vectors:
> set.seed(42)
> x.matrix<-matrix(0, nrow=27, ncol=5)
> for(i in 1:27){
+ x.matrix[i, ] <- rnorm(5, mu[i], sig[i])
+ }
>
> set.seed(42)
> xmat <- matrix(rnorm(27*5, rep(mu, each=5), rep(sig, each=5)),
+ nrow=27, byrow=TRUE)
>
On 03/28/2014 06:33 AM, Prabhakar Ghorpade wrote:
Hi,
when typing code I get like this
x<-(1,2,
+
+
+
Can I delete + and comma after 2 and go back to end of comma as x<-c(1,2)
Hi Prabhakar,
I don't think so, as the command line reading won't pass your backspace.
I'd say just press Ct
The hpd functions in the TeachingDemos package do not currently take
weights, so they will not work with the results of importance
sampling.
On Wed, Mar 26, 2014 at 10:35 PM, Baran rad wrote:
> hi :)
> if i using importance sampling for obtain postreior density, can i use
> teachingdemos for obta
Or, to show the "power of the R language", so to speak, and assuming that
Dave is correct, there is no need to use a for() loop:
xmat <- matrix( rnorm( 27*5, rep(mu,5), rep(sig,5) ) , nrow=27)
To verify that I've created the vectors in the right order, and converted
to a matrix in the right orde
Hi,
when typing code I get like this
x <-(1,2,
+
+
+
Can I delete + and comma after 2 and go back to end of comma as x <-c(1,2)
Thanks.
Prabhakar
--
Dr.Ghorpade Prabhakar B.
Ph.D. Scholar ( Animal Biochemistry)
Indian Veterinary Research Institute.
India
[[alternative HTML vers
On 03/28/2014 12:53 AM, Manuel Carona wrote:
Hi,
I have implemented a therapeutic intervention on two groups (one is a
control group) and tested them in two moments using some assessment
tools (with normative data). Now I want to compare the experimental
group with the control group using clinic
On Mar 27, 2014, at 8:14 AM, Hermann Norpois wrote:
> Hello,
>
> My main question is wheter my data is distributed normally. As the
> shapiro.test doesnt work for large
> data sets I prefer the ks.test.
> But I have some problems to understand the completely different p-values:
>
>> ks.test (te
On Mar 27, 2014, at 8:53 AM, Manuel Carona wrote:
> Hi,
>
> I have implemented a therapeutic intervention on two groups (one is a
> control group) and tested them in two moments using some assessment
> tools (with normative data). Now I want to compare the experimental
> group with the control
On Mar 27, 2014, at 7:48 AM, Prabhakar Ghorpade wrote:
> Hi,
> How do I find help on %%
?'%%' # Need to quote "specials"
Also look at:
?'<-'
?'%in%
?'function'
--
David.
>
> Regards,
> Prabhakar
>
>
> On Thu, Mar 27, 2014 at 4:56 PM, Jorge I Velez
> wrote:
>
>> Hi there,
>>
>> Tr
Perhaps have a look at the sp package and especially the over function.
Br. Frede
Sendt fra Samsung mobil
Oprindelig meddelelse
Fra: Leonid Shvartser
Dato:27/03/2014 20.27 (GMT+01:00)
Til: r-help@r-project.org
Emne: [R] index polygons
Hello,
I need to compare a polygon with
On Mar 27, 2014, at 4:31 AM, IZHAK shabsogh wrote:
> i try to generate 27*5 matrix of observation using the following code but is
> given error
> kindly assist and correct the problem
>
>
> x1<-c(-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1)
> x2<-c(-1,-1,-1,0,0,0,1,1,1,-1,
On Mar 27, 2014, at 6:53 AM, Manuel Carona wrote:
> Hi,
>
> I have implemented a therapeutic intervention on two groups (one is a
> control group) and tested them in two moments using some assessment
> tools (with normative data). Now I want to compare the experimental
> group with the control g
Thanks, your solution using ave() works perfectly.
/johannes
-Ursprüngliche Nachricht-
Von: Bert Gunter
An: Johannes Radinger
Cc: R help
Gesendet: Mittwoch, 26. März 2014 16:45:43 GMT+00:00
Betreff: Re: [R] dataframe calculations based on certain values of a column
I believe this wil
First, read the Introduction to R to understand what dataframes are and how
to access data within them. Then correct you code so it does not have
error: 'x' has a trailing comma and 'y' was not call the 'c' function.
Once you have done that, you might get the following:
> x<-c(5,9,8,3,4,5,6,7
It would have been helpful it you had posted the error message you got.
Here is what I got running your code:
Error in x.matrix[i, j] <- rnorm(5, mu[i], sig[i]) : object 'j' not found
No suitable frames for recover()
It seems you have not defined 'j' that you trying to index with.
Jim Holtman
Hello,
I need to compare a polygon with a set of other polygons.
I'd like to index polygons with a grid in order not to compare definitely
distant ones.
How can I do this?
I need
1. Transform polygons to cells
2. Find cell neighbors (defined cell distance) for a polygon
Thanks
Leoni
ok
btw thanks
On Thu, Mar 27, 2014 at 5:55 PM, Hadley Wickham wrote:
> It's rude to ask a question both on r-help and on stackoverflow
> (http://stackoverflow.com/questions/22685896), because people might
> spend their time answering your question when it's already been
> answered elsewhere.
>
Here is one way. I took your data and applied the 'approx' function to get
evenly spaced values and then 'filter' to create a moving average of a
window of size 5.
set.seed(1)
x = c(0,1, 3,3.4, 5, 10, 11.23)
y = x**2 + rnorm(length(x))
rbind(x, y)
# create 'even' spacing using the 'approx' func
Hello,
If I have x<-c(5,9,8,3,4,5,6,7,)
and y <-("a", "b","c", "d","e","f", "g","h")
and D <- data.frame(x,y)
I did this max(x)
and My expectede anwer is b.
How do I select element in y which correspond to maximum number in x?
thanks.
regards,
Prabhakar
[[alternative HTML version d
Hi Kevin,
thanks for the advice! The code is now reproducible. Sorry!
Actually I wanted to print only one x-label ("All Postdoctorates") in bold
face, not all of them. This would be easily achieved, if "All
Postdoctorates" appeared always in the same row of each panel, but this is
not the case he
Hi,
I have implemented a therapeutic intervention on two groups (one is a
control group) and tested them in two moments using some assessment
tools (with normative data). Now I want to compare the experimental
group with the control group using clinical equivalence testing. To do
this I need to sp
i try to generate 27*5 matrix of observation using the following code but is
given error
kindly assist and correct the problem
x1<-c(-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1)
x2<-c(-1,-1,-1,0,0,0,1,1,1,-1,-1,-1,0,0,0,1,1,1,-1,-1,-1,0,0,0,1,1,1)
x3<-c(-1,-1,-1,-1,-1,-1,-1,
Hello,
My main question is wheter my data is distributed normally. As the
shapiro.test doesnt work for large
data sets I prefer the ks.test.
But I have some problems to understand the completely different p-values:
> ks.test (test, pnorm, mean (test), sd (test))
One-sample Kolmogorov-Smirno
Hi,
How do I find help on %%
Regards,
Prabhakar
On Thu, Mar 27, 2014 at 4:56 PM, Jorge I Velez wrote:
> Hi there,
>
> Try
>
> X[X %% 3 == 0]
>
> HTH,
> Jorge.-
>
>
> On Thu, Mar 27, 2014 at 6:46 PM, Prabhakar Ghorpade <
> dr.prabhaka...@gmail.com> wrote:
>
>> Hi,
>> here's my code
>>
>> X <- 1:
Pascal, Jorge, Berend, Pikal and Rui. Thanks all it helps.
On Thu, Mar 27, 2014 at 8:18 PM, Prabhakar Ghorpade <
dr.prabhaka...@gmail.com> wrote:
> Hi,
> How do I find help on %%
>
> Regards,
> Prabhakar
>
>
> On Thu, Mar 27, 2014 at 4:56 PM, Jorge I Velez
> wrote:
>
>> Hi there,
>>
>> Try
>>
>
Thank you! Works like a charm!
On Thu, Mar 27, 2014 at 12:19 PM, David Carlson wrote:
> That only requires two small changes in Sarah's first solution:
>
> > Finaldata[, !colnames(Finaldata) %in% V$v2[V$v1 == 1]]
> Length Rate
> 1 0.53607323 0.01739951
> 2 0.15405615 0.11837908
> 3
On Thu, 27 Mar 2014, Rich Shepard wrote:
s95ec.z <- read.zoo("s95ec.dat", sep = ".", format = "%Y-%m-%d"
^
) didn't
copy but was present on the command line.
Tryi
On Thu, 27 Mar 2014, Rich Shepard wrote:
I'll report back on whether that avoids the errors.
Changing from ts.union() to ts.intersect() did not make a difference. So,
I went back to bare metal by using awk to generate two-column text files,
each consisting of a date and an interger or real v
On Mar 27, 2014, at 11:58 AM, Martin Maechler
wrote:
>> Marc Schwartz
>>on Wed, 26 Mar 2014 16:25:08 -0500 writes:
>
>> On Mar 26, 2014, at 4:14 PM, David Winsemius wrote:
>
>>>
>>> On Mar 25, 2014, at 5:31 PM, Rolf Turner wrote:
>>>
On 26/03/14 12:51, David Winsemius wro
> Marc Schwartz
> on Wed, 26 Mar 2014 16:25:08 -0500 writes:
> On Mar 26, 2014, at 4:14 PM, David Winsemius
wrote:
>>
>> On Mar 25, 2014, at 5:31 PM, Rolf Turner wrote:
>>
>>> On 26/03/14 12:51, David Winsemius wrote:
On Mar 25, 2014, at 9:5
On 27/03/2014 12:46 PM, Jannis wrote:
On 27.03.2014 15:58, Duncan Murdoch wrote:
> I would use a slightly less dirty hack: call globalVariables() to
> declare that i is global.
>
> The foreach() function is using nonstandard evaluation to make this
> work, and codetools (that does the checks) do
On 27.03.2014 15:58, Duncan Murdoch wrote:
I would use a slightly less dirty hack: call globalVariables() to
declare that i is global.
The foreach() function is using nonstandard evaluation to make this
work, and codetools (that does the checks) doesn't know all the ins
and outs of it. (for
Dear all,
I have a vector x and a vector y = f(x).
e.g.
x = c(0,1, 3,3.4, 5, 10, 11.23)
y = x**2 + rnorm(length(x))
I would like to apply a moving average to y, knowing that the x values
are not uniformly spaced. How can I do this in R?
What alternatives I have to filter out the noise from Y??
T
That only requires two small changes in Sarah's first solution:
> Finaldata[, !colnames(Finaldata) %in% V$v2[V$v1 == 1]]
Length Rate
1 0.53607323 0.01739951
2 0.15405615 0.11837908
3 0.04542388 0.53702702
4 0.15633703 0.68870041
5 0.35293973 0.38258981
I have tried to read the source code but since I am not a computer engineer nor
a computer programmer I was not able to fully understand it. I wonder if I
should look for somebody here on campus (Washington State University) who may
be able to read it for me. In any case, I think that David Carl
Hi Sarah,
Thanks! Do agree its over complicated.
However looking at the solutions I think I did not state my problem
completely.
V provides choices for only certain set of columns in Finaldata.
So v2 may not represent all columns of Finaldata.
I want to retain columns not provided as a choice for
Here is how you can do it with data.table:
> x <- read.table(text = " author_id paper_id prob
+731249431
+731249431
+731 688974 1
+731 964345.8
+731 1201905.9
+731 12679921
+736249 .2
+736 68
On Thu, 27 Mar 2014, William Dunlap wrote:
Why aren't you using ts.intersect instead of ts.union?
Bill,
I was wondering the same thing. The reason (perhaps excuse would be the
more appropriate term) is my emulating the example in Paul Cowpertwait and
Andrew Metcalfe's "Introductory Time Ser
There are many ways. You're making it overly complicated
Here, in an actual reproducible example (as you were requested to submit):
V <- data.frame(v1=c(1,0,0), v2=c("Shape", "Length", "Rate"),
stringsAsFactors=FALSE)
Finaldata <- data.frame(Shape = runif(5), Length = runif(5), Rate = runif(5))
Why aren't you using ts.intersect instead of ts.union?
Bill Dunlap
TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of Rich Shepard
> Sent: Thursday, March 27, 2014 5:45 AM
> To: r-help@r-proje
Hi all!
I am trying to drop columns from a data frame dynamically depending on user
input. The dataframe whose columns need to be dropped is called "Finaldata"
So here is what I do:
V is a dataframe with columns v1 and v2 as follows
v1 v2
1 1 Shape
2 0 Length
3 0
On 27/03/2014 10:29 AM, Jannis wrote:
Dear R users,
I have created a package and would like to check it for CRAN/R-Forge
submission. I use some parallelized code with the help of foreach and doMC.
R CMD check now gives me a warning even though my code is correctly
programmed (at least I think
A reproducible example would help us help you, but you probably need to use
the "font" argument, something along the lines of this...
xlab = list("some axis label",cex=1.5,font=4) # 2=bold, 4 = bold italic
Kevin
On Thu, Mar 27, 2014 at 3:57 AM, el_alisio wrote:
> Dear R-users,
>
> I would li
Dear R users,
I have created a package and would like to check it for CRAN/R-Forge
submission. I use some parallelized code with the help of foreach and doMC.
R CMD check now gives me a warning even though my code is correctly
programmed (at least I think that it is). The following dummy fun
Dear Frede and pascal,
Thankyou very much. I am so glad that you replied positively.
Thankyou very once again.
Eliza
> From: kri...@ymail.com
> Date: Thu, 27 Mar 2014 22:38:42 +0900
> Subject: Re: [R] reading dataset
> To: fr...@vestas.com
> CC: eliza_bo...@hotmail.com; r-help@r-project.org
>
> Hi
Hi,
Maybe I wasn't clear enough.
Day 01 precip map of 180x140
Day 01 rstn map of 180x140
Day 01 flag map of 180x140
Day 02 precip map of 180x140
And so on
HTH
Pascal
On Thu, Mar 27, 2014 at 10:14 PM, Frede Aakmann Tøgersen
wrote:
> No you're not right as far as I can tell f
No you're not right as far as I can tell from the read_v1100r2.f90 fortran code
you can find in the download folder.
For day 1 in 1961 I think that prcp = ccc[1:25200], rstn = ccc[25201:50400],
and rsnw = ccc[50401:75600] and the same rule applies to the following days
with appropriate indices.
On Wed, 26 Mar 2014, Rich Shepard wrote:
According to the ts.union help, "‘ts.union’ pads with ‘NA’s to the total
time coverage" but the problem seems to be that one series has a single NA
while the other has 2 NAs. Both series have 792 rows of (date, value).
I should have written in the fi
It's rude to ask a question both on r-help and on stackoverflow
(http://stackoverflow.com/questions/22685896), because people might
spend their time answering your question when it's already been
answered elsewhere.
Hadley
On Thu, Mar 27, 2014 at 6:01 AM, Rohit Gupta wrote:
> I have a data A whi
Dear Pascal,
Thanks for your reply. From your answer I perceived that if followings are
first three elements of a file
> dput(ccc[1:3])
c(0.15912090241909, 0.167244642972946, 0.192471280694008)
then 0.15912090241909 is precipitation magnitude , 0.167244642972946 is RSTN
and 0.192471280694008 is f
Hi All,
I am using R for the purpose of multilevel modelling for the first time. I am
trying to examine individuals interpersonal changes in the dependent variable
over time and how this varies between groups.
I am using the following code:
treat.lme1<-lme(DependentVariable~Treatment*I(Time-1),
Hello,
Inline.
Em 27-03-2014 11:32, Pascal Oettli escreveu:
Hello,
Something like that?
R> X <- 1:100
R> Y <- X %% 3
R> Y <- ifelse(Y==0, TRUE, FALSE)
The ifelse here is not needed, you can simply do
Y <- Y == 0
Rui Barradas
R> X[Y]
[1] 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48
Excellent Arun, it worked, many thanks.
E.
On Thu, Mar 27, 2014 at 11:55 AM, arun wrote:
> HI Elio,
>
> May be this helps:
> indx <- order(gsub("\\d+","",colnames(res)))
> res1 <- res[indx,indx]
> A.K.
>
>
>
>
>
>
>
> On Thursday, March 27, 2014 6:24 AM, Elio Shijaku
> wrote:
>
> Hi Arun,
>
Hello,
Something like that?
R> X <- 1:100
R> Y <- X %% 3
R> Y <- ifelse(Y==0, TRUE, FALSE)
R> X[Y]
[1] 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75
[26] 78 81 84 87 90 93 96 99
HTH,
Pascal
On Thu, Mar 27, 2014 at 6:23 PM, Prabhakar Ghorpade
wrote:
> Hi,
> here's
Hi
One option is
seq(3,100,3)
or you can use rounded numbers for comparison
?round
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Prabhakar Ghorpade
> Sent: Thursday, March 27, 2014 8:47 AM
> To: r-help@r-project.or
Hi there,
Try
X[X %% 3 == 0]
HTH,
Jorge.-
On Thu, Mar 27, 2014 at 6:46 PM, Prabhakar Ghorpade <
dr.prabhaka...@gmail.com> wrote:
> Hi,
> here's my code
>
> X <- 1:100
>
> I want to select number divisible by 3 out of them how can I select it?
>
> ( I tried following
> X <- 1:100
> DIV <- Y -
On 27-03-2014, at 10:23, Prabhakar Ghorpade wrote:
> Hi,
> here's my code
>
> X <- 1:100
>
> I want to select number divisible by 3 out of them how can I select it?
>
> ( I tried following
> X <- 1:100
> DIV <- Y - > X/3
>
> But I am getting whole number and number with fractions. WHole intg
Hi,
here's my code
X <- 1:100
I want to select number divisible by 3 out of them how can I select it?
( I tried following
X <- 1:100
DIV <- Y - > X/3
But I am getting whole number and number with fractions. WHole intgers are
my number of interest from original X. How can I traceback to number
d
I have a data A which looks like
author_id paper_id prob
731249431
731249431
731 6889741
731 964345.8
731 1201905.9
731 12679921
736249 .2
736 6889 1
736 94345.7
7
Hi,
here's my code
X <- 1:100
I want to select number divisible by 3 out of them how can I select it?
( I tried following
X <- 1:100
DIV <- Y - > X/3
But I am getting whole number and number with fractions. WHole intgers are
my number of interest from original X. How can I traceback to number
d
Dear R-users,
I would like to change the font of one specific axis label to bold face in
each panel. Taking one of Deepayan's plots as an example, how do I change
"All Postdoctorates" to bold face?
dotplot(prop.table(postdoc, margin = 1), groups = FALSE, index.cond =
function(x, y) median(x), xla
HI Elio,
May be this helps:
indx <- order(gsub("\\d+","",colnames(res)))
res1 <- res[indx,indx]
A.K.
On Thursday, March 27, 2014 6:24 AM, Elio Shijaku wrote:
Hi Arun,
Thanks a lot for your continuous help. I am attaching a sample with the row /
column names. Basically, the order should
Hi,
Try:
A <- as.data.frame(matrix(1:20,byrow=TRUE,ncol=4))
B <- as.data.frame(matrix(21:40,byrow=TRUE,ncol=4))
AB <- rbind(A,B)
#if the ?row.names of both datasets are from 1:nrow(dataset)
AB1 <- AB[order(as.numeric(c(row.names(A),row.names(B,]
#or
AB2 <- AB[c(matrix(seq(dim(AB)[1]),nrow=2,b
HI Eliza,
May be this helps:
set.seed(42)
el <- matrix(sample(1:20,327*365,replace=TRUE),ncol=327)
colnames(el) <- as.character(interaction(df1,sep="*"))
el1 <- el[,colnames(el) %in% as.character(interaction(df2,sep="*"))]
dim(el1)
#[1] 365 5
A.K.
On Tuesday, March 25, 2014 4:15 PM, eliza
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