Sorkin, John wrote/hat geschrieben on/am 01.07.2024 17:54:
#I am trying to write code that will create a matrix with a variable number of
columns where the #number of columns is 1+Grps
#I can do this:
NSims <- 4
Grps <- 5
DiffMeans <- matrix(nrow=NSims,ncol=1+Grps)
DiffMeans
#I have a problem w
I think you should reconsider your goal. Matrices must have all elements of the
same type, and in this case you seem to be trying to mix a number of something
(integer) with mean values (double). This would normally be stored together in
a data frame or separately in a vector for counts and a ma
Às 16:54 de 01/07/2024, Sorkin, John escreveu:
#I am trying to write code that will create a matrix with a variable number of
columns where the #number of columns is 1+Grps
#I can do this:
NSims <- 4
Grps <- 5
DiffMeans <- matrix(nrow=NSims,ncol=1+Grps)
DiffMeans
#I have a problem when I try to
Às 16:54 de 01/07/2024, Sorkin, John escreveu:
#I am trying to write code that will create a matrix with a variable number of
columns where the #number of columns is 1+Grps
#I can do this:
NSims <- 4
Grps <- 5
DiffMeans <- matrix(nrow=NSims,ncol=1+Grps)
DiffMeans
#I have a problem when I try to
NSims <- 4
Grps <- 5
DiffMeans <- matrix(nrow=NSims,ncol=1+Grps)
DiffMeans
#I have a problem when I try to name the columns of the matrix. I want the
first column to be NSims, #and the other columns to be something like Value1,
Value2, . . . Valuen where N=Grps
Colnames <- as.vector("NSims")
num
quot;Smith" "Smith" "Smith" "Smith" "Jones" "Jones" "Jones"
"Gunter"
If that remains unclear, I am doing something wrong today.
It is nice to see alternatives but some people just want one answer. Read my
first mes
"If you want a much more compact solution that handles arbitrary pairs of
"what to copy", number_of_copies, you can write a function that evaluates
two arguments at a time or takes two vectors as arguments like this one I
wrote quickly and crudely:"
Please! -- The "times" argument of rep can be a
For the particular example you asked for, consider the "each" you can use
with rep()
rep(1:13, each=84)
This is what it does for a shorter version of 4 each:
> rep(1:13, each=4)
[1] 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6
7 7 7 7 8 8 8 8 9 9 9 9 10 10
Bert Gunter
Sent: Thursday, June 13, 2024 9:13 PM
To: Ebert,Timothy Aaron
Cc: Francesca PANCOTTO ; r-help@r-project.org
Subject: Re: [R] Create a numeric series in an efficient way
[External Email]
Nope. She would have wanted the 'each' argument = 84. See ?rep.
-- Bert
On Thu, Jun 13,
t; The second line sorts, but that may not be needed depending on
> application. The object class is also different in the sorted solution.
>
> Tim
> -Original Message-
> From: R-help On Behalf Of Francesca
> PANCOTTO via R-help
> Sent: Thursday, June 13, 2024 2:22 PM
rom: R-help On Behalf Of Francesca PANCOTTO via
R-help
Sent: Thursday, June 13, 2024 2:22 PM
To: r-help@r-project.org
Subject: Re: [R] Create a numeric series in an efficient way
[External Email]
I apologize, I solved the problem, sorry for that.
f.
Il giorno gio 13 giu 2024 alle ore 16:42 Fra
I apologize, I solved the problem, sorry for that.
f.
Il giorno gio 13 giu 2024 alle ore 16:42 Francesca PANCOTTO <
francesca.panco...@unimore.it> ha scritto:
> Dear Contributors
> I am trying to create a numeric series with repeated numbers, not
> difficult task, but I do not seem to find an e
Dear Bert,
Many thanks for your suggestion! I am reading the section to
understand more about this topic. It is highly relevant to what I plan
to work on.
Regards,
Shu Fai
On Thu, Oct 26, 2023 at 5:38 AM Bert Gunter wrote:
>
> As you seem to have a need for this sort of capability (e.g. bquote)
As you seem to have a need for this sort of capability (e.g. bquote),
see Section 6: "Computing on the Language" in the R Language
Definition manual. Actually, if you are interested in a concise
(albeit dense) overview of the R Language, you might consider going
through the whole manual.
Cheers,
B
Dear Iris,
Many many thanks! This is exactly what I need! I have never heard
about bquote(). This function will also be useful to me on other
occasions.
I still have a lot to learn about the R language ...
Regards,
Shu Fai
On Wed, Oct 25, 2023 at 5:24 PM Iris Simmons wrote:
>
> You can try ei
You can try either of these:
expr <- bquote(lm(.(as.formula(mod)), dat))
lm_out5 <- eval(expr)
expr <- call("lm", as.formula(mod), as.symbol("dat"))
lm_out6 <- eval(expr)
but bquote is usually easier and good enough.
On Wed, Oct 25, 2023, 05:10 Shu Fai Cheung wrote:
> Hi All,
>
> I have a pro
Sorry for a typo, regarding the first attempt, lm_out2, using
do.call(), I meant:
'It does have the formula, "as a formula": y ~ x1 + x2.
However, the name "dat" is evaluated. ...'
Regards,
Shu Fai
On Wed, Oct 25, 2023 at 5:09 PM Shu Fai Cheung wrote:
>
> Hi All,
>
> I have a problem that may h
This seems to work. A couple of fine points, including handling duplicated Pct
values right, which is easier if you do the reversed cumsum.
> dd2 <- dummydata[order(dummydata$Pct),]
> dd2$Cum <- rev(cumsum(rev(dd2$Totpop)))
> use <- !duplicated(dd2$Pct)
> approx(dd2$Pct[use], dd2$Cum[use], ctof,
Sorry, misstatements. It should (of course) read:
If one makes the reasonable assumption that Pct is much larger than
Cutoff, sorting Pct is the expensive part e.g O(nlog2(n) for
Quicksort (n = length Pct). I believe looping is O(n^2).
etc.
On Mon, Oct 16, 2023 at 7:48 AM Bert Gunter wrote:
>
>
If one makes the reasonable assumption that Pct is much larger than
Cutoff, sorting Cutoff is the expensive part e.g O(nlog2(n) for
Quicksort (n = length Cutoff). I believe looping is O(n^2). Jeff's
approach using findInterval may be faster. Of course implementation
details matter.
-- Bert
On Mo
Dear Jason,
The code could look something like:
dummyData = data.frame(Tract=seq(1, 10, by=1),
Pct = c(0.05,0.03,0.01,0.12,0.21,0.04,0.07,0.09,0.06,0.03),
Totpop = c(4000,3500,4500,4100,3900,4250,5100,4700,4950,4800))
# Define the cutoffs
# - allow for duplicate entries;
by = 0.03; # by
Dear Jason,
I do not think that the solution based on aggregate offered by GPT was
correct. That quasi-solution only aggregates for every individual level.
As I understand, you want the cumulative sum. The idea was proposed by
Bert; you need only to sort first based on the cutoff (e.g. usin
After I sent this, a colleague referred me to the GPT-4 interface on Bing. I
entered the exact email query below and it provided the following solution,
which worked for the toy example and was successfully adapted to my application:
# Define the cutoffs
cutoffs <- seq(0, 0.15, by = 0.01)
# Cr
Pre-compute the per-interval answers and use findInterval to look up the
per-row answers...
dat <- read.table( text=
"Tract Pct Totpop
1 0.054000
2 0.033500
3 0.014500
4 0.124100
5 0.
Well, here's one way to do it:
(dat is your example data frame)
Cutoff <- seq(0, .15, .01)
Pop <- with(dat, sapply(Cutoff, \(p)sum(Totpop[Pct >= p])))
I think there must be a more efficient way to do it with cumsum(), though.
Cheers,
Bert
On Sat, Oct 14, 2023 at 12:53 AM Jason Stout, M.D. wrot
lf Of Sorkin, John
Sent: Tuesday, July 4, 2023 12:17 AM
To: Rolf Turner ; Bert Gunter
Cc: r-help@r-project.org (r-help@r-project.org) ;
Achim Zeileis
Subject: Re: [R] Create a variable lenght string that can be used in a
dimnames statement
My life is complete.
I have inspired a fortune!
John
(Sorry for the double post.)
On Tue, 4 Jul 2023 10:14:43 +0300
Ivan Krylov wrote:
> Try replacing the _second_ paste() in the example above with a c().
What I had forgotten to mention is that you also need to replace the
initial assignment
> string=""
with the following:
string = charact
On Mon, 3 Jul 2023 20:08:06 +
"Sorkin, John" wrote:
> # create variable names xxx1 and xxx2.
> string=""
> for (j in 1:2){
> name <- paste("xxx",j,sep="")
> string <- paste(string,name)
> print(string)
> }
> # Creation of xxx1 and xxx2 works
> string
You need to distinguish between a s
My life is complete.
I have inspired a fortune!
John
From: Rolf Turner
Sent: Monday, July 3, 2023 6:34 PM
To: Bert Gunter
Cc: Sorkin, John; r-help@r-project.org (r-help@r-project.org); Achim Zeileis
Subject: Re: [R] Create a variable lenght string that
At least on my system string has a single value of " xxx1 xxx2" not "xxx1" and
"xxx2".
The variable zzz has two values: "J K xxx1" and "J K xxx2"
What you want is "J", "K", "xxx1", "xxx2"
If I cheat everything works. So then the goal is to rewrite the program so
cheating is not needed.
# cr
On Mon, 3 Jul 2023 13:40:41 -0700
Bert Gunter wrote:
> I am not going to try to sort out your confusion, as others have
> already tried and failed.
Fortune nomination!!!
cheers,
Rolf Turner
--
Honorary Research Fellow
Department of Statistics
University of Auckland
Stats. Dep't. (secreta
I am not going to try to sort out your confusion, as others have already
tried and failed. But I will point out that "string" of variables is pretty
much nonsense in R. A "character vector"/"vector of strings" is probably
what you mean and want to provide column names // names for the second
compon
quot; to the string of column names
zzz <- paste("j","k",string)
zzz
# assign column names, j, k, xxx1, xxx2 to the matrix
# create column names, j, k, xxx1, xxx2.
dimnames(myvalues)<-list(NULL,c(zzz))
colnames(myvalues)<-zzz
____
F
aste("j","k",string)
zzz
# assign column names, j, k, xxx1, xxx2 to the matrix
# create column names, j, k, xxx1, xxx2.
dimnames(myvalues)<-list(NULL,c(zzz))
colnames(myvalues)<-zzz
____
From: Jeff Newmiller
Sent: Monday, July 3, 2023
umn names, j, k, xxx1, xxx2 to the matrix
# create column names, j, k, xxx1, xxx2.
dimnames(myvalues)<-list(NULL,c(zzz))
colnames(myvalues)<-zzz
____
From: Jeff Newmiller
Sent: Monday, July 3, 2023 2:45 PM
To: Sorkin, John
Cc: r-help@r-project.org
Subj
I really think you should read that help page. colnames() accesses the second
element of dimnames() directly.
On July 3, 2023 11:39:37 AM PDT, "Sorkin, John"
wrote:
>Jeff,
>Thank you for your reply.
>I should have said with dim names not column names. I want the Mateix to have
>dim names, no
Jeff,
Thank you for your reply.
I should have said with dim names not column names. I want the Mateix to have
dim names, no row names, dim names j, k, xxx1, xxx2.
John
John David Sorkin M.D., Ph.D.
Professor of Medicine
Chief, Biostatistics and Informatics
University of Maryland School of Medici
Às 19:00 de 03/07/2023, Sorkin, John escreveu:
I am trying to create an array, myvalues, having 2 rows and 4 columns, where the column
names are j,k,xxx1,xxx2. The code below fails, with the following error, "Error in
dimnames(myvalues) <- list(NULL, zzz) :
length of 'dimnames' [2] not equal
?colnames
On July 3, 2023 11:00:32 AM PDT, "Sorkin, John"
wrote:
>I am trying to create an array, myvalues, having 2 rows and 4 columns, where
>the column names are j,k,xxx1,xxx2. The code below fails, with the following
>error, "Error in dimnames(myvalues) <- list(NULL, zzz) :
> length of '
but they too take one approach to solve a problem rather
> than "here is a problem" and "these are all possible solutions." I
> appreciate seeing alternative solutions.
>
> Tim
>
> -Original Message-
> From: R-help On Behalf Of Richard O'Kee
Hana, the "right" answer depends on exactly what you need. Here are three
correct solutions. They use the same basic strategy to give different results.
There are also other approaches in R to get the same outcome. You could use
data_catigocal[i,j] and some for loops.
size1 <-5
ngroup <- 1
Hello,
Replace the for loop by a lapply loop and assign the names after, with
sprintf("Corr%d", 1:10)
Is out a data.frame? Can you post dput(head(out))?
Hope this helps,
Rui Barradas
Às 18:33 de 14/05/2022, Sorkin, John escreveu:
I am trying to create distinct variable names of the form Co
Thank you for the clarification.
On Sat, Oct 23, 2021 at 11:42 PM Bert Gunter wrote:
>
> You appear to be bombarding the list with statistics questions. Please note
> per the posting guide linked below:
>
> "Questions about statistics: The R mailing lists are primarily intended for
> questions
You appear to be bombarding the list with statistics questions. Please note
per the posting guide linked below:
"*Questions about statistics:* The R mailing lists are primarily intended
for questions and discussion about the R software. However, questions about
statistical methodology are sometime
rner
> Subject: Re: [R] Create a function problem
>
> Hi Rolf,
> I am a beginner for R.
> I have a date frame raw. it contents the fields of pedigree.name, UPN,
> Test.Result.tr_Test.Result1, Result.tr_gene1,
> Test.Result.tr_Variant..nucleotide.1 .. Test.Result.
Have you tried putting the a,b, and c column names you are passing in quotes
i.e. as strings? Currently your function is expecting separate objects with
those names.
The select function itself can accept unquoted column names, as can others in
R, because specific processing they do in the backg
On Sat, 15 May 2021 00:55:08 + (UTC)
Kai Yang wrote:
> Hi Rolf,
> I am a beginner for R.
Then I suggest that you spend some time learning basic R syntax, with
the help of some of the excellent online tutorials. "An Introduction
to R" from https://cran.r-project.org/manuals.html would be a g
Hi Rolf,
I am a beginner for R.
I have a date frame raw. it contents the fields of pedigree.name, UPN,
Test.Result.tr_Test.Result1, Result.tr_gene1,
Test.Result.tr_Variant..nucleotide.1 .. Test.Result.tr_Test.Result20,
Result.tr_gene20, Test.Result.tr_Variant..nucleotide.20
Basically, I wa
On Fri, 14 May 2021 17:42:12 + (UTC)
Kai Yang via R-help wrote:
> Hello List, I was trying to write a function. But I got a error
> message. Can someone help me how to fix it? Many thanks,Kai
> > k_subset <- function(p, a, b, c){
> + p <- select(raw
> + ,Pedigree.name
> +
Thanks for the reprex. I think this is one way to do what you want:
dt$flag2 <- 0 + with(dt,Item == "DESK" & check %in% code2)
> dt$flag2 <- 0 + with(dt,Item == "DESK" & check %in% code2)
> dt
name Item check flag2
1A DESK NORF 0
2B RANGE GARRA 0
3C CLOCK PALM 0
4
"I thought: why make this overly complicated,..."
Indeed, though "complicated" is in the eyes of the beholder.
One wonders whether any of this is necessary, though: see ?apply , as in
apply(a, 1, whatever...)
to do things rowwise.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind i
On 2020-04-09 18:00 +, aiguo li wrote:
| That is awesome! Thanks.
Dear AiGuo,
I thought: why make this overly
complicated, when this is also
possible:
a <- matrix(LETTERS[1:16], nrow=4)
X <- split(x=a[,-1], f=a[,1])
lapply(X=X, FUN=as.factor)
Best,
Rasmus
_
On 2020-04-09 18:00 +, aiguo li wrote:
| That is awesome! Thanks.
I'm glad this was helpful for you!
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
On 2020-04-09 18:50 +0100, Rui Barradas wrote:
| Hello,
|
| Your post is unreadable, please repost in
| *plain text*, not HTML.
Hi! It was not so bad? I was able to
extract out the core parts at least to
prepare an answer ... maybe a bit hard with
no line breaks, but ...
Best,
Rasmus
___
Hello,
Your post is unreadable, please repost in *plain text*, not HTML.
Rui Barradas
Às 16:00 de 09/04/20, aiguo li via R-help escreveu:
Hello allI need to create a r list with each row as a list object and named with the
element in the first column. Illustrated below:> a<-
as.data.frame(m
On 2020-04-09 15:00 +, aiguo li via R-help wrote:
| Hello allI need to create a r list with
| each row as a list object and named with
| the element in the first column.
Dear aiguo,
Perhaps this fits your bill?
a <- matrix(LETTERS[1:16], nrow = 4)
FUN <- function(x) { as.factor(x[-1]
Hi Marna,
I was able to have another look at it. I think this is what you want
except for the column names (I'm lazy):
dAT<-structure(list(Id = structure(c(1L, 2L, 3L, 4L, 1L, 2L, 5L, 4L,
1L, 6L, 6L, 3L, 4L, 1L), .Label = c("a", "b", "c", "e", "f",
"m"), class = "factor"), date = structure(c(1L, 1
Hi Marna,
You can get the information you need with this:
dAT$date<-as.Date(dAT$date,"%d-%b-%y")
diffs<-function(x,maxn) return(diff(x)[1:maxn])
initdate<-function(x) return(min(x))
datediffs<-aggregate(dAT$date,list(dAT$Id),diffs,3)
I can't do the manipulation of the resulting values at the mome
Just use a temporary variable!
## You should apply is.na() only to the numeric columns you need, not the
whole data frame
## see ?'[' for why.
vnew <- vdat[,3:5]
vnew[is.na(vnew)] <- 0
vdat$xy <- as.matrix(vnew) %*% c(2, 5, 3)
vdat
I still question whether this is wise; but that's for you to dete
Sorry for the confusion, my sample data does not represent the
actual data set.
The range of value can be from -ve to +ve values and 0 could be a
true value of an observation. So, instead of replacing missing value
by zero, I want exclude them from the calculation.
On Sat, Apr 13, 2019 at
Looks to me like your initial request contradicts your clarification. Can you
explain this discrepancy?
On April 13, 2019 8:29:59 PM PDT, Val wrote:
>Hi Bert and Jim,
>Thank you for the suggestion.
>However, those missing values should not be replaced by 0's.
>I want exclude those missing values
Hi Bert and Jim,
Thank you for the suggestion.
However, those missing values should not be replaced by 0's.
I want exclude those missing values from the calculation and create
the index using only the non-missing values.
On Sat, Apr 13, 2019 at 10:14 PM Jim Lemon wrote:
>
> Hi Val,
> For this pa
If the NA's are really 0's, replace them with 0 before doing the
calculation. (see ?is.na).
If they are not 0's, think again about doing this as the results would
probably mislead.
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
--
Hi Val,
For this particular problem, you can just replace NAs with zeros.
vdat[is.na(vdat)]<-0
vdat$xy <- 2*(vdat$x1) + 5*(vdat$x2) + 3*(vdat$x3)
vdat
obs Year x1 x2 x3 xy
1 1 2001 25 10 10 130
2 2 2001 0 15 25 150
3 3 2001 50 10 0 150
4 4 2001 20 0 60 220
Note that this is not a gen
2019 8:21 AM
> To: bienvenidoz...@gmail.com
> Cc: R-help
> Subject: Re: [R] Create a sequence
>
>
>
> > On Apr 9, 2019, at 9:07 AM, bienvenidoz...@gmail.com wrote:
> >
> > how to create
> > u = (1, −1, 2, −2, . . . , 100, −100) in r
> >
> > Tha
--Original Message-
From: R-help On Behalf Of Marc Schwartz via
R-help
Sent: Tuesday, April 9, 2019 8:21 AM
To: bienvenidoz...@gmail.com
Cc: R-help
Subject: Re: [R] Create a sequence
> On Apr 9, 2019, at 9:07 AM, bienvenidoz...@gmail.com wrote:
>
> how to create
> u = (1, −1, 2
> On Apr 9, 2019, at 9:07 AM, bienvenidoz...@gmail.com wrote:
>
> how to create
> u = (1, −1, 2, −2, . . . , 100, −100) in r
>
> Thanks
> Bienvenue
Hi,
See ?seq and ?rep
> rep(seq(100), each = 2) * c(1, -1)
[1]1 -12 -23 -34 -45 -56 -67
[14] -7
I an finding your question very opaque. For one thing, why aren't you
specifying any edges? Since I don't do this kind of analysis normally I don't
know what your combined graph should look like. Can you provide a link you what
you hope to end up with?
Note that the Posting Guide emphasizes tha
Thanks for your reply but it does not solve the problem. Since I have 7 text
rows in my df then I have to run something like
net2=graph(c("account","block","block","solve","solve","problem")) for each,
I'm looking for a way to bring all the relations in one single plot
On Thursday, January 3
BBB <- lapply( CCC, function( v ) v[ 0 wrote:
>Hi All--
>
>I have a list which contain variables 0s at the end of each vector on
>the isit. I want to create a new list with only numbers > 0.
>It seems simple, but i tried several option, none of which worked.
>
>CCC <- list(A=c(1,2,3,0,0,0,0), B=c(
Dear Petr, thank you for the guidance.
A colleague managed to solve it
I'll definitely use "dput" for future postings.
Regards
--
Kevin Wamae
On 17/01/2019, 03:57, "PIKAL Petr" wrote:
Hi
Instead of attachment which is usually removed you should use dput
Hi
Instead of attachment which is usually removed you should use dput
Something like output from
dput(head(yourdata,30))
To remove duplicate values see
unique or duplicated
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of Kevin Wamae
> Sent: Thursday, January 17, 2019 1:2
Hi David,
Thank you so much for your kind suggestion and fixing my code!! I learn a lot
from you today.
Ding
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Thursday, April 19, 2018 6:58 PM
To: Ding, Yuan Chun
Cc: r-help@r-project.org
Subject: Re: [R
> On Apr 19, 2018, at 1:22 PM, David Winsemius wrote:
>
>
>> On Apr 19, 2018, at 11:20 AM, Ding, Yuan Chun wrote:
>>
>> Hi All,
>>
>> I want to create a categorical variable, cat.pfoa, in the file of pfas.pheno
>> (a data frame) based on log2pfoa values. I can do it using the following
>>
> On Apr 19, 2018, at 11:20 AM, Ding, Yuan Chun wrote:
>
> Hi All,
>
> I want to create a categorical variable, cat.pfoa, in the file of pfas.pheno
> (a data frame) based on log2pfoa values. I can do it using the following code.
>
> pfas.pheno <-within(pfas.pheno, {cat.pfoa<-NA
> cat.pfoa[pf
Hi Rui,
Thank you very much for your help!! It works very well, I got it.
Ding
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: Thursday, April 19, 2018 11:35 AM
To: Ding, Yuan Chun ; r-help@r-project.org
Subject: Re: [R] create multiple categorical
Hello,
When programming it is better to use dat[["variable"]] than dat$variable.
So your code could be
pfas.pheno[[cat.var]] <- NA
pfas.pheno[[cat.var]][pfas.pheno[,i] <= quantile(pfas.pheno[,i],0.25,
na.rm =T)] <- 0
etc.
Untested.
Hope this helps,
Rui Barradas
On 4/19/2018 7:20 PM, Ding
Here is perhaps a better sawtooth wave generator:
sawtooth <- function(freq, duration = samp.rate, from = 0, samp.rate =
44100, bit = 1,
stereo = FALSE, xunit = c("samples", "time"), reverse
= FALSE, ...){
xunit <- match.arg(xunit)
durFrom <- preWaveform(freq = freq
Ccing the maintainer if the tuneR package.
Looks to me like sawtooth (and square) don't behave as expected when using
xunit="samples". Workaround is to use xunit="time" instead:
sawtooth(110,duration=1/100,samp.rate=sample_rate,xunit="time")
I looked at the code but found it to be opaque.
--
My tuneR sawtooth wave function generator is broken.
When I use the sine function, I get exactly what I expect: a sine wave whose
frequency is defined by the freq parameter. In particular, higher frequencies
have shorter wavelengths (more cycles per second means shorter waves). When I
create
you sould look at "boot" package. also search "bootstrap R" keywords in
google.
20 Eki 2017 23:12 tarihinde "Marna Wagley" yazdı:
> Hi R Users,
> I do have very big data sets and wanted to run some of the analyses many
> times with randomization (1000 times).
> I have done the analysis using an
> On May 6, 2017, at 8:14 AM, David Winsemius wrote:
>
>>
>> On May 6, 2017, at 7:47 AM, Elahe chalabi via R-help
>> wrote:
>>
>> Hi all,
>>
>> I have a text classification task which is classification of a Control group
>> and Alzheimer group texts. I have generated DocumentTermMatrix for
> On May 6, 2017, at 7:47 AM, Elahe chalabi via R-help
> wrote:
>
> Hi all,
>
> I have a text classification task which is classification of a Control group
> and Alzheimer group texts. I have generated DocumentTermMatrix for both
> groups and then created a list with one extra element showi
Hi Val,
How about this:
cinmat<-
matrix(c(mydat$x1>0,mydat$x1==0&mydat$x2==0,mydat$x1==0&mydat$x2>0),
ncol=3)
mydat$x4<-rowSums(mydat[,c("x1","x3","x2")]*cinmat)
Jim
On Sat, Mar 25, 2017 at 2:56 PM, Val wrote:
> Hi all,
>
>
> I have several variables in a group and one group contains three
?ifelse
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Fri, Mar 24, 2017 at 8:56 PM, Val wrote:
> Hi all,
>
>
> I have several variables in a g
For those interested, I figured out a way using "convert" on the linux
command line.
Thanks
On Tue, Feb 14, 2017 at 6:02 PM, Bert Gunter wrote:
> Ulrik:
>
> Sheepishly Nitpicking (only because you are a regular and wise R-help
> contibutor):
>
> gganimate is not a library, it's a package.
>
> N
Ulrik:
Sheepishly Nitpicking (only because you are a regular and wise R-help
contibutor):
gganimate is not a library, it's a package.
No need to reply.
Best,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Be
On 14/02/2017 5:53 AM, Shane Carey wrote:
Hi,
I have many png files that I would like to stitch together, in order to
make a gif file.
Any ideas how I would do this?
ImageMagick is good for this. The magick package in R makes most (all?)
of its features available. The image_animate() funct
Hi Ulrick,
I created the png's in R and was hoping there would be a way of stitching
them together in R. Is there a different forum I should be posting to?
Thanks
On Tue, Feb 14, 2017 at 11:33 AM, Ulrik Stervbo
wrote:
> Hi Shane,
>
> Wrong forum. This might be what you are looking for
>
> ffm
Hi Shane,
Wrong forum. This might be what you are looking for
ffmpeg -i %03d.png output.gif
Or use the library gganimate.
Best
Ulrik
Shane Carey schrieb am Di., 14. Feb. 2017, 12:08:
> Hi,
>
> I have many png files that I would like to stitch together, in order to
> make a gif file.
>
> Any
On 06 Dec 2016, at 11:17 , Jim Lemon wrote:
> Hi Paul,
> The easy to understand way is:
>
> n <- c(1:10)
> # Create empty list to store vectors
> list_of_vecs <- list()
>
> # Create n vectors of random numbers - length 10. This works ok.
> for (i in n){
> list_of_vecs[[i]]<-rnorm(10,0,1)
> }
Hi Paul,
The easy to understand way is:
n <- c(1:10)
# Create empty list to store vectors
list_of_vecs <- list()
# Create n vectors of random numbers - length 10. This works ok.
for (i in n){
list_of_vecs[[i]]<-rnorm(10,0,1)
}
If you really want to use "assign":
for (i in n){
vecname<-paste('v
Thank you Jim.
On 6 December 2016 at 9:17:21 pm, Jim Lemon (drjimle...@gmail.com) wrote:
Hi Paul,
The easy to understand way is:
n <- c(1:10)
# Create empty list to store vectors
list_of_vecs <- list()
# Create n vectors of random numbers - length 10. This works ok.
for (i in n){
> On Oct 19, 2016, at 9:41 AM, Jeff Newmiller wrote:
>
> When I replace length with nchar, it works fine for me without mapply.
>
> substr( rep( s, nchar(s) ), 1, seq.int( nchar(s) ) )
I failed to make the second `nchar` -> `length` substitution. It now works for
me as well.
--
David
> --
When I replace length with nchar, it works fine for me without mapply.
substr( rep( s, nchar(s) ), 1, seq.int( nchar(s) ) )
--
Sent from my phone. Please excuse my brevity.
On October 19, 2016 9:36:25 AM PDT, David Winsemius
wrote:
>
>> On Oct 19, 2016, at 8:44 AM, Jeff Newmiller
> wrote:
>>
> On Oct 19, 2016, at 8:44 AM, Jeff Newmiller wrote:
>
> These don't look like "suffixes" to me, but whatever.
>
> s <- "abc"
> substr( rep( s, length(s) ), 1, seq.int( length(s) ) )
I suspect that `nchar` was meant instead of `length` but it still failed. How
about:
lets <- paste0(letters,
These don't look like "suffixes" to me, but whatever.
s <- "abc"
substr( rep( s, length(s) ), 1, seq.int( length(s) ) )
--
Sent from my phone. Please excuse my brevity.
On October 19, 2016 8:01:10 AM PDT, Witold E Wolski wrote:
>Is there a build in function, which creates n suffixes of length
purrr::map(paste0(letters, collapse=""), ~purrr::map2_chr(.,
1:nchar(.), ~substr(.x, 1, .y)))[[1]]
seems to crank really fast at least on my system
what did you try that was slow?
On Wed, Oct 19, 2016 at 11:01 AM, Witold E Wolski wrote:
> Is there a build in function, which creates n suffixes o
> On Oct 12, 2016, at 11:20 AM, Ashta wrote:
>
> Hi David and all,
>
> I want run the following script in a loop but faced difficulty.
>
> trt=c(1,2,2,4,5,6,7,8)
> for(i in 1:length (trt))
> {
> try[i] <- (select trt, date1, date2, datediff(date1,date2) as
> d12diff [i] from
>
Hi David and all,
I want run the following script in a loop but faced difficulty.
trt=c(1,2,2,4,5,6,7,8)
for(i in 1:length (trt))
{
try[i] <- (select trt, date1, date2, datediff(date1,date2) as
d12diff [i] from
dateTable where trt=[i]")
}
I would appreciate if you point me
1 - 100 of 704 matches
Mail list logo
