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
d some error checking like making sure the first parameter is a
matrix and the other two parmeters are appropriate strings.
If the function will only be called once or twice it might be simpler to not
use a function.
Tim
-Original Message-
From: R-help On Behalf Of Sorkin, John
Sent: M
#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 name the columns of the matrix. I want the
firs
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
rep_many(1:13, rep(84,13))
>
>
> The same ideas can be used using a data.frame or functional programming
> methods but the above is simple enough to flexibly create two vectors
> specifying how much of each.
>
> You said you found a solution, so you may want to share what you chose
o you may want to share what you chose
already.
-Original Message-
From: R-help On Behalf Of Francesca PANCOTTO
via R-help
Sent: Thursday, June 13, 2024 10:42 AM
To: r-help@r-project.org
Subject: [R] Create a numeric series in an efficient way
Dear Contributors
I am trying to create a nume
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 Contributors
I am trying to create a numeric series with repeated numbers, not difficult
task, but I do not seem to find an efficient way.
This is my solution
blocB <- c(rep(x = 1, times = 84), rep(x = 2, times = 84), rep(x = 3, times
= 84), rep(x = 4, times = 84), rep(x = 5, times = 84), re
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
Hi All,
I have a problem that may have a simple solution, but I am not
familiar with creating calls manually.
This is example calling lm()
``` r
set.seed(1234)
n <- 10
dat <- data.frame(x1 = rnorm(n),
x2 = rnorm(n),
y = rnorm(n))
lm_out <- lm(y ~ x1 + x2, dat
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
the result
result
So thanks to all for considering this query�we're in a brave new world of
AI-generated coding.
Message: 3
Date: Fri, 13 Oct 2023 20:13:56 +
From: "Jason Stout, M.D."
To: "r-help@r-project.org"
Subject: [R] Create new data frame with conditi
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
This seems like it should be simple but I can't get it to work properly. I'm
starting with a data frame like this:
Tract Pct Totpop
1 0.054000
2 0.033500
3 0.014500
4 0.124100
5 0.21
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
c("j","k",string) # assign column names, j, k, xxx1, xxx2 to the
matrix # create column names, j, k, xxx1, xxx2.
dimnames(myvalues)<-list(NULL,c(zzz))
Regards,
Tim
-Original Message-
From: R-help On Behalf Of Sorkin, John
Sent: Monday, July 3, 2023 4:08 PM
T
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
Colleagues,
I am sending this email again with a better description of my problem and the
area where I need help.
I need help creating a string of variables that will be accepted by the
dimnames function. The string needs to start with the dimnames j and k followed
by a series of dimnames xxx1
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 '
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 to array extent"
Please help me get the code t
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
categorical2 <- cbind(option2, data_catigocal)
option3 <- rep(group_vector, group_size)
option3a <- sample(option3, size1, replace=FALSE)
data_categorical3 <- cbind(option3a, data_catigocal)
Tim
-Original Message-
From: R-help On Behalf Of anteneh asmare
Sent: Tuesday, June 14, 2022
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
Hello,
I am using the ksvm function from the library kernlab to generate an
SVM classification. I am running the model with k-mean
cross-validation, thus obtaining different accuracy.
Is it possible to merge the different models obtained with the
separate data set to generate a kind of median model
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
> +
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
+ ,UPN
+ ,a
+ ,b
+ ,c
+
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
Hi all, I have a sample of data as shown below,
dt <-read.table(text="name Item check
A DESK NORF
B RANGE GARRA
C CLOCKPALM
D DESK RR
E ALARMDESPRF
H DESK RF
K DESK CORR
K WARF CORR
G NONE RF ",header=TRUE, fill=T)
I want create another
Indeed, replacing ReporteRs functionality was the topic of this thread. I
know... it can be a bit much to read an entire thread, but reading is needed if
you want to avoid non-sequitur, especially when the thread was not on topic for
this mailing list to begin with.
On April 15, 2020 10:33:05 A
However, the Reporters github page clearly says that the package has
been removed from CRAN and it has been replaced by the officer
package.
Note that ReporteRs has been removed from CRAN the 16th of July 2018
and is not maintained anymore. please migrate to officer.
https://cran.r-project.org/we
"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]
Many thanks ! Bert Gunter's answer is of course correct. The maintainer
suggests Stackoverflow or commercial advice - which is worrying as I have
some further questions concerning lists in [officer} - but first I'll do
some more research !
On Thu, Apr 9, 2020 at 5:12 PM Jeff Newmiller
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. Illustrated below:> a<-
as.data.frame(matrix(LETTERS[1:16],nrow = 4))> a V1 V2 V3 V41 A E I M2 B
F J N3 C G K O4 D H L P
I want the list looks like$A[1] E I MLevels
Rmarkdown is my first choice as well... but the options available there for
controlling PDF output via LaTeX macros are incredibly powerful, while the
options available for controlling Word output from Rmarkdown are extremely
limited by comparison. ReporteRs and officeR provide considerably more
The OP clearly hasn't followed the excellent documentation... this is not a
maintainer problem. It is the act of printing the completed object that puts a
file on disk.
However, the package author recommends in the README that help be requested
from stackoverflow, as the mailing list Posting Gu
But the OP explicitly notes that the read_docx package does *not* write
files and asks whether there are *other* packages or functions that provide
that functionality. It still seems to me that the ReporteR maintainer might
be the best place to go for that info, although there is certainly no
assur
But before hassling the maintainer the OP should read the package vignettes and
run some examples... read_docx does not write to any files, so complaining that
it doesn't will be fruitless.
On April 8, 2020 9:31:41 AM PDT, Bert Gunter wrote:
>This sounds like the sort of specialized question t
This sounds like the sort of specialized question that should be directed
to the maintainer (?maintainer) rather than to a general Help list such as
this.
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathe
Dear All,
Mac Catalina - R 3.6.3 - all up-to-date packages.
I would like to re-create the functionality which was found in the package
{ReporteRs} by creating a .docx file in a folder on my computer from within
R - which can subsequently be used by the {officer} function read_docx.
The function re
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$Id),diffs,3)
I can't do the manipulation of the resulting values at the mome
Hi R User.
I have a date set in which I wanted to find the duration (period) between
released time and detection time for each individual. Is there any simplest
way to create a matrix?
I appreciate your help.
Thanks,
MW
--
Here is the example data,
dAT<-structure(list(Id = structure(
Just use a temporary variable!
## You should apply is.na() only to the numeric columns you need, not the
whole data frame
## see ?'[' for why.
vnew <- vdat[,3:5]
vnew[is.na(vnew)] <- 0
vdat$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
Hi All,
I have a data frame with several columns and I want to create
another column by using the values of the other columns. My
problem is that some the row values for some columns have missing
values and I could not get the result I waned .
Here is the sample of my data and my attem
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
how to create
u = (1, −1, 2, −2, . . . , 100, −100) in r
Thanks
Bienvenue
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PLEASE do read the
;data.frame", row.names = c(NA,
>-7L))
>
>#And I ran this
>
>library(igraph)
>net2=graph(c("account","block","block","solve","solve","problem"))
>plot(net2)
>
># I do not get 7 plots, only one? See attached
his
library(igraph)
net2=graph(c("account","block","block","solve","solve","problem"))
plot(net2)
# I do not get 7 plots, only one? See attached.
WHP
From: R-help On Behalf Of Elahe chalabi via
R-help
Sent: Wednesday, January 30, 2019
Hi all,
I have a small dataframe and I would like to show in a network plot how words
are related to the word "problem" with arrows (keeping the order of the words
in sentences).
Here's the df:
dput(df)
structure(list(text = structure(c(1L, 7L, 3L, 4L, 5L, 6L, 2L), .Label =
c(
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(
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(2,3,4,5,0,0,0,0,0,0))
for (i in 1:length(CCC)) {
f
nuary 17, 2019 1:29 AM
> To: r-help@r-project.org
> Subject: [R] create groups from data with duplicates, such that each
group has
> a duplicate represented once
>
> Hi, I have a sequencing run with ~3000 samples (attached dataset). The
> samples were initiall
7, 2019 1:29 AM
> To: r-help@r-project.org
> Subject: [R] create groups from data with duplicates, such that each group has
> a duplicate represented once
>
> Hi, I have a sequencing run with ~3000 samples (attached dataset). The
> samples were initially tagged and amplified by PCR
Hi, I have a sequencing run with ~3000 samples (attached dataset). The samples
were initially tagged and amplified by PCR in duplicate. The tags used range
from MID01 to MID26.
MID01-MID13 were used for pair 1 while MID14-MID26 were used for pair 2. The
tags are re-used to allow samples to be p
Hi all,
I have the following data for which I create a document term matrix first and
then I add the time available to the dtm. In order to see the correlations to
the term "updat" in the different years, I would like to have a heat-map for
findassoc in a way that x-axis shows the time.
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
>>
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