> On Jul 6, 2017, at 2:19 PM, Dai, Shengyu wrote:
>
> Hi R helpers,
>
>
>
> I hope this email finds you well.
>
>
>
> I am having trouble with R loop function in Tableau.
>
There is no "R loop function". (There is an R `for`-function.) If you are
having trouble with code then it should
Hi R helpers,
I hope this email finds you well.
I am having trouble with R loop function in Tableau. Please see the attachment
of a simple dataset. The problem is to test whether the value of columns is
match.
Because Tableau do not have iteration function, I coded “if statement” in
Hi tan sj,
It is by no means easy to figure out what you want without the code,
but If I read your message correctly, you can run the loops either
way. When you have nested loops producing output, it is often a good
idea to include the parameters for each run in the output as well as
the result so
hi, i am new in this field.
I am now writing a code in robustness simulation study. I have written a brief
code "for loop" for the factor (samples sizes d,std deviation ) , i wish to
test them in gamma distribution with equal and unequal skewness, with the above
for loop in a single code if pos
Hi,You could use:
dat1 <- structure(list(Haplotype = c("H1", "H1", "H1", "H2", "H2", "H2",
"H3", "H3", "H3", "H4", "H4", "H4", "H4", "H4", "H4"), Frequency = c(0.8278,
0.02248, 0.1494, 0.8238, 0.02248, 0.1497, 0.1497, 0.02248, 0.8244,
0.628, 0.02248, 0.1483, 0.1637, 0.01081, 0.01798)), .Names =
I've also tried to make the function work for one particular group, then
apply the same function to the whole data frame with all groups using by()
or lapply() as follow. But I'm still receiving error messages. Could someone
please explain what is happening here?
dfdfdf <- function(localdata){
ord
Hi,
I'm working on a loop function for a large dataset which contains 1000
different groups. I would like to reconstruct the order of events within
each group by using a loop function in R. (Currently the order of events are
based on the ascending order of prev_event within the group)
A demo dat
Worked like a charm!
Thanks a lot Jean V Adams
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From:
testrider
To:
r-help@r-project.org
Date:
08/04/2011 07:21 AM
Subject:
[R] R loop problem
Sent by:
r-help-boun...@r-project.org
I have run into a spee
A subset of actual data and what you would expect as a result would be
very helpful. All you say is that p.q are vectors, but it would
appear that they are character vectors, but the content is unknown.
Also will the expression "q[i==p]<-z[i==y]" have the same length on
each side; the vectors app
I have run into a speed issue, and given the size of the problem it feels
like there should be an easy solution. Here is the problem statement with
some arbitrary numbers added.
#p,q: vector with length(q)==length(p)==1 and length(levels(p))==3000
#y,z: vectors with length(levels(y))=length(y)
Hi Gustaf
Thank you for the help, but I have over 1000 records, I can’t write it all
in the code I didn’t get the code I’m sorry.
Regards
Mohamad Al-shammari
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On Fri, Apr 23, 2010 at 11:14 AM, mhalsham wrote:
>
> Hi
> Yes I have managed to read the file (Table2.txt)
> The command I have used
> a<- read.table("table3.txt", fill=TRUE, header=FALSE)
> If I read the first row the result output will be like that.
> a[1,]
>
> Result would be
>
> V1 V
Hi
Yes I have managed to read the file (Table2.txt)
The command I have used
a<- read.table("table3.txt", fill=TRUE, header=FALSE)
If I read the first row the result output will be like that.
a[1,]
Result would be
V1 V2 V3V4V5 V6 V7 V8V9 V10 V11 V12
1
On Thu, Apr 22, 2010 at 7:20 PM, mhalsham wrote:
>
> Ok sorry for bad explanation from my side
> What I want. I have a txt file name is (table3.txt) this file contains 1293
> rows and some of these row will have 1 column and some of them will have up
> to may be 40 column. For example
>
> A
Ok sorry for bad explanation from my side
What I want. I have a txt file name is (table3.txt) this file contains 1293
rows and some of these row will have 1 column and some of them will have up
to may be 40 column. For example
You did not give a good description of your problem, or what you wanted as a
solution (what is a 'button'), or a reasonable idea of what you data is.
Here is one solution
x <- read.table(textConnection("Deafness, EYA4, DIAPH1, MYO7A, TECTA,
COL11A2, POU4F3, MYH9, ACTG1, MYO6"),
as.is=TRUE, se
Thanks a lot for the help, but it’s not what I needed.
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On Apr 16, 2010, at 11:52 AM, Thomas Stewart wrote:
I'm not sure I completely understand your question, but I think the
solution
to your problem is the reshape function in the reshape package.
Except there is no reshape function in the reshape package. Your code
works because the reshape
I'm not sure I completely understand your question, but I think the solution
to your problem is the reshape function in the reshape package. Here is a
silly example of how it would work:
> V<-matrix(rbinom(15,4,.5),nrow=3)
> X<-data.frame(A=c("A","B","C"),V=V)
> X
A V.1 V.2 V.3 V.4 V.5
1 A 1
Hi every one I’m new to R and I cant figure our how to use the loop to do the
following task, any help would be very kind of every one.
I have a file called (table3.txt) that contains over 1000 row and over 40
columns.
So for example first row would look like that
Deafness, EYA4, DIAPH1,
On Apr 15, 2010, at 4:24 PM, Changbin Du wrote:
HI, Dear community,
I am building the following loop,
ww<-function(file) {
lossw<-vector()
for (x in seq(0.1, 0.9, by=0.1)) {
cat('xweight ', x, '\n')
lossw[i] <- cross.validation(file, x)$avg
}
return(los
Thanks so much, Marius! It works!
On Thu, Apr 15, 2010 at 1:35 PM, Marius 't Hart wrote:
> lossw[i] <- cross.validation(file, x)$avg
>
> change to:
>
> lossw <- append(lossw,cross.validation(file, x)$avg)
>
>
>
>
> Changbin Du wrote:
>
>> HI, Dear community,
>>
>> I am building the following
HI, Dear community,
I am building the following loop,
ww<-function(file) {
lossw<-vector()
for (x in seq(0.1, 0.9, by=0.1)) {
cat('xweight ', x, '\n')
lossw[i] <- cross.validation(file, x)$avg
}
return(lossw) }
MY question is how to index the lossw[
Hi there,
That's exactly what I want.
I have checked ?combn out,
but I could get the following,
suppose that I want ALL possible combinations of them,
as this
==
apply(
combn(paste('x', 1:4, sep =""), 2), 2,
function(v) get(v[1])*get(v[2])
An aside to the main question:
I don't think that
i+1:3
is doing what you think it is.
On 26/03/2010 23:01, casperyc wrote:
Hi,
I am tring to write a loop to compute this,
==
x1=c(
rep(-1,4),
rep(1,4)
)
x2=c(
rep(c(-1,-1,1,1),2)
Hi casperyc,
Here is a suggestion:
# all at once
apply(combn(paste('x', 1:3, sep =""), 2), 2,
function(v) get(v[1])*get(v[2]) )
# step by step
thex <- paste('x', 1:3, sep ="")
thex
combs <- combn(thex, 2)
combs
apply(combs, 2, function(v) get(v[1])*get(v[2]) )
x1, x2 and x3 correspo
Hi,
I am tring to write a loop to compute this,
==
x1=c(
rep(-1,4),
rep(1,4)
)
x2=c(
rep(c(-1,-1,1,1),2)
)
x3=c(
rep(c(-1,1),4)
)
x1*x2
x1*x3
x2*x3
suppose i have x1,x2,x3
Note that "%Y-%m-%d" is the default format so you can omit it.
Perhaps you want something along these lines (modify formula
to get the precise numbering you like):
a$wk <- julian(a$date, as.Date("2008-01-01")) %/% 7
Also note %W can be used as a format string. See
?strptime for the codes.
R Ne
008-01-01")] doesn't
make much sense. But if you do this you will have problem changing the
appropriate value of a$wk.
--- On Fri, 6/6/08, Andrew McFadden <[EMAIL PROTECTED]> wrote:
> From: Andrew McFadden <[EMAIL PROTECTED]>
> Subject: [R] R loop
> To: r-help@
ent: Thu 6/5/2008 9:01 PM
To: Andrew McFadden; r-help@r-project.org
Subject: Re: [R] R loop
The bolding is lost in plain-text email,
but some things to note:
:: for(i in a$date){ - this will make i take on date values,
you probably meant i to index the rows of a[]
:: you missed inde
ssage-
From: [EMAIL PROTECTED] on behalf of Andrew McFadden
Sent: Thu 6/5/2008 8:46 PM
To: r-help@r-project.org
Subject: [R] R loop
I all
I am just trying to get the idea of a loop if statement for another
purpose. I was going to label a new variable based on date ie first week
1, second wee
I all
I am just trying to get the idea of a loop if statement for another
purpose. I was going to label a new variable based on date ie first week
1, second week 2 etc. My code in bold is wrong but seems logical to me,
could someone help.
x <-
rep(c(2660156,2663703,2658165,2659303,2661531,2660914
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