On 05-11-2014, at 14:41, Noha Osman wrote:
> Hi Folks
>
> Iam a new user of R and I have a question . Hopefully anyone help me in that
> issue
>
>
> I have that dataset as following
>
> Sample Population Species Tissue R GB
> 1 Bari1_062-1 Bari1 ret
While you should definitely read the tutorial that Don is referring
to, I'd recommend you take a different approach and use more R
idiomatic code here.
In base R, this could be addressed with few approaches. Look for help
on the following functions:
* tapply
* by
* aggregate
I'd rather rec
On 11/5/2014 5:41 AM, Noha Osman wrote:
Hi Folks
Iam a new user of R and I have a question . Hopefully anyone help me in that
issue
I have that dataset as following
Sample Population Species Tissue R GB
1 Bari1_062-1 Bari1 ret seed 94.52303 80.7
Have you read the tutorial that comes with the R distribution? This is a very
basic database calculation that you will
encounter (or some slight variation of it) over and over. The solution is a
few lines of code, and someone may write it
out for you, but if no one does
You have 20 population
That code will not work. get() and assign() are troublesome for a
variety of reasons. E.g.,
* adding made-up names to the current environment is dangerous. They
may clobber fixed names in the environment. You may be confused about
what the current environment is (especially when refactoring co
It's a great method, but there is a memory problem, DFS would occupy a large
memory. So from this point of view, i prefer the loop.
>> for (i in 1 : nrow(unique)){
>> tmp=get(past0("DF",i))[1,]
>> assign(paste0("df",i),tmp)
>> dfi=dfi[,1:3]
>> names(dfi)=names(tmp[c(1,4,5)])
>> dfi=rbind(dfi,tm
> I was able to create 102 distinct dataframes (DFs1, DFs2, DFs3, etc) using
> the assign() in a loop.
The first step to making things easier to do is to put those data.frames
into a list. I'll call it DFS and your data.frames will now be DFs[[1]],
DFs[[2]], ..., DFs[[length(DFs)]].
DFs <- la
On Jun 20, 2014, at 9:18 AM, Bea GD wrote:
> I've tried several things but I don't manage to get this plot right. Any help
> greatly appreciated!
> I'm running a for loop to produce 4 plots at once. Each plot should only show
> data points for a specific group (i.e. 4 plots for groups 1 to 4).
Hi Arun,
Thanks a lot for your script. Ill work on it tomorrow.
Cheers
On 24/04/2014, at 11:54 AM, "arun kirshna [via R]"
wrote:
> HI,
> I guess you got an output like this using my script:
> ##Please use ?dput() to show the example data.
>
> FA <- structure(list(Sample = c("L1 Control",
Hi Arun,
Your script works but it does not do what I was after. To be a bit more
specific, this the table FA in which Im working on ( but the original one
has 34 fatty acids instead of 3: C14.0, C15.0, and C15.1).
SampleC14:0C15:0C15:1
L1 Control0.4565091920.4695626870.774909216
L1 Control0.51398
HI,
I guess you got an output like this using my script:
##Please use ?dput() to show the example data.
FA <- structure(list(Sample = c("L1 Control", "L1 Control", "L1 Control",
"BBM Control", "BBM Control", "BBM Control", "L1 Ash", "L1 Ash",
"L1 Ash", "BBM Ash", "BBM Ash", "BBM Ash"), C14.0 = c
Thank you Arun for your quick answer.
Ill try it and ill let you know.
Cheers
On 21/04/2014, at 11:30 PM, "arun kirshna [via R]"
wrote:
> Hi,
>
> Using the example data from library(gvlma)
>
> library(gvlma)
> data(CarMileageData)
> CarMileageNew <- CarMileageData[,c(5,6,3)]
> lst1 <-
Hi,
Using the example data from library(gvlma)
library(gvlma)
data(CarMileageData)
CarMileageNew <- CarMileageData[,c(5,6,3)]
lst1 <- list()
y <- c("NumGallons", "NumDaysBetw")
for(i in seq_along(y)){
lst1[[i]] <- gvlma(lm(get(y[i])~MilesLastFill,data=CarMileageNew))
lst1}
pdf("gvlmaplot.pdf
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of bett kimutai
> Sent: Wednesday, June 26, 2013 3:17 AM
> To: Law, Jason; r-help@r-project.org
> Subject: Re: [R] Loops
>
> Thanks for you response.
rectify my loop to run
for all the pipes and then create a table to hold these values. Sorry for the
syntax errors, I guess I will get them with time.
thanks again..
From: "Law, Jason"
ect.org>
Sent: Tuesday, June 25, 2013 3:32 PM
Subject: RE: [R
Not sure what you're trying to do, but it looks like most of what you're
attempting to do in the code can be done just using vectors rather than loops,
at least the inner loop. For example:
k <- 1.15
l <- exp((1 / k) * (7.16 - 0.44 + 0.12 - 0.016))
z <- (log(1 / p) * l)^k
See ifelse for how to
Dear All,
I have spent most of my time trying to figure out how to simulate the number
of breaks in a pipe using monte carlo simulation.
i have 20,000 individual pipes that i have to run, and for each pipe i have to
run 1000 times while checking some conditions and therefore, i have to use a
The comments on StackOverflow are fair, I believe...
Please dput() your matrices, so that your code becomes reproducible!
On Wednesday 12 June 2013 11:14:35 maggy yan wrote:
> I have to use a loop (while or for) to return the result of hadamard
> product. now it returns a matrix, but when I use i
gt; 25)+1),as.matrix),c(2,3),function(x) x))
> res<-rbind(t(pnew),colSums(p))
> row.names(res)<-1:nrow(res)
> res<- 100-100*abs(res/rowSums(res)-(1/3))
> A.K.
>
>
> - Original Message -
> From: Rui Barradas
> To: Francesca
> Cc: r-help@r-project.org
)-(1/3))
A.K.
- Original Message -
From: Rui Barradas
To: Francesca
Cc: r-help@r-project.org
Sent: Sunday, January 27, 2013 6:17 AM
Subject: Re: [R] Loops
Hello,
I think there is an error in the expression
100-(100*abs(fa1[i]/sum(fa1[i])-(1/3)))
Note that fa1[i]/sum(fa1[i]) is always 1. If it&
On 27.01.2013 12:50, Richard D. Morey wrote:
Dear Contributors,
I am asking help on the way how to solve a problem related to loops for
that I always get confused with.
I would like to perform the following procedure in a compact way.
Consider that p is a matrix composed of 100 rows and three
> Dear Contributors,
> I am asking help on the way how to solve a problem related to loops for
> that I always get confused with.
> I would like to perform the following procedure in a compact way.
>
> Consider that p is a matrix composed of 100 rows and three columns. I need
> to calculate the su
Hello,
I think there is an error in the expression
100-(100*abs(fa1[i]/sum(fa1[i])-(1/3)))
Note that fa1[i]/sum(fa1[i]) is always 1. If it's fa1[i]/sum(fa1), try
the following, using lists to hold the results.
# Make up some data
set.seed(6628)
p <- matrix(runif(300), nrow = 100)
idx <- se
You can use the assign() function. E.g., assign("a", 3,)
But the more R thing to do is to create a list of your results (which
you can name) and to handle that whole object. Then if you want to do
things to each object you create, it's as simple as lapply
Michael
On Tue, Jan 24, 2012 at 7:27 PM,
Hi
>
> Dear R help listers,
>
> I am trying to replicate results in Gelman and Hill's book (Chapter 3
> in regressions and multilevel models). Below I estimated two models
> (chp3.1 and chp3.3 in R codes) with the same data and dependent
> variable but different independent variables. I have been
Thanks a lot,
it was very helpful, I did something like that:
EV <- ifelse(sprd > mediaSDP & sprd_d <= mediaSDP_d, -1, 0)
SV <- ifelse(sprd > media & sprd_d <= media_d, -17, 0)
EC <- ifelse(sprd < mediaSDN & sprd_d >= mediaSDN_d,1,0)
SC <- ifelse(sprd < media & sprd_d >= media_d,17,0)
in order t
Look at ?ifelse. You'll need to nest them however. If you are feeling a little
more ambitious this can be done with ?switch, but for clarity, I'd rather see
pre-allocation followed by assignments Something along this pattern:
out <- numeric(50)
x <- sample(3, 50, TRUE)
out[x==3] <- "A"
out[x==2
Thank you so much!
foreL<-8
b0f<-matrix(nrow=9, ncol=foreL)
ct<-1 ### use this as the index of b0f
for(ar.ord in 1:3){
for(ma.ord in 1:3){
b0f[ct,]<-c(predict(arima(para_qtr[1:(n-8),1],order=c(ar.ord,1,ma.ord)),
n.ahead=foreL)$pred)
ct<-ct+1 ### increment the counter
}
}
this one works! Best rega
Look at the combinations function in gtools and index by the rows of that
output for a single loop. Pass values as the three parameters.
Sorry for being terse - writing on my phone.
Michael
On Sep 14, 2011, at 5:36 PM, Fred wrote:
> Dear forum,
>
> I would like to forecast e.g. with the ar
A little bit change of your code may do the work
foreL<-8
b0f<-matrix(nrow=9, ncol=foreL)
ct<-1 ### use this as the index of b0f
for(ar.ord in 1:3){
for(ma.ord in 1:3){
b0f[ct,]<-c(predict(arima(para_qtr[1:(n-8),1],order=c(ar.ord,1,ma.ord)),
n.ahead=foreL)$pred)
ct<-ct+1 ### increment the counter
The first thing I'd mention is that *apply are really just well implemented
loops themselves so you can't expect true magic from them. However, if you
really want to use them rather than just a regular loop (which seems more
intuitive for a request like this), this might do what you are looking for
Hi Michael,
First of all thanks for your response.
I do know that if I make my stimation on a single data the regression has no
sense but it will be getting sense in the growing next estimations.
I change my asking doubt.
I want to use this regressions as a first filter. Only this.
Can anyone
I may be totally off base with this, but I'm wondering what exactly this
would suggest or why you want to do it. Specifically "multiple regression
with only intercept" -- how is it multiple if you don't have any regressors?
Furthermore, you want to run a "regression" on a single data point --
reall
Hi:
Try this:
## Function that takes a data frame as input and outputs a data frame:
chrSumm <- function(d) { # d is a data frame
colnames(d) <- c("chr","start","end","base1","base2",
"totalreads","methylation","strand")
TR <- nrow(d)
RG1 <- sum(d['totalreads']
Tena koe
Try something along the following lines:
chrData <- vector('list', 22)
names(chrData) <- paste('chr', 1:22, sep='')
for (i in 1:length(chrData))
{
chrData[[i]] <- read.table(file=paste('chr', i, '.out.txt', sep=''), header=F)
...
}
HTH
Peter Alspach
> -Original Message--
Whoa!
1. First and most important, there is very likely no reason you need
to do this. R can handle multiple groupings automatically in fitting
and plotting without creating artificial labels of the sort you appear
to want to create. Please read an "Intro to R" and/or get help to see
how.
2. The
How about this?
> indx <- unique(cbind(Dates, Groups))
> indx
DatesGroups
[1,] "12/10/2010" "A"
[2,] "12/10/2010" "B"
[3,] "13/10/2010" "A"
[4,] "13/10/2010" "B"
[5,] "13/10/2010" "C"
> indx <- data.frame(indx, id=1:nrow(indx))
> indx
Dates Groups id
1 12/10/2010
Dear Chandra,
You're on the wrong track. You don't need for loops as you can do this
vectorised.
as.numeric(interaction(data$Groups, data$Dates, drop = TRUE))
Best regards,
Thierry
> -Oorspronkelijk bericht-
> Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> N
On Wed, Jul 20, 2011 at 10:43 PM, David Winsemius
wrote:
>
> On Jul 21, 2011, at 1:04 AM, Daniel Malter wrote:
>
>> http://mlg.eng.cam.ac.uk/dave/rmbenchmark.php
>>
>> I haven't ever tried it myself, but online sources suggest that Matlab
>> possibly gains speed by internally avoiding loops rather
rent.
--
David.
Daniel
From: David Winsemius [dwinsem...@comcast.net]
Sent: Wednesday, July 20, 2011 9:01 AM
To: Daniel Malter
Cc: r-help@r-project.org
Subject: Re: [R] loops and simulation
On Jul 20, 2011, at 1:34 AM, Daniel Malter wrote:
snipped
requests,
aniel
From: David Winsemius [dwinsem...@comcast.net]
Sent: Wednesday, July 20, 2011 9:01 AM
To: Daniel Malter
Cc: r-help@r-project.org
Subject: Re: [R] loops and simulation
On Jul 20, 2011, at 1:34 AM, Daniel Malter wrote:
snipped
> requests, except that
On Jul 20, 2011, at 1:34 AM, Daniel Malter wrote:
snipped
requests, except that you were referring to SAS and had heard that R
"does
not like loops." (This is factually wrong. But R can be slow looping).
Where did you hear this? Can you cites any references?
--
David Winsemius, MD
West Ha
Yes, there are in Europe. And there are summer classes in the US, as well.
And no, this list is not so much about helping beginners to learn R. For
that, there is a myriad of online sources. Rather, this list is for people
who have exhausted their ability to (elegantly) solve a problem.
Also, it s
I dare the conjecture that if you had written the code, you would know how to
do this. This suggests that you are asking us to do your homework, which is
not the purpose of this list. A simple inclusion of the code in a for or
while loop and storing the estimated parameters with the index of the
it
I would prefer version 1.
Version to creates a global variable R which you do not really need
since it contains the same values as d$r.
In option 2, you should probably remove the variable r itself
after it has been appended to d.
On 4/18/2010 5:23 PM, Laura Ferrero-Miliani wrote:
> Thanks for co
Thank you so much all!
How a simple problem can get complicated by not having enough
knowledge, but hopefully I am learning
Have a great Sunday!
L
On Sun, Apr 18, 2010 at 4:56 PM, jim holtman wrote:
> You are working with a matrix, so the "$" operator is not allowed (e.g.,
> d$c).
>
> Also in
Hey Laura,
Just to add a cautionary note, in
> a <- c(1:4)
> b <- c("meep", "foo", "meep", "foo")
> d <- cbind(a, b)
d is a matrix and will only be one type of matrix. Since you have
both integer (a) and character (b) data, it has to be at the character
level. From the help for cbind:
"The t
You are working with a matrix, so the "$" operator is not allowed (e.g.,
d$c).
Also in your test, you have to test against the second column (e.g., d[i,
2])
try this:
> a <- c(1:4)
> b <- c("meep", "foo", "meep", "foo")
> d <- cbind(a, b)
>
>
> for(i in seq(along=d[,2])) {if (d[i,2]=="meep") { p
Hi Garth,
Your code is really confusing! You should start by reading the help file
on the for() function and understanding what it does:
?"for"
Your line
for(i in 1:nboot){
}
is simply starting a loop around the variable 'i', which will change
values following the sequence 1:nboot.
It seems
49 matches
Mail list logo
