Nice suggestion for the extra "Time" column.
But I think I didn't ask clear enough my problem.
My main problem is to find a way to "classify" the rrt's, so that we don't have
to check each dataframe by our selfs.
So I need a function that fills in the extra "Time" column by taking a look at
the
Hi,
If variable `V1` is factor and it follows the same pattern as in your example:
test<- structure(list(V1 = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L,
1L, 2L, 2L, 2L), .Label = c("a", "b"), class = "factor")), .Names = "V1", class
= "data.frame", row.names = c("1",
"2", "3", "4", "5", "6", "7
Dear Michael,
Thanks for your codes. However, lapply does not work in my case since I've
some files missing in the data (say, the file data101.dat). Do you have any
suggestions on this?? Thank you very much.
Best Wishes,
Ray
On Wed, Jan 23, 2013 at 8:15 PM, R. Michael Weylandt <
michael.weyla...
Hi,
x<-1:80
y<- x[-(1:77)]
y
#[1] 78 79 80
#or
?tail() #already suggested
If you want only the last element,,
library(pastecs)
last(x)
#[1] 80
A.K.
- Original Message -
From: hp wan
To: r-help@r-project.org
Cc:
Sent: Thursday, January 24, 2013 7:23 PM
Subject: [R] How to extrac
Thanks so much. It greatly enrich my mind.
2013/1/25 arun
>
>
> Hi,
> x<-1:80
> y<- x[-(1:77)]
> y
>
> #[1] 78 79 80
> #or
>
> ?tail() #already suggested
>
> If you want only the last element,,
> library(pastecs)
> last(x)
> #[1] 80
> A.K.
>
>
>
>
> - Original Message -
> From: hp wan
This issue is very platform-specific. you should post on R-sig-mac.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
On Jan 24, 2013, at 4:23 PM, hp wan wrote:
> Hi all mailing listers,
>
> I wanna get the last several elments of vector.
> e.g. x <- c(1,2,3,.,78, 79, 80)
>
> How can I implement to assign last three elements to y, y <- c(78, 79, 80)
> ?
?tail
> In Matlab, It can easily achieve by y=x(e
On Jan 24, 2013, at 4:26 PM, heatherchapman wrote:
> Hi,
> I need some help figuring out how to save or copy graphs I've created in the
> Quartz screen. I'm using a Mac, Mountain Lion OS, and have version 2.15.2.
>
> When I save a graphic, the process either crashes R, or if it saves the
> file,
I ran across this page for C, Java, etc. No R.
http://pastebin.com/
It looks similar and more than what I was looking for, just saying.
Mike
On Wed, Jan 23, 2013 at 9:34 PM, Mark Lamias wrote:
> Are you only interested in formatting code from copy and pasting to/from
> email? If you are int
That is to say, there is no variable in R corresponding to "end" iterm in
Matlab which means the sum of all elments. Maybe we can reshape the vector
and then get the first several elements. It is easy to achieve no matter
which means used.
Thank Daniel J. Nordlund
Huaping Wan
2013/1/25 Nordlu
rle() or c(TRUE, cumsum( v[-length(v)] != v[-1] )) will work on any
type for which != is defined. The ones involving diff() require numeric inputs.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-p
Hi,
I need some help figuring out how to save or copy graphs I've created in the
Quartz screen. I'm using a Mac, Mountain Lion OS, and have version 2.15.2.
When I save a graphic, the process either crashes R, or if it saves the
file, when I open it, it is a blank file or I get an error message tha
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of hp wan
> Sent: Thursday, January 24, 2013 4:23 PM
> To: r-help@r-project.org
> Subject: [R] How to extract elements from vector in reverse order?
>
> Hi all mailing listers,
>
Hi all mailing listers,
I wanna get the last several elments of vector.
e.g. x <- c(1,2,3,.,78, 79, 80)
How can I implement to assign last three elements to y, y <- c(78, 79, 80)
?
In Matlab, It can easily achieve by y=x(end-2:end)
Thanks
Huaping Wan
[[alternative HTML version
Thank you for the tips. I should mention that the V1 is numeric in this case
but I am trying to automate the overall procedure so it may not be numeric
always. Would that matter with these functions?
Sent from my iPhone
On Jan 24, 2013, at 6:34 PM, arun wrote:
> Hi,
> You could do this:
> tes
Hi,
I guess you meant:
c(TRUE, cumsum( v[-length(v)] != v[-1] ))
[1] 1 0 0 1 1 1 2 2 2 3 3 3
this:
cumsum(c(TRUE, v[-length(v)] != v[-1] ))
# [1] 1 1 1 2 2 2 3 3 3 4 4 4
A.K.
- Original Message -
From: William Dunlap
To: arun ; Jeffrey Fuerte
Cc: R help
Sent: Thursday, January 24,
Hi Don,
Thanks for the advise.
I followed this video: http://www.youtube.com/watch?v=uw4NtcTD4iM
and achieved my goal: i.e. by using 'matrix' function and 'c' function as
shown below:
> tail_clusters <-
matrix(c(input_tail[,1],input_tail[,2],clustering_tail$cluster),ncol=3)
and separating them
Yes.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: arun [mailto:smartpink...@yahoo.com]
> Sent: Thursday, January 24, 2013 3:53 PM
> To: William Dunlap
> Cc: R help
> Subject: Re: [R] ifelse to speed up loop?
>
> Hi,
>
> I guess you meant:
> c(TRUE
I like
v <- test[,1]
c(TRUE, cumsum( v[-length(v)] != v[-1] ))
(R's arithmetic on logicals treats TRUE as 1 and FALSE as 0.)
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
>
HI,
This might be better.
test$group<-cumsum(abs(c(1,diff(test[,1]
A.K.
- Original Message -
From: Jeffrey Fuerte
To: r-help@r-project.org
Cc:
Sent: Thursday, January 24, 2013 4:20 PM
Subject: [R] ifelse to speed up loop?
Hello,
I'm not sure how to explain what I'm looking for
I should add that if your V1 column is numeric, fiddling around with
?diff or equivalent will also identify the groupings for you and may
be faster than rle().
-- Bert
On Thu, Jan 24, 2013 at 3:32 PM, Bert Gunter wrote:
> Your query is a little unclear to me, but I suspect
> ?rle
> is what you w
Hi,
You could do this:
test<-read.table(text="
V1
1 1
2 1
3 1
4 2
5 2
6 2
7 1
8 1
9 1
10 2
11 2
12 2
",sep="",header=T)
test$Group<-cumsum(abs(c(0,diff(test[,1]+1
test
# V1 Group
#1 1 1
#2 1 1
#3 1 1
#4 2 2
#5 2 2
#6 2 2
#7 1 3
# 1
Your query is a little unclear to me, but I suspect
?rle
is what you want.
-- Bert
On Thu, Jan 24, 2013 at 1:20 PM, Jeffrey Fuerte wrote:
> Hello,
>
> I'm not sure how to explain what I'm looking for but essentially I have a
> test dataset that looks like this:
>
> test:
>V1
> 1 1
> 2 1
Hello
I am using library(spatstat). I managed to test of the random distribution of a
collection of sampled trees in the forest. But what I want now is to test if
their diameter is randomly distributed across the forest, for example the
bigger trees are not clustered together.
So far I have no
Hello,
I'm not sure how to explain what I'm looking for but essentially I have a
test dataset that looks like this:
test:
V1
1 1
2 1
3 1
4 2
5 2
6 2
7 1
8 1
9 1
10 2
11 2
12 2
And what I want to be able to do is create another column that captures a
"grouping" variable t
HI,
You can get the plot in the same window, but the width seems to be a problem.
stest<- read.table(text="
site year conc
south 2001 5.3
south 2001 4.67
south 2001 4.98
south 2002 5.76
south 2002 5.93
north 2001 4.64
north 2001 6.32
north 2003 11.5
north 200
Hi and thanks. This unfortunately doesn't work because:
"ggplot2 does not currently support free scales with a non-cartesian coord or
coord_flip."
-Original Message-
From: arun [mailto:smartpink...@yahoo.com]
Sent: January-24-13 1:49 PM
To: Gorczynski, George
Cc: R help
Subject: Re: [R]
After you get that message, use
conflicts()
(and read the help page for the conflicts function)
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 1/23/13 10:04 PM, "Fumie Sugahara" wrote:
>Hi
>
>The message occurred
You find the element of clustering_tail that indicates which which point
is in which cluster (the help page for kmeans tells you). Then you use
that element to subset your input data (1.tsv). Then you save each subset
to a separate folder.
By "save to a folder" I would assume you mean write a tsv
Or just use aggregate()
# Redefine match_df to match your data.frame
> match_df <- data.frame(Seq = 1:5, criteria = letters[c(3, 5, 3, 3, 4)])
> aggregate(mat, list(match_df$criteria), sum)
Group.1 V1 V2 V3
1 a 7 17 27
2 c 2 7 12
3 d 6 16 26
# If you really want a matrix:
>
Dear Ellison,
thanks a lot for your reply. Your explanation makes things much clearer.
Sincerely,
f.
On 24 January 2013 05:58, S Ellison wrote:
>
>
> On 23 Jan 2013, at 21:36, "Francesco Sarracino"
> wrote:
>
> > what I meant refers to the fact that I've read on "an R and
> > S-plus co
dat3 is the dataframe where there are some rrt values merged, which is actually
the problem: how on Earth discide which rows van be merged
Thanks for your input!
Bart
-Original Message-
From: arun
Sent: 24 Jan 2013 20:18:28 GMT
To: Bart Joosen
Cc: R help
Subject: Re: [R] sorting/gr
HI,
If I understand your question, "dat3" is not you wanted.
Is it something like this you wanted?
library(reshape2)
dcast(dat,rrt~Mnd,value.var="Result")
# rrt 0 3 6 9
#1 0.35 NA NA 0.05 NA
#2 0.36 NA NA NA 0.06
#3 0.44 NA NA 0.40 NA
#4 0.45 0.1 0.2 NA 0.60
#5 0.46 N
Hi,
You could also use:
mat <- matrix(1:15, 5)
set.seed(5)
match_df <- data.frame(Seq = 1:5, criteria = sample(letters[1:5], 5, replace =
T),stringsAsFactors=F)
library(plyr)
res<-daply(as.data.frame(mat),.(match_df$criteria),colSums)
res
#match_df$criteria V1 V2 V3
#
Hi,
I'm a database admin for a database which manage chromatographic results of
products during stability studies.
I use R for the reporting of the results in MS Word through R2wd.
But now I think I need your help:
suppose we have the following data frame:
ID rrt Mnd Result
1 0.45 0
thanks, Brian. I remember sending code to the R list, it was about the
lattice package some time around June 2012. And someone else reformatted,
and it my surprise it was eye-opening.
Anyways, I will dig around on this list.
Thanks,
Mike
On Thu, Jan 24, 2013 at 2:42 PM, Brian Diggs wrote:
> O
Hello,
Something like this?
do.call(rbind, lapply(split(as.data.frame(mat), match_df$criteria),
colSums))
Hope this helps,
Rui Barradas
Em 24-01-2013 19:39, Christofer Bogaso escreveu:
Hello again,
Ley say I have 1 matrix and 1 data frame:
mat <- matrix(1:15, 5)
match_df <- data.frame(
On 1/23/2013 6:40 PM, C W wrote:
Hey Mark,
I am aware of this package. The situation is,
1. I am emailing my code from my machine to a public machine. Installation
is a hassle.
2. Copy pasting for a website.
For your second use case, there is the Pretty R syntax highlighter at
http://www.insi
Hello again,
Ley say I have 1 matrix and 1 data frame:
> mat <- matrix(1:15, 5)
> match_df <- data.frame(Seq = 1:5, criteria = sample(letters[1:5], 5, replace
> = T))
> mat
[,1] [,2] [,3]
[1,]16 11
[2,]27 12
[3,]38 13
[4,]49 14
[5,]5 10 15
> m
Hello,
To get the months names you can try
format(data[,1], "%B")
As for the axis, see the help page ?axis. Argument 'labels'.
Hope this helps,
Rui Barradas
Em 24-01-2013 17:34, Grigory Fateyev escreveu:
Hello!
I have data.frame with column class POSIXct:
data[, 1] <- as.POSIXct(data[, 1
Please define 'mean probabilities'.
To compute the C-index or Dxy you need anything that is monotonically
related to the prediction of interest, including the linear combination
of covariates ignoring all intercepts. In other words you don't need
to go to the trouble of computing probabiliti
On 24-01-2013, at 18:56, hp wan wrote:
> Hi mailing listers,
>
> Sorry, I made a little mistake in the previous mail. B^{1} should be B^{-1}.
>
> Is there certain function in R deal with how to compute generalized
> eigenvalues, that is the problem: A*x* = ëB*x *? When I use
> eigen(B^{-1}*A),
Hello,
Sorry, it's RIGHT parenthesis.
Rui Barradas
Em 24-01-2013 18:54, Rui Barradas escreveu:
Hello,
Your function declaration has a syntax error, one left parenthesis too
much. Corrected it would be
dummyfunction <- function(filters = function(x) {b = 0; x > b} ){
# rest of code here
Hi,
In the ?cut(), you can specify `labels`.
#or, you can try:
test.df<-read.table(text="
Var1 Freq
1 (0,0.05] 68
2 (0.05,0.1] 87
3 (0.1,0.15] 85
4 (0.15,0.2] 72
5 (0.2,0.25] 65
6 (0.25,0.3] 73
7 (0.3,0.35] 74
",sep="",header=T,stringsAsFactors=F)
test.df[,1]<-as.numeric(gsub("
Hi,
set.seed(25)
Z<-rnorm(50,0.5,1)
res<- as.data.frame(table(cut(Z,seq(0,1,by=0.05),labels=seq(0.05,1,by=0.05))) )
head(res)
# Var1 Freq
#1 0.05 1
#2 0.1 1
#3 0.15 0
#4 0.2 0
#5 0.25 0
#6 0.3 1
A.K.
- Original Message -
From: Wim Kreinen
To: r-help@r-project.
Thank you, Jeff. I think I should have been more clear. I don't want to
modify my data. I need to create a series of box plots for different sites
(north and south in my example but there are many more) but I don't want
empty rows in my plot - for years when data was not collected. I tried to
make
Hi Ehsan,
What do you mean by percent correctly predicted? Could you provide more info
about your model? It might be helpful to get a clue.
ya
From: ehsan rahimi
Date: 2013-01-24 12:52
To: r-help
Subject: [R] mlogit Package
I have problem as follows:
I use "mlogit package" to develop a multi
Hello,
Your function declaration has a syntax error, one left parenthesis too
much. Corrected it would be
dummyfunction <- function(filters = function(x) {b = 0; x > b} ){
# rest of code here
filters # this returns a function, don't need return()
}
x <- -5:5
f <- dummyfunction() # this
Hi all,
I have some questions about the predicted HR in coxph function including
psline covariate.
If just fitting covariate as linear form, by default, the reference value
for each of predictions type (linear predictor, risk and terms) is the mean
covariate within strata.
If the psline is speci
On 24/01/2013 1:32 PM, Jannis wrote:
Dear R community,
I have a problem when I use functions as default values for argumnents
in other functions. When I use curly brackets { here, I can not create a
package with inlinedocs. It will give me the error when using
package.skeleton() in my package st
Hello,
The question is a bit confusing, but for what I could understand, try
changing to
rr <- getrange(dataw, mydatao, method = method)
Hope this helps,
Rui Barradas
Em 24-01-2013 18:26, Tammy Ma escreveu:
HI,
I guess I did sth wrong with the optional argument in the program. but I do
Hello,
try the following.
test.df$Var1 <- seq(0,1, by=0.05)[-1]
test.df
Hope this helps,
Rui Barradas
Em 24-01-2013 17:39, Wim Kreinen escreveu:
Hello,
I have a dataframe (test.df) with intervals that were generated by table
(see below). I would prefer a dataframe labeled like this:
tes
The eigenvalue problem is not unique to R. This is an R mailing list. What is
your question about R?
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live G
Dear R community,
I have a problem when I use functions as default values for argumnents
in other functions. When I use curly brackets { here, I can not create a
package with inlinedocs. It will give me the error when using
package.skeleton() in my package structure:
Error in parse(text = ut
On Jan 24, 2013, at 10:16 AM, Bert Gunter wrote:
> You need to get familiar with BioConductoR and post on that list.
> Almost certainly you are trying to reinvent the wheel, and may even be
> constructing it square.
>
> Cheers,
> Bert
>
> On Thu, Jan 24, 2013 at 8:25 AM, Benjamin Ward (ENV) wr
HI,
I guess I did sth wrong with the optional argument in the program. but I don't
know how should I make it.
in my main program:
source("getelasticity_overPB.r")
crossprice<-getelasticity(mydata,mydatao,"nest")$cross_p
is same with
crossprice<-getelasticity(mydata,mydatao)$cross_p
I don't k
You need to get familiar with BioConductoR and post on that list.
Almost certainly you are trying to reinvent the wheel, and may even be
constructing it square.
Cheers,
Bert
On Thu, Jan 24, 2013 at 8:25 AM, Benjamin Ward (ENV) wrote:
> Dear all,
>
> I'm trying to write a function, that will take
At 22:06 23/01/2013, Alma Wilflinger wrote:
Hi Michael,
The supervisor for my Master's Thesis told me that my means are the
effect size and cause of this I have to take figure 1 for all
standard deviations. So I hope that was the right information.
Alma
There is a fairly comprehensive list
Dear useRs,
Sorry for such a stupid question. But i cant live without asking it. Actually,
I was going through "copula" package in R. I needed to know that if there is a
way in R to calculate the emipirical copula density between two random
variables and plot it like the way we do in "levelpl
Hello!
I have data.frame with column class POSIXct:
data[, 1] <- as.POSIXct(data[, 1], format = "%d.%m.%Y %H:%M")
I need to write custom axis for plot where can be ex.: January (mean
of january's temperature). List of month I get as table(months(data[,
1])). What is next step?
Thanks.
--
Всег
Dear all,
I'm trying to write a function, that will take as an argument, some aligned
genome sequences, and using a sliding window, do pairwise comparisons of
sequence similarity. Coding the sliding window I think I can manage but what
I'm trying to get to grips with is getting it so as every p
HI,
If you use ?join(), it will preserve the order of the first dataframe.
library(plyr)
df3<-rbind.fill(crosspries[[1]],crosspries[[2]])
template2<-read.table(text=outer(unique(df3[,1]),unique(df3[,2]),FUN=paste),sep="",stringsAsFactors=F)
names(template2)<-names(df3[1:2])
res1<-join(template2,df
Hi mailing listers,
Sorry, I made a little mistake in the previous mail. B^{1} should be B^{-1}.
Is there certain function in R deal with how to compute generalized
eigenvalues, that is the problem: A*x* = ëB*x *? When I use
eigen(B^{-1}*A), error happened. It displays there are many Inf element
Hi mailing listers,
Is there certain function in R deal with how to compute generalized
eigenvalues, that is the problem: A*x* = ëB*x *? When I use eigen(B^{1}*A),
error happened. It displays there are many Inf elements in B^{1}.
Thanks
Huaping Wan
[[alternative HTML version deleted]]
Thanks a lot, Duncan, that solved it!
Cheers,
Marius
Duncan Murdoch writes:
> On 13-01-24 2:09 AM, Marius Hofert wrote:
>> Dear Daniel,
>>
>> That's exactly what I also suspected (last post). The question now seems how
>> to
>> correctly convert .Random.seed from signed to unsigned so that i
> I'm trying to iterate model fits while holding one variable
> (l,a,b or s) constant. In order to do this, each time I need
> to re-run the mle2 fit with one variable held constant,
Have you read the help page closely enough, I wonder?
If I look at the mle2 help page, I see an argument called
Hi the list,
As far as I understand, a possibility to create a "scene" is to provide a
list of triangles to the function makeTriangles. Is there also a way to turn
the data that are used by "plot3d" into a scene? I would like to write:
--- 8< -
M <- matrix(runif(12),4)
plot3d(M)
### Some
I'm still having trouble, I think that I may have poorly explained the
problem I'm running into.
The code I'm working with looks like this:
mle.nb<-function(l,a,b,s){
t1pred=(data$T*l)/(1+a*data$T)^b
-sum(dnbinom(x=data$T.1, mu=t1pred, size=s,log=TRUE))
}
fit.mle.nb=mle2(mle.nb, start=lis
Hi,
Another way might be:
df1 <- merge(crosspries[[1]], crosspries[[2]], all=TRUE)
template1<-read.table(text=outer(unique(df1[,1]),unique(df1[,2]),FUN=paste),sep="",stringsAsFactors=F)
res<-merge(df1,template1,by.x=c("Product","Year_Month"),by.y=c("V1","V2"),all=TRUE)
res1<-do.call(rbind,split(
Dear R Helpers
I am having difficulty understanding how to use the penalty matrix for the
nomROC function in package 'nonbinROC'.
The documentation says that the values of the penalty matrix code the
penalty function L[i,j] in which 0 <= L[i,j] <= 1 for
j > i, but gives no other constraints. It
Hi
Please excuse this naive question, but I have search the FAQs and lists
and web for an answer but not found one.
I have a text file of several hundred thousand rows and somewhere in the middle
there is a malformed row.
Doing:
mydata<-read.table("mydata.dat", header=FALSE, sep=" ",
col
HI,
You could use:
library(reshape)
res<-melt(dat,id.var=c("region","state"))
names(res)[3:4]<-c("species","presence")
res<-res[rev(order(res$region,res$state)),]
row.names(res)<- 1:nrow(res)
res
# region state species presence
#1 sydney nsw species3 1
#2 sydney nsw species2
Dear Dr Harrell,
Thank you very much for your answer. Actually I also tried to found the C index
by hand on these data using the mean probabilities and I found 0.968, as you
just showed.
I understand now why I had a slight difference with the outpout of lrm. I am
thus convinced that this result
Hi Ben,
Thanks for info. That is definitely a viable solution for the example I
provided. It is more common that I have larger files with more edits to
make.
The main reason I went with a data frame methodology is becuase of the file
types I have. Essentially, I have an XML 50+ "rows" by 10 "co
Hello Adam,
I had a similar problem with a big dataframe, and building an xmlTree in
the clean way was extremely slow; so i resorted to manual method. Not
tested, but if your dataframe is my_df, then something like the
following should do:
buildEntry <- function(x) {
cat(paste('\n', sep=""))
Tammy,
# use the function dput() to provide code for us to easily recreate your
example data
crosspries <- list(structure(list(Product = c("A", "B", "C"), Year_Month =
c(201208L,
201208L, 201208L), prod1 = c(1L, 2L, 1L)), .Names = c("Product",
"Year_Month", "prod1"), class = "data.frame", row.nam
Dear list,
I simply can't figure out how to append an BSON array in MongoDB (with
package 'rmongodb') using either the '$push' or '$addToSet' operator.
It's sort of the last missing puzzle piece regarding CRUD operations, so
any hint will be greatly appreciated! Below you'll find a link to an
lrm does some binning to make the calculations faster. The exact calculation
is obtained by running
f <- lrm(...)
rcorr.cens(predict(f), DA), which results in:
C IndexDxy S.D. nmissing
0.96814404 0.93628809 0.0380833632.
In a Cox model, the baseline hazard takes the place of an intercept.
Try the following:
dummy <- rnorm(nrow(goodexp))
testfit <- lm(dummy ~ ExpTemp + Stability + Period, data=goodexp, x=T)
Now look at testfit$x, which is the design matrix for the linear model with an
intercept.
Examinat
Dear R
I am trying to understand and use the flexible parametric survival model
suggested by Royston and Parmar.
However I am stuck trying to plot the adjusted survival curves for
different covariates in the following code:
library(flexsurv)
library(graphics)
spl <- flexsurvspline(Surv(futime, fu
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of bsm2
> Sent: 23 January 2013 20:00
> To: r-help@r-project.org
> Subject: [R] Pasting a list of parameters into a function
>
> I need to repeat a function many times, with diff
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and
Hello!
I have csv file (attached) with any data, where first column is time.
When I read into data.frame become temperature values. What it can be?
data <- read.csv("rp5_vladimir2010.csv", colClasses = "character", sep
= ";") names(data)
data$Time
data[,1]
--
Всего наилучшего!
greg [at] dobroe
I have a code using the gRain package and there cptable is used to get the
conditional probability tables. My problem is that when I want to use this
table I get an error message that the file is too long to keep source. I have
tried to change the option keep.source but without luck. Anyone expe
I have problem as follows:
I use "mlogit package" to develop a multinomial logit model but I can not find
Percent Correctly predicted. would you help me?
Thanks
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ES function gives the below error.
> ES(sim, p=.95, method=c("modified"),portfolio_method=c("component"),
> weights=w1)
/Error in checkData(R, method = "xts", ...) : The data cannot be converted
into a time series. If you are trying to pass in names from a data object
with one column, you should
I have two directories
https://echange-fichiers.inra.fr/get?k=AcHKdNI4No44GEsj7PK with 12 (global
maps)binary files in each.I used the code given below to calculate the
spatial correlation between these files and it worked well(the output is a
global correlation map). I wonder if there is a simple
On 13-01-24 2:09 AM, Marius Hofert wrote:
Dear Daniel,
That's exactly what I also suspected (last post). The question now seems how to
correctly convert .Random.seed from signed to unsigned so that it is accepted by
the rlecuyer package.
The rlecuyer package assumes that seed values are positi
HI,
I have the following list:
crosspries
$crosspries[[1]]
Product Year_Month prod1
A 2012081
B 2012082
C 2012081
$crosspries[[2]]
Product Year_Monthprod1
Hi, Daisy,
try
dat2 <- reshape( dat, varying = c( "species1", "species2", "species3"),
v.name = "presence", timevar = "species",
times = c( "species1", "species2", "species3"),
direction="long")
dat2[ rev( order( dat2$region)), ]
Hth -- Ger
Hello,
I tried using reshape to rearrange my data to long format but I could
not get the output the table they way I wanted it. Anyway I came up
with a hack that does works, but I still would like to know if I can
do it with reshape.
Here is my code and a dummy set of data. It returns the data in
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