plot(fit,all.terms=TRUE,select=3,xlabs="???")
On 20/12/12 02:06, Andrew Crane-Droesch wrote:
Dear List,
plot.gam appears to be having trouble communicating its xlab and ylab
information to termplot. A simple example:
library(mgcv)
x = 1:1000
y = runif(1000)*x^.5
z = rnorm(1000)*y
other = sin(
On 12/19/2012 11:18 PM, maxbre wrote:
Given this example
library(plotrix)
a<-c(34,10,70)
b<-c(33,10,20)
c<-c(33,80,10)
test<-data.frame(A=a,B=b,C=c)
triax.plot(test,
main ="title",
at=seq(0.25,0.75,by=0.25),
tick.labels=list(l=seq(0.25,0.75,by=0.25),
Dear R-listers,
I am a newbie with R and I am struggling with something I consider very
basic. I wish to produce a table (to import in a latex file) of summary
statistics, but for as much as I've been looking around and trying various
alternatives (plyr, reporttools, pastecs and Hmisc) I haven't f
On 12-12-20 6:45 AM, Francesco Sarracino wrote:
Dear R-listers,
I am a newbie with R and I am struggling with something I consider very
basic. I wish to produce a table (to import in a latex file) of summary
statistics, but for as much as I've been looking around and trying various
alternatives
thanks jim,
I had to move the legend to a different place (as the default) because
for some reasons it appeared to me "sliced" on the left side (but I'm
not sure if that was due to my own configuration, anyway...); I think
that the possibility to control the size of the point labels would be a
Following problem:
Say you have a bunch of parameters and want to produce results for all
combinations of those:
height<-c("high","low")
width<-c("slim","wide")
then what i used to do was something like this:
l<-list()
for(h in height){
l[[h]]<-list()
for(w in width){
Dear all,
I have 2 vectors of 10 elements, each of them contains either 0 or 1, like
following:
> Null<-rep(0,10)
> Null
[1] 0 0 0 0 0 0 0 0 0 0
> One<-rep(1,10)
> One
[1] 1 1 1 1 1 1 1 1 1 1
How can I obtain a matrix rows of which can take all possible combinations?
e.g.
0 0 0 0 0 0 0 0 0 0
On 12/20/2012 02:13 PM, arun wrote:
> library(gtools)
> permutations(2,10,0:1,repeats.allowed=TRUE)
Thanks, this is exactly what I need.
Regards,
Olga
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
you might also try (names modified a bit since R already has a NULL object)
> Zeros <- rep(0,10)
> Ones <- rep(1,10)
> expand.grid(Map(c, Zeros, Ones))
On Thu, Dec 20, 2012 at 9:24 PM, Olga Lyashevska wrote:
> On 12/20/2012 02:13 PM, arun wrote:
>> library(gtools)
>> permutations(2,10,0:1,repea
On 12/20/2012 02:13 PM, arun wrote:
> library(gtools)
> permutations(2,10,0:1,repeats.allowed=TRUE)
This does what I need, but if I increase a number of permutations (59
instead of 10), then I get an error.
Error: cannot allocate vector of size 1.5 Gb
Is there a smart way of increasing the maxim
Dear Jessica
Aggregate is a function that allows you to perform loops across multiple
variables.
tempData <- data.frame(height = rnorm(20, 100, 10),
width = rnorm(20, 50, 5),
par1 = rnorm(20))
tempData$htfac <- cut(tempData$height, c(0, 100, 200))
tempData$wdfac <- cut(te
Dear:
I am having problem with simulation. Sometimes for 100 iteration, my
program works but sometimes give error message regarding some "built-in
function" which I never used
in my code. I have no idea about the problem. Any suggestion, comments or
idea regarding this would be highly appreciable.
HI,
You could replace the times=. in res1 with:
res1<-na.omit(cbind(ID=rep(dat2[,2][grep("ID",dat2$variable)],times=length(grep("p",names(dat1,dat2[grep("p",dat2$variable),],dat2[grep("lat",dat2$variable),],dat2[grep("long",dat2$variable),]))
A.K.
- Original Message -
From: arun
HI,
library(gtools)
If you need ?combinations()
combinations(2,10,0:1,repeats.allowed=TRUE)
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#[1,] 0 0 0 0 0 0 0 0 0 0
#[2,] 0 0 0 0 0 0 0 0 0 1
#[3,] 0 0 0 0 0
Hi all,
I'm trying to perform a nested ANOVA, in which "Rewetted" is the main
factor and "Position" is a fixed factor nested in "Location". I'm
interested in the effect of "Rewetting" and "Position" on the dependent
variable (in this case called N), and also in the interaction between these
factor
On 20-12-2012, at 15:49, Rajibul Mian wrote:
> Dear:
>
> I am having problem with simulation. Sometimes for 100 iteration, my
> program works but sometimes give error message regarding some "built-in
> function" which I never used
> in my code. I have no idea about the problem. Any suggestion, c
Aggregate is highly confusing (and i would have appreciated if you used my
example instead, i don't get it to do anything sensible on my stuff).
And this seems not what i asked for anyway. This may be a named list but not
named and structured as i want it at all.
happy Christmas too
On 20.12.2
I think your immediate problem is that you are asking this particular question
in the wrong forum. It appears that you are following instructions (in a book?
a class?) and you need to direct your questions to the author of those
instructions.
You may find it helpful to know:
a) The load functi
Christofer Bogaso gmail.com> writes:
>
[snip]
> I was trying to load a .r file using source() and load() functions.
> Although I could be able to load the codes (mainly few user defined
> functions) written in that .r. file correctly, however getting
> following error:
>
> > source("D:/Book
Hello all, i'm a newbie to R and now I get stuck by the question, would
anybody help me please?
Suppose, I got the following
> test <- data.frame(test1=c(c(0.00, 0.005, 0.01, 0.015, 0.02)))
> test
test1
1 0.000
2 0.005
3 0.010
4 0.015
5 0.020
Now I want another column inside test, named test2
Dear R forum
I have a following vector of random no.s
x = runif(100, 0.01, 0.99)
[1] 0.47212037 0.77867992 0.33947474 0.93369035
[5] 0.03720073 0.79307831 0.81801835 0.92710688
.
I need to count the random no. falling in the range (0 - 0.10)
Dear R-Community,
I'm writting my diploma-thesis this days. Therefore I have to estimate a few
coefficients with a Survreg function (library=Survival). My Problem is that
the estimated coefficients doesn't make sense.
*My Questions:*
- What role plays the scale-Value for the coefficient-estimati
Have considered "cut" and "table" ?
On Dec 20, 2012 12:08 PM, "Katherine Gobin"
wrote:
> Dear R forum
>
>
> I have a following vector of random no.s
>
> x = runif(100, 0.01, 0.99)
>
>
> [1] 0.47212037 0.77867992 0.33947474 0.93369035
> [5] 0.03720073 0.79307831 0.81801835 0.92710688
> ...
Hello all, i'm a newbie to R and now I get stuck by the question, would
anybody help me please?
Suppose, I got the following
> test <- data.frame(test1=c(c(0.00, 0.005, 0.01, 0.015, 0.02)))
> test
test1
1 0.000
2 0.005
3 0.010
4 0.015
5 0.020
Now I want another column inside test, named test2
I don't know of one. If building your own you could use rpart with the "maxdepth=1" as
the tool to find the best split at each node.
Terry Therneau
On 12/20/2012 05:00 AM, r-help-requ...@r-project.org wrote:
Hi,
I've searched R-help and haven't found an answer. I have a set of data from
whi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Yanyuan Zhu
> Sent: Thursday, December 20, 2012 7:48 AM
> To: r-help@r-project.org
> Subject: [R] an entry level (stupid) question
>
> Hello all, i'm a newbie to R and now I get
Hi,
test$test2<-ifelse(test$test1!=0,39,1/test$test1)
A.K.
- Original Message -
From: Yanyuan Zhu
To: r-help@r-project.org
Cc:
Sent: Thursday, December 20, 2012 10:45 AM
Subject: [R] an entry level (stupid) question
Hello all, i'm a newbie to R and now I get stuck by the question, w
Hi Yanyuan
Check out:
?"=="
?"["
?"<-"
You should be able to assign the column quickly with no more than two lines of
code using square brackets and assign.
Or to literally use if, look at:
?"ifelse"
Happy Christmas
Chris
Chris Campbell
Tel. +44 (0) 1249 705
Hello,
I want to plot 13 barplots in a 4x4 matrix-window (1st layer).
So that's no Problem, it worked out.
The problem is now, that I want to add two additional layers to this
graphic.
They contain similar barplots, just with another color for the second
bar of each barplot (2nd layer) and a
Hi,
When I was using the regular plot() function, I added this:
--8<---cut here---start->8---
if (!is.null(file)) {
do.call(tools::file_ext(file),list(file = file))
on.exit(dev.off())
cat("writing",file,"\n")
}
--8<---cut here
Arranging data as a list of lists of lists of lists [...] of scalar values
generally
will lead to slow and hard-to-read R code, mainly because R is designed to
work on long vectors of simple data. If you were to start over, consider
constructing
a data.frame with one column for each attribute.
Really must have been unclear at some point, sorry.
William, thats interesting, but not really helping the main problem, which is:
how to do
> l[[ as.character(grid[1, ]) ]] <- 1
without having initialized the list in the loop before.
Well, or how to initialize it without having to do the lo
Jessica
This is super ugly but assigns as you requested in the original post:
height<-c("high","low")
width<-c("slim","wide")
doSomething <- function() {
cat("done...\n\n")
return(list(a = list(1), b = list(2), c = list(3))) }
tempList <- list()
for(h in height){
If you really want to use the nested lists and use things like
lst[[ c("high", "north", "unmarried") ]] <- value
when lst[["high"]] or lst[["high"]][["north"]] does not exist,
then I think you will have to make a class that does this.
Another advantage of writing a class is that it could be rew
Hello,
Inline.
Em 20-12-2012 15:47, Yanyuan Zhu escreveu:
Hello all, i'm a newbie to R and now I get stuck by the question, would
anybody help me please?
Suppose, I got the following
test <- data.frame(test1=c(c(0.00, 0.005, 0.01, 0.015, 0.02)))
test
test1
1 0.000
2 0.005
3 0.010
4 0.015
Also note that a column of a data.frame can be a list of complicated things.
E.g.,
> d <- expand.grid(am=c(0, 1), gear=c(3,4,5))
> d$results <- I(lapply(seq_len(nrow(d)), function(i)try(lm(mpg~wt,
> subset=gear==d$gear[i] & am==d$am[i], data=mtcars
> d[ d$am==1 & d$gear==5, "results"
Hi, how are you?
I have the following truncated R code:
fileobs <- "MaxFloodDepth_Observed.txt"
file1 <- "MaxFloodDepth_1.txt"
file2 <- "MaxFloodDepth_2.txt"
...
file54 <- "MaxFloodDepth_54.txt"
observeddepth = as.matrix(read.ascii.grid(fileobs)$data)
observeddepth[observeddepth!=0]<-1
model
Hi,
Try this:
range1<-cut(x,breaks=c(0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1),labels=paste("(",c("0-0.1","0.1-0.2","0.2-0.3","0.3-0.4","0.4-0.5","0.5-0.6","0.6-0.7","0.7-0.8","0.8-0.9","0.9-1"),")",sep=""))
table(range1)
#range1
# (0-0.1) (0.1-0.2) (0.2-0.3) (0.3-0.4) (0.4-0.5) (0.5-0.6) (0.6-0.7)
Is it possible to control formatting of the numbers which go along the
axes in plots?
e.g.
plot(x=1:100,y=1:100)
will label the X axis as "0d+00", "2e+05" &c.
I want that to read 0, 200k, 400k &c.
I know of the function axis(), but it offers far too much control for
this simple task.
thanks
It is possible, but only by using axis() since you can specify axis breaks
in a plot command, but not the labels. You can ignore most of the axis()
options so the commands are pretty simple:
plot(x=c(1, 100), y=c(1, 100), xlab="x", ylab="y",
xaxt="n", yaxt="n", las=2)
pos <- c(0, 200
See FAQ 7.22 and fortune("line 800").
cheers,
Rolf Turner
On 21/12/12 06:29, Sam Steingold wrote:
Hi,
When I was using the regular plot() function, I added this:
--8<---cut here---start->8---
if (!is.null(file)) {
do.call(tools::file_ex
> * David L Carlson [2012-12-20 13:58:00 -0600]:
>
> It is possible, but only by using axis() since you can specify axis breaks
> in a plot command, but not the labels. You can ignore most of the axis()
> options so the commands are pretty simple:
>
> plot(x=c(1, 100), y=c(1, 100), xlab="x
Jonas,
I think f2si(0) should be "0", not "" as it is now.
Thanks.
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http://www.childpsy.net/ http://ffii.org http://mideasttruth.com
http://thereligionofpeace.com http://iris.org.il http://truepeace.org
Type louder,
By "too much control" do you mean that axis requires too many inputs?
You can use axTicks to get the positions of the tick marks that would have
been drawn and create labels based on those positions. E.g.,
> myAxis <- function (side) {
at <- axTicks(side = side)
lab <- ifelse(abs(at)>=1e
On Dec 20, 2012, at 10:01 AM, Jessica Streicher wrote:
> Really must have been unclear at some point, sorry.
Hasn't it become abundantly clear that this would have progress farther had you
post a complete example?
--
David.
>
> William, thats interesting, but not really helping the main prob
Hello ,
My optmatch package is loaded on R and otherwise running fine.
I get an error after lcds successfully completes a logistic regression and
I then try to obtain a propensity score:
pdist <- pscore.dist(lcds)
Error: could not find function "pscore.dist"
Does anyone know if pscore.dist
Hi everyone, I have received assistance from my previous post
(comparison of large data set), but I still need some assistance.
This is the revised code (along with some new code that I am requesting
assistance with correcting):
files <- paste0("MaxFloodDepth_", 1:54, ".txt")
fileobs <- "MaxFlood
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of MA
> Sent: Thursday, December 20, 2012 3:09 PM
> To: r-help@r-project.org
> Subject: [R] Optmatch Package Question
>
> Hello ,
>
> My optmatch package is loaded on R and otherwi
On 12/20/2012 10:45 PM, Francesco Sarracino wrote:
Dear R-listers,
I am a newbie with R and I am struggling with something I consider very
basic. I wish to produce a table (to import in a latex file) of summary
statistics, but for as much as I've been looking around and trying various
alternativ
On 12/21/2012 04:13 AM, Anne-Marie Brigitte Gallrein wrote:
Hello,
I want to plot 13 barplots in a 4x4 matrix-window (1st layer).
So that's no Problem, it worked out.
The problem is now, that I want to add two additional layers to this
graphic.
They contain similar barplots, just with another
Hello,
A reproducible example would be nice.
Anyway, with "lattice::xyplot", you can store, then print the figure, in
order to write it into a file:
> xyObj <- xyplot(...)
> postscript(...)
> print(xyObj)
> dev.off()
Regards,
Pascal
Le 21/12/2012 02:29, Sam Steingold a écrit :
Hi,
When I w
Hello ,
My optmatch package is loaded on R and otherwise running fine.
I get an error after lcds successfully completes a logistic regression and
I then try to obtain a propensity score:
pdist <- pscore.dist(lcds)
Error: could not find function "pscore.dist"
Does anyone know if pscore.di
Hey,
I have an double loop like this:
chunk <- list(1:10, 11:20, 21:30)
for(k in 1:length(chunk)){
print(chunk[k])
DummyCatcher <- NULL
for(i in chunk[k]){
print("i load something")
dummy <- 1
print("i do something")
Hey,
it seems like it is indeed:
library(sos)
findFn(pscore.dist)
Best, Peter
source: http://stackoverflow.com/a/10553795/1144966
Am 21.12.2012 00:08, schrieb MA:
Hello ,
My optmatch package is loaded on R and otherwise running fine.
I get an error after lcds successfully completes a logi
> Jeffrey Dick
> on Thu, 20 Dec 2012 21:40:15 +0800 writes:
> you might also try (names modified a bit since R already has a NULL
object)
>> Zeros <- rep(0,10)
>> Ones <- rep(1,10)
>> expand.grid(Map(c, Zeros, Ones))
Wow -- really neat!
Thank you, Jeff
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