Hello,
I want to create and save objects in a loop, but this is precluded by the
following obstacle:
this part of the script fails to work:
assign(x=paste("a", 1, sep=""), value=1);
save(paste("a", 1, sep=""), file=paste(paste("a", 1, sep=""), ".RData", sep=""))
Do you know any workaround? I a
All,
Thanks. That works.
D.
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Is that what you mean:
> 2 + 3
[1] 5
> .Last.value
[1] 5
Rgds,
Rainer
(from R-intro.pdf, page 7, 2nd footnote)
Original-Nachricht
> Datum: Thu, 19 Jul 2012 22:12:22 -0700 (PDT)
> Von: darnold
> An: r-help@r-project.org
> Betreff: [R] Last answer
> Hi,
>
> In Matlab, I can
My apologies. I stand corrected. Thanks Michael.
Josh
On Thu, Jul 19, 2012 at 10:29 PM, R. Michael Weylandt
wrote:
> See: https://stat.ethz.ch/pipermail/r-help/2012-February/303110.html
>
> Michael
>
> On Fri, Jul 20, 2012 at 12:19 AM, Joshua Wiley wrote:
>> Hi David,
>>
>> No, but you can st
See: https://stat.ethz.ch/pipermail/r-help/2012-February/303110.html
Michael
On Fri, Jul 20, 2012 at 12:19 AM, Joshua Wiley wrote:
> Hi David,
>
> No, but you can store the results and access that.
>
> ## parentheses to force printing
> (x <- 2 + 3)
>
> x
>
> Cheers,
>
> Josh
>
>
> On Thu, Jul 1
Hi David,
No, but you can store the results and access that.
## parentheses to force printing
(x <- 2 + 3)
x
Cheers,
Josh
On Thu, Jul 19, 2012 at 10:12 PM, darnold wrote:
> Hi,
>
> In Matlab, I can access the last computation as follows:
>
>>> 2+3
>
> ans =
>
> 5
>
>>> ans
>
> ans =
>
Hi,
In Matlab, I can access the last computation as follows:
>> 2+3
ans =
5
>> ans
ans =
5
Anything similar in R?
David
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My amateur approach:
I put your data in a dataframe called t:
> head( t )
Date Score
1 2008-05-01 08:58:0080
2 2008-05-01 13:31:0011
3 2008-05-01 16:35:0081
4 2008-05-01 23:20:00 152
5 2008-05-02 01:01:00 130
6 2008-05-02 03:35:00 122
Then I created a vector wit
> dtf <- read.table(text="y A B C
+ 0 11 2
+ 0 12 1
+ 1 11 2
+ 0 11 2
+ 1 11 2
+ 1 12 1
+ 0 12 2",
+ header=TRUE)
> dtagroup <- aggregate(y~A+B+C, dtf, sum)
# Gets you the groups. If you need the column/row order:
> dtag
Hi,
I have my data the following way:
y A B C
0 11 2
0 12 1
1 11 2
0 11 2
1 11 2
1 12 1
0 12 2
.
.
.
And so on. How can I make my data look like the following:
y A B C
2 1 1 2
1 1 2 1
0 1 2 2
.
.
.
Hello,
Sorry, forgot about that. It's trickier to write code without a dataset
to test it.
Try
pattern <- "L[1-8][12]"
and after the grep print nms to see if it's right.
Rui Barradas
Em 20-07-2012 00:33, Lib Gray escreveu:
I'm getting this error message:
nms<-names(data)[grep(vars,names(
Hello,
I guess so, and I can save you some typing.
vars <- sort(apply(expand.grid("L", 1:8, 1:2), 1, paste, collapse=""))
Then use it and see the result.
Rui Barradas
Em 20-07-2012 00:00, Lib Gray escreveu:
The variables are actually L11, L12, L21, L22, ... , L81, L82. Would just
creating a
Hello,
Follow this example.
# First see the regular expression
# at work without changing any disk file
x <- c("file0.1_data.RData", "file0.2_data.RData")
sub("^file0\\.", "file", x)
# Now change them
from.files <- list.files(pattern = "^file0\\.._data.Rdata$")
to.files <- sub("^file0\\.", "fi
On 12-07-19 4:00 PM, Jan van der Laan wrote:
When the length of the end result is not known, doubling the length of
the list is also much faster than increasing the size of the list with
single items.
f <- function(n, preallocate) {
v <- if(preallocate) vector("list",n) else list() ;
Look at kruskalmc() in package pgirmess.
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.o
Dear All,
I want to change file names.
I have file0.1_data.RData (I have several files whose names are
file0.x_data.Rdata)
I want to rename it to file1_data.RData
How can I do it?
Thank you for saving my time!
[[alternative HTML version deleted]]
__
Tena koe Anna
Yes: see https://stat.ethz.ch/pipermail/r-help/2006-August/111234.html which
includes an excellent description on this written by Bill Venables back in 1997.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org
Your question is not entirely clear to me, but I think you might want:
plot(X, Y, xlim=c(0, 25), ylim=c(0, 25))
Sarah
On Thu, Jul 19, 2012 at 5:31 PM, Lekgatlhamang, lexi Setlhare
wrote:
> Dear all,
>
> I have a challenge with a supposedly simple graph (a scatter). I wanted to
> use R to creat
Dear all,
I have a challenge with a supposedly simple graph (a scatter). I wanted to use
R to create a plot/graph with a positive y-axis intercept but it does not seem
to give me what I expect to see. As an example, I used the following values
> X<- c(0, 4, 8, 11)
> Y<- c(8, 12, 15, 22)
then
Short answer:
as.POSIXct(0, origin = ISOdatetime(1970,1,1,10,0,0, tz = "GMT"))
Long answer -- as.POSIXct goes through a somewhat crazy chain of
method dispatch to work, and only sometimes cares about the tz
argument.
For your case
as.POSIXct(x) -- calls -->
as.POSIXct.numeric(x) -- calls -->
The following three calls all produce the same result (my machine is in EST):
> as.POSIXct(0, tz="", origin=ISOdatetime(1970,1,1,10,0,0))
[1] "1970-01-01 10:00:00 EST"
> as.POSIXct(0, tz="EST", origin=ISOdatetime(1970,1,1,10,0,0))
[1] "1970-01-01 10:00:00 EST"
> as.POSIXct(0, tz="GMT", origin=IS
On Thu, Jul 19, 2012 at 2:41 PM, uday wrote:
> Hi,
>
> Recently I have installed R version 2.14.1, after installation I am trying
> to install some packages and I get error message. even I tried
> install.packages("Rcmdr")
Why would that help?
Also: don't double post. Sarah already told you how
Hello,
Try the following. The data is your example of Patient A through E, but
from the output of dput().
dat <- structure(list(Patient = structure(c(1L, 1L, 1L, 1L, 1L, 2L,
2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 5L, 5L, 5L), .Label = c("A",
"B", "C", "D", "E"), class = "factor"), Cycle = c(1L, 2
Same post on Stack Overflow (again):
http://stackoverflow.com/q/11567745/271616
--
Joshua Ulrich | FOSS Trading: www.fosstrading.com
On Thu, Jul 19, 2012 at 11:15 AM, cursethiscure
wrote:
> I think the code is part of the RTAQ package but is not included in it, as I
> obtained it from
> https:
Hi,
A client has a consumption measure on each of four products. The sample
size is 75. The consumption distributions are highly skewed for each
product. He would like a pairwise comparison test of the products, much
like Tukey's HSD but using medians rather than means. Is there such a
me
> On Thu, Jul 19, 2012 at 3:00 PM, Jan van der Laan wrote:
>
> When the length of the end result is not known, doubling the length of the
> list is also much faster than increasing the size of the list with single
> items.
>
> [snip]
>
> What causes these differences? I can imagine that the time n
Cross-posted on Stack Overflow:
http://stackoverflow.com/q/11567745/271616
--
Joshua Ulrich | FOSS Trading: www.fosstrading.com
On Thu, Jul 19, 2012 at 12:23 PM, cursethiscure
wrote:
> I am working with xts dependent data, and my code is as follows (the problem
> is explained throughout):
>
>
On 2012-07-19 11:05, Julie Shoemaker wrote:
Hi all,
I'm attempting to gap-fill a dataset, replacing the missing values with
each month's day or night median value.
The problem is that my code results in some, but not all the NA's being
replaced and I cannot figure out how this is possible. When
I think the code is part of the RTAQ package but is not included in it, as I
obtained it from
https://r-forge.r-project.org/scm/viewvc.php/pkg/RTAQ/R/HAR_model.R?view=markup&root=blotter&sortby=author&pathrev=1028.
It is not my code and I make no claim to other's good work, and apologize if
I sh
Dear list,
I am using the np package. With the npindex function I estimate a
semiparametric single index model using the method of Klein-Spady.
P(Z=1|X) = G(X’b)
I don’t have any problems to calculated the fitted values and standard
errors X’b:
bw = npindexbw(xdat=x, ydat=y_bi, method="kleins
On Thu, Jul 19, 2012 at 04:12:07AM -0700, arunkumar wrote:
> hi
>
> My inputs is min=(10,10,10,10,10) and max=(100,100,100,100,100)
> total = 300
> i have to generate 5 numbers between min and max and those numbers should
> sum upto total
Hi.
Try the following.
while (1) {
x <- 10 +
- Forwarded Message -
From: arun
To: Abraham Mathew
Cc: R help
Sent: Thursday, July 19, 2012 3:58 PM
Subject: Re: [R] Removing values from a string
Hi,
Try this:
one = data.frame(keyword=c("|auto", "NA|auto|insurance|quote",
"NA|auto|insurance",
"NA|ins
Hi,
Try this:
one = data.frame(keyword=c("|auto", "NA|auto|insurance|quote",
"NA|auto|insurance",
"NA|insurance", "NA|auto|insurance", ""))
onenew<-data.frame(keyword=gsub("(NA){0,1}\\|","",one$keyword))
onenew1<-data.frame(keyword=gsub("(){0,1}","",onenew$keyword))
Hi, everyone.
I have some questions about quantile regression in R.
I am running an additive quantile regression first for a complete matrix and
then with some selected rows.
I am doing the following:
datos <-read.table("Regresion multiple.txt",header=T)
Fit<-rqss(datos$campings
~datos$Cobarb
Hi all,
I'm attempting to gap-fill a dataset, replacing the missing values with
each month's day or night median value.
The problem is that my code results in some, but not all the NA's being
replaced and I cannot figure out how this is possible. When I look at
the individual line's where th
I would like to use a A priori mean comparison/orthogonal contrast test on my
data.
Im new to using R and I would like to know if its possible to perform this
test in R and how can be done.
/Anna
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Thank you very much, Michael!
That was exactly the hint I needed. I ended up using "try" as below, in case
anyone happens to read this later on...
Works perfectly fine.
So yes, it _is_ amazing, what one can do with R ;-)
Regards,
Berry
internet <- try(read.table("http://www...";), silent=T )
Jan:
Point taken.
However, if possible, as Bill Dunlap indicated, it still may make
sense to create an oversized list first and then populate what you
need of it with your loop.
Note that a lot of this can be finessed with lapplyand friends
anyway, letting R worry about the details of creating a
When the length of the end result is not known, doubling the length of
the list is also much faster than increasing the size of the list with
single items.
f <- function(n, preallocate) {
v <- if(preallocate) vector("list",n) else list() ;
for(i in seq_len(n)) {
v[[i]] <- i
Hi,
Recently I have installed R version 2.14.1, after installation I am trying
to install some packages and I get error message. even I tried
install.packages("Rcmdr") but still I am unable to fix this problem.
I would be very grateful if somebody can help me to fix this problem.
install.packa
Hello,
Try this:
#Generate 4 random numbers.
set.seed(1)
rnorm(4,60)
#[1] 59.37355 60.18364 59.16437 61.59528
> sum(rnorm(4,60))
#[1] 240.7348
fifthnumber<-300-240.7348
fifthnumber
#[1] 59.2652
A.K.
- Original Message -
From: arunkumar
To: r-help@r-project.org
Cc:
Sent: Thu
I couldn't find anything in the chron or timeDate packages, and a good search
yielded rounding to the nearest half hour, which I don't want.
The data:
structure(list(Date = structure(c(1209625080, 1209641460, 1209652500,
1209676800, 1209682860, 1209692100, 1209706980, 1209722580, 1209726300,
12
There are a couple of ambiguities in your request, but this should get
you started:
> one$keyword <- gsub("NA\\|", "", one$keyword)
> one$keyword <- gsub("^\\|", "", one$keyword)
> one
keyword
1 auto
2 auto|insurance|quote
3 auto|insurance
4insurance
On 07/19/2012 06:11 PM, Bert Gunter wrote:
Hadley et. al:
Indeed. And using a loop is a poor way to do it anyway.
v <- as.list(rep(FALSE,dotot))
is way faster.
-- Bert
I agree that not using a loop is much faster, but I assume that the
original question is about the situation where the siz
On 2012-07-19 07:10, Bart Ferket wrote:
Dear professor Harrell,
I probably have the same problem as Haleh Ghaem Maralani.
I am using the rms package and the rcspline.plot function to assess the
relation of a continuous predictor to the log hazard function.
I would like to use the "adj" stateme
So I have the following data frame and I want to know how I can remove all
"NA" values from each string, and also
remove all "|" values from the START of the string. So they should
something like "auto|insurance" or "auto|insurance|quote"
one = data.frame(keyword=c("|auto", "NA|auto|insurance|quot
On Thu, Jul 19, 2012 at 10:24:17AM -0700, Linh Tran wrote:
> Hi fellow R users,
>
> I am desperately hoping there is an easy way to do this in R.
>
> Say I have three functions:
>
> f(x) = x^2
> f(y) = 2y^2
> f(z) = 3z^2
>
> constrained such that x+y+z=c (let c=1 for simplicity).
>
> I want to
Can someone please give me an example of how to enter starting values for a
GRM (IRT) model using the ltm package?
The instructions from the ltm manual are below, but when I create either a
list of the values or a matrix I get the error, "start.val not of proper
type." I can find n
On 2012-07-19 05:56, penguins wrote:
Thanks William, that works fantastically!
I had a quick play with my data and have realised a potential problem in
that if an individual ends the series at home it records an additional
trip-no when one wasnt made. I was wondering whether you could think of a
On 07/19/2012 05:50 PM, Hadley Wickham wrote:
On Thu, Jul 19, 2012 at 8:02 AM, Jan van der Laan wrote:
The following function is faster than your g and easier to read:
g2 <- function(dotot) {
v <- list()
for (i in seq_len(dotot)) {
v[[i]] <- FALSE
}
}
Except that you don't need
Hi Linh,
Here is an approach:
f <- function(v) {
v <- v/sum(v)
(v[1]^2) + (2 * v[2]^2) + (3*v[3]^2)
}
(res <- optim(c(.6, .3, .1), f))
res$par/sum(res$par)
This is a downright lazy way to implement the constraint. The main
idea is to combine all three functions into one function that take
Hi fellow R users,
I am desperately hoping there is an easy way to do this in R.
Say I have three functions:
f(x) = x^2
f(y) = 2y^2
f(z) = 3z^2
constrained such that x+y+z=c (let c=1 for simplicity).
I want to find the values of x,y,z that will minimize f(x) + f(y) + f(z).
I know I can use th
On Thu, Jul 19, 2012 at 11:11 AM, Bert Gunter wrote:
> Hadley et. al:
>
> Indeed. And using a loop is a poor way to do it anyway.
>
> v <- as.list(rep(FALSE,dotot))
>
> is way faster.
>
> -- Bert
>
Its not entirely clear to me what we are supposed to conclude about this.
I can confirm Bert's cla
On Thu, Jul 19, 2012 at 9:21 AM, William Dunlap wrote:
> Preallocation of lists does speed things up. The following shows
> time quadratic in size when there is no preallocation and linear
> growth when there is, for size in the c. 10^4 to 10^6 region:
Interesting, thanks! I wish there was a be
Hello,
Try the following.
d <- read.csv(text="
Patient, Cycle, Variable1, Variable2
A, 1, 4, 5
A, 2, 3, 3
A, 3, 4, NA
B, 1, 6, 6
B, 2, NA, 6
C, 1, 6, 5
C, 3, 2, 2
", header=TRUE)
d
compl <- lapply(split(d, d$Patient), function(x) if(all(diff(x$Cycle) ==
1)) x)
holes <- lapply(split(d, d$Patie
You should include your data using dput(mymatrix) to make things easier
structure(list(Label = structure(c(4L, 1L, 2L, 3L), .Label = c("de 31 a 40",
"de 41 a 50", "Mayor de 51", "Menor de 30"), class = "factor"),
Blogs = c(57.14, 63.83, 72.64, 62.07), Wikis = c(28.57, 61.7,
70.75, 58.62
On Thu, Jul 19, 2012 at 04:12:07AM -0700, arunkumar wrote:
> hi
>
> My inputs is min=(10,10,10,10,10) and max=(100,100,100,100,100)
> total = 300
> i have to generate 5 numbers between min and max and those numbers should
> sum upto total
Hi.
If we subtract the minimum from each number, then
Thanks, Bill, but:
> system.time(z1 <-f(n=1e5,pre=TRUE))
user system elapsed
0.320.000.32
> system.time(z2 <- as.list(seq_len(1e5)))
user system elapsed
0 0 0
> identical(z1,z2)
[1] TRUE
So the point is not to use an R level loop at all.
-- Bert
On Thu, Jul
Those restrictions you have given do not define a unique distribution!
so you need to think better
about what you need. For instance, if you want a uniform distribution
between min and max with n=5
independent observations from that, but conditional upon sum=total.
For that, you could use rejectio
Preallocation of lists does speed things up. The following shows
time quadratic in size when there is no preallocation and linear
growth when there is, for size in the c. 10^4 to 10^6 region:
> f <- function(n, preallocate) { v <- if(preallocate)vector("list",n) else
> list() ; for(i in seq_len(n
also rep.int()
> system.time(for (i in 1:1000) x <- rep.int(FALSE, 10))
user system elapsed
0.290.020.29
> system.time(for (i in 1:1000) x <- rep(FALSE, 10))
user system elapsed
1.960.082.05
On Thu, Jul 19, 2012 at 9:11 AM, Bert Gunter wrote:
> Hadley et. al
> When i make Boxplots with a lot of boxes, the names of them
> get only written down every second "column".
> Since they aren't in any way ordered, you don't see anymore
> to what they belong.
Jessica,
Another possibility if the names are long is to use abbreviated factor levels.
The labels a
Hadley et. al:
Indeed. And using a loop is a poor way to do it anyway.
v <- as.list(rep(FALSE,dotot))
is way faster.
-- Bert
On Thu, Jul 19, 2012 at 8:50 AM, Hadley Wickham wrote:
> On Thu, Jul 19, 2012 at 8:02 AM, Jan van der Laan wrote:
>> Johan,
>>
>> Your 'list' and 'array doubling' code
The error message is here: configure: error: Can't find HDF5
The hdf5 package is an interface to the HDF5 library, so you need to
install hdf5 and hdf5-devel through your package manager (yum or
apt-get or whatever you use to install things on your linux distro).
Or you can get it straight from t
Dear Jessica,
You might try par(las=2) to rotate the tick labels to be perpendicular to the
axes.
I hope this helps,
John
John Fox
Sen. William McMaster Prof. of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Ca
No you are not correct. The kde function estimates the density of 1 to 6
dimensions. To visualize, this try plotting the density of just the data
(using density() instead of kde):
plot(density(elevation$data))
rug(elevation$data)
The elevations are plotted along the x-axis and their density is pl
On Thu, Jul 19, 2012 at 8:02 AM, Jan van der Laan wrote:
> Johan,
>
> Your 'list' and 'array doubling' code can be written much more efficient.
>
> The following function is faster than your g and easier to read:
>
> g2 <- function(dotot) {
> v <- list()
> for (i in seq_len(dotot)) {
> v[[
Hi
Thanks for reply
after usinginstall.packages("hdf5")
I get error
{Installing package(s) into ‘/home/uday/R/x86_64-pc-linux-gnu-library/2.14’
(as ‘lib’ is unspecified)
trying URL
'http://cran.revolutionanalytics.com/src/contrib/hdf5_1.6.9.tar.gz'
Content type 'application/x-gzip' length 50870 b
Dear Chris,
many thanks! This is just what I had in mind! (namely the 'sapply' solution).
Thank you and best regards!
Zdenek
-Original Message-
From: Chris Campbell [mailto:ccampb...@mango-solutions.com]
Sent: Thursday, July 19, 2012 4:11 PM
To: Skála, Zdeněk (INCOMA GfK)
Cc: r-help@
HI,
Possibly check.names=FALSE issue.
Try this:
dat1<-read.table(text="
2.5a 3.6b 7.1c 7.9d
100 3 4 2 3
200 3.1 4 3 3
300 2.2 3.3 2 4
",sep="",header=TRUE)
dat1
#You can get rid of those X by either using check.names=FALSE while reading the
data
#wi
Hello,
I have a double matrix that I want to represent in a line chart. Although I
have seen some examples I still don't manage to get it. My data is this (a
double matrix called mymatrix) :
Blogs Wikis Redes Etiq. SPC LMS
Menor de 30 57.14 28.57 14.29 28.57 57.14 28.57
de 31 a 4
Sorry, I just found this to be a common "problem" of tri.mesh:
I had to "jitter" one of my first three coords in the point set:
x[2] <- x[2] + 0.01
Though, that does not seem to sound clean. Is there a better way?
2012/7/19 Erdal Karaca
> I am trying to triangulate a point set as follows:
>
>
Hello,
I didn't give enough information when I sent an query before, so I'm trying
again with a more detailed explanation:
In this data set, each patient has a different number of measured variables
(they represent tumors, so some people had 2 tumors, some had 5, etc). The
problem I have is that
I am trying to triangulate a point set as follows:
> head(cbind(x,y))
x y
[1,] -78.1444 -60.4424
[2,] -78.1444 -58.4424
[3,] -78.1444 -56.4424
[4,] -78.1444 -54.4424
[5,] -76.1444 -60.4424
[6,] -76.1444 -58.4424
> length(x)
[1] 5000
> tri <- tri.mesh(x, y)
Fehler in tri.mesh(x, y) : error in trm
My data:
> head(cbind(x,y,z))
x y z
[1,] -78.1444 -60.4424 -10.09
[2,] -78.1444 -58.4424 -10.26
[3,] -78.1444 -56.4424 -10.45
[4,] -78.1444 -54.4424 -10.64
[5,] -76.1444 -60.4424 -10.19
[6,] -76.1444 -58.4424 -10.34
> tris <- deldir(x, y)
> triangs <- triang.list(tris)
> head(tris$delsgs)
x1 y1
Is this homework?
runif() is one way to generate random numbers, but there are others
depending on the distribution desired. And of course the fifth number
is deterministic.
Sarah
On Thu, Jul 19, 2012 at 7:12 AM, arunkumar wrote:
> hi
>
> My inputs is min=(10,10,10,10,10) and max=(100,100,1
I have this problem as well
Could you please kindly let me know what s is please. Is s equals to the
value of lambda?
Many thanks
Lini
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Thanks William, that works fantastically!
I had a quick play with my data and have realised a potential problem in
that if an individual ends the series at home it records an additional
trip-no when one wasnt made. I was wondering whether you could think of a
way to alter it slightly so that the i
Dear professor Harrell,
I probably have the same problem as Haleh Ghaem Maralani.
I am using the rms package and the rcspline.plot function to assess the
relation of a continuous predictor to the log hazard function.
I would like to use the "adj" statement, for example using this test
dataset:
hi
My inputs is min=(10,10,10,10,10) and max=(100,100,100,100,100)
total = 300
i have to generate 5 numbers between min and max and those numbers should
sum upto total
Can anyone help?
-
Thanks in Advance
Arun
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If you need to do this for a lot of dates findInterval() will do
it faster. E.g.,
> f <- function(dates, data) {
+ i <- findInterval(dates, data$date)
+ i[i==0] <- NA # before first data$date
+ data[i, ]
+ }
> xx <- as.Date(c("2010-10-06", "2010-10-25", "2009-01-01", "201
You might alternatively find the horizontal = TRUE with las=2 to be useful; e.g.
dat <- data.frame(val=rnorm(100),
grp=rep(apply(matrix(sample(letters,100,rep=TRUE),nr=5),2,paste,collapse=""),5))
boxplot(val~grp,horizontal=TRUE,data=dat,las=2)
## Note that las=2 might also help with horizonta
You will probably need to write a custom function, but it could
use the built-in write.dcf() or formatDL() functions. E.g.,
> # options(width=50)
> write.dcf(list(One=paste(1:50,collapse=" "), Two=paste(state.abb[1:5],
collapse=", ")))
One: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
17
Johan,
Your 'list' and 'array doubling' code can be written much more efficient.
The following function is faster than your g and easier to read:
g2 <- function(dotot) {
v <- list()
for (i in seq_len(dotot)) {
v[[i]] <- FALSE
}
}
In the following line in you array doubling function
Copied the wrong lines, sry
l<-rep(list(1:5),20); boxplot(l,names=sample(1:20,20))
of course.
thanks for the answer
.
On 19.07.2012, at 16:17, Peter Ehlers wrote:
> On 2012-07-19 06:58, Jessica Streicher wrote:
>> When i make Boxplots with a lot of boxes, the names of them get only written
>>
You're right - easily tested by just re-sizing the graphics box - sort of
counter-intuitive until I remember the clue is in the name - this generates a
CIRCLE, come what may...
Stuart
From: Sarah Goslee [mailto:sarah.gos...@gmail.com]
Sent: 19 July 2012 11:18
To: Stuart Leask
Cc: r-help@r-proje
Great, Thanks! This is really helpful!
On Thu, Jul 19, 2012 at 12:25 AM, ilai wrote:
> Maybe I'm missing something too but from your example seems like you are
> looking for
>
> xyplot(rnorm(12) ~ 1:12 , type="l",
> scales=list(x=list(at=seq(2,12,2),labels=c(1, ' ', 3 , ' ' , 5 , ' ' ))),
> par.s
On 2012-07-19 06:58, Jessica Streicher wrote:
When i make Boxplots with a lot of boxes, the names of them get only written down every
second "column".
Since they aren't in any way ordered, you don't see anymore to what they belong.
example:
l<-rep(list(1:5),20); boxplot(l,names=sample(20,1:20)
Dear Zdenek
You could generate this file using a loop.
# the data
abc <- list(one=(1:2), two=(1:5))
# create a connection
sink("aa.txt", append=T, split=T)
# for each element in the list, print
for (list_name in names(abc)) {
cat(list_name,
When i make Boxplots with a lot of boxes, the names of them get only written
down every second "column".
Since they aren't in any way ordered, you don't see anymore to what they belong.
example:
l<-rep(list(1:5),20); boxplot(l,names=sample(20,1:20))
Is there a way to show them all, or do i have
?read.table
Using my normal file path and th data as you supplied it.
x <- read.csv("/home/john/rdata/ages.csv", sep = " ", header = TRUE)
Change the file path to whatever yours is.
Example might be
x <- read.csv("C:/mydata/ages.csv", sep = " ", header = TRUE)
in Windows.
Note that you c
As far as I can see those instructions are not all that helpful.
The answer below, from R-sig-Debian worked very nicely for me a little over a
month ago.
> -Original Message-
> From: marutter@gmail.com
> Sent: Tue, 12 Jun 2012 16:56:16 -0400
> To: jrkrid...@inbox.com
> Subject: Re: [
Code?
Sample data?
John Kane
Kingston ON Canada
> -Original Message-
> From: rshep...@appl-ecosys.com
> Sent: Wed, 18 Jul 2012 14:30:24 -0700 (PDT)
> To: r-help@r-project.org
> Subject: [R] cenbox(): Changing Default x-axis Group Labels
>
>I've looked at the lattice book and the 'R
This should definitely be posted on the r-sig-mixed-models list, not here.
-- Bert
On Thu, Jul 19, 2012 at 6:16 AM, Yolande Tra wrote:
> Hello,
>
> I have the following design, counts were collected at different transects,
> different depths and different sites at different times. Time is conti
Hello,
I have the following design, counts were collected at different transects,
different depths and different sites at different times. Time is continuous
and assumed to be random, all the others are categorical fixed where
transect is nested within depth which is nested within site.
I would
As Duncan Murdoch pointed out, the example data frame that you provided
doesn't give very interesting results (all the shock values are zero), so
I created a different shock variable for illustration. I suggest using
the interp() function in the R package akima.
df <- structure(list(c = 1:6, z
On 12-07-19 2:01 AM, Akhil dua wrote:
Hello every one
can any one tell me how to draw contour with this data set
c zshock
1 0.45450237 0
2 0.02663337 0
3 -2.08444556 0
4 -0.12715275 0
5 0.67066360 0
6 -0.73540081 0
I want to dr
Hello,
There is a mistake. Type
> install.packages("hdf5")
instead of
> installed.packages("hdf5")
in order to install a package.
Best Regards,
Passcal
Le 19/07/2012 17:02, uday a écrit :
Hi,
Recently I have installed R in my Linux operating system , after
installation I was trying to in
Hi Marion,
as stated in the help file ?apply coerces a data.frame object into an
array. Since an array has only one type of data, this coercion turns all
your variables into strings (because this data type can hold all
information given without loss).
If it happens that your data.frame consists on
At 22:27 18/07/2012, Chet Seligman wrote:
This doesn't work, what should I do?
Try reposting on the r-sig-debian list.
It might help to tell them what version of Ubuntu you are using and
to paste in your sources.list
sudo apt-get install r-base
[sudo] password for cseligman:
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