On 21/08/2012 22:46, Marc Schwartz wrote:
On Aug 21, 2012, at 3:39 PM, Bennet Fauber wrote:
As a follow-up to my prior post, if I remove --with-blas
--with-lapack, then the stats test passes:
...
Testing examples for package ‘stats’
comparing ‘stats-Ex.Rout’ to ‘stats-Ex.Rout.save’ ... OK
.
Hi All,
If I want to read some info from a website, which login is required, do you
know how to do that? For example,
html =
"http://www.linkedin.com/search-fe/group_search?pplSearchOrigin=GLHD&keywords=international";
tmp = readLines(html)
Error in file(con, "r") : cannot open the connection
The pooled standard deviation is calculated exactly how you describe it. There
are, however, two differences between your approach and the way escalc()
computes the SMD and corresponding sampling variance:
1) Instead of the equation for approximating the correction factor (J),
escalc() uses the
On Aug 21, 2012, at 3:24 PM, Sapana Lohani wrote:
Hello,
Re framing my previous question, my dataframe (Soil Type) looks
something like below. I want the code to find "/" first and split
the string into 2 or 3 depending on the number of "/". I don't want
to specify the number of column
On Aug 21, 2012, at 10:39 AM, Peter Ehlers wrote:
On 2012-08-21 08:28, Brian Kriegler wrote:
All of these suggestions are great. Let me add one other dimension.
Suppose I want to add curly quotations, italicize, underline, or
bold
some of the words. I can get this to work:
plot(-15:15,1
HI,
Try this:
gsub("^-\\d(\\d{4}.).*","\\1",a)
#[1] "1020."
gsub("^.*(.\\d{5}).","\\1",a)
#[1] ".90920"
A.K.
- Original Message -
From: Dr. Holger van Lishaut
To: "r-help@r-project.org"
Cc:
Sent: Tuesday, August 21, 2012 3:24 PM
Subject: [R] Regular Expressions in grep
Dear r-help m
Hello,
Re framing my previous question, my dataframe (Soil Type) looks something like
below. I want the code to find "/" first and split the string into 2 or 3
depending on the number of "/". I don't want to specify the number of columns
after split. How can I do that ?
Granitic Hills 16-20
Hello,
I recently started using the metafor package (version 1.6-0) in R (2.15.1,
64-bit Windows 7) and noticed that I was getting slightly different values
when I manually calculated the standardized mean difference versus what
escalc was giving me. Here's a very simple example:
escalc(measur
Hello,
Try the following.
library(reshape2)
# Don't need 'cbind'
mat <- data.frame(sp1=c(rep(0, 5), rep(1, 5)),
sp2=sample(c(rep(0, 6),rep(1, 4))),
fac=c(rep("a", 3), rep("b", 3), rep("c", 4)))
mat.new <- matrix (c(0, 1, 1, 1, 1, 1), nrow=2, ncol=3)
rownames(mat.new) <- c("sp1
On Aug 21, 2012, at 3:39 PM, Bennet Fauber wrote:
> As a follow-up to my prior post, if I remove --with-blas
> --with-lapack, then the stats test passes:
>
> ...
> Testing examples for package ‘stats’
> comparing ‘stats-Ex.Rout’ to ‘stats-Ex.Rout.save’ ... OK
> ...
>
> Perhaps this is now a qu
Dear R users and developers,
I have a dataset containing 34 variables measured in a survey, which has
some missing items. I would like to conduct a factor analysis of this
data. I tested mi, Amelia, and MissForest as alternative packages in
order to impute the missing data. I now have 5 separa
As a follow-up to my prior post, if I remove --with-blas
--with-lapack, then the stats test passes:
...
Testing examples for package ‘stats’
comparing ‘stats-Ex.Rout’ to ‘stats-Ex.Rout.save’ ... OK
...
Perhaps this is now a question about building R with the Intel MKL
libraries instead of one a
Dear list,
I want to update my prabclus package which I haven't done for quite a
while.
In the previous version, I had .dat files in my data subdirectory, which I
read using .R files. Now R CMD check gives me a warning that .dat files
are no longer accepted there.
So I changed my filenames
Hello,
Any one who can help me with the problems regrarding segmented
regression.
1.When I run the code as following:
library(segmented)
n <-100
S<-200
x1 <- seq(-0.99,0.99,length=29)
MSEPtrm<-matrix(0,S,1)
MAEtrm<-matrix(0,S,1)
TRM<-matrix(0,S,length(x1))
for (s in seq(1:S)){
set.seed(123*s
On Aug 21, 2012, at 3:40 AM, Nanthinee Jeevanandam wrote:
I have run this model for my data set on insect survival time for
different temperature treatments in R. I understand the Wald statistic
reports on whether the variable
makes a difference to the model but there is also a p-value so i am
Dear list,
I'm trying to create a matrix by combining the sites that species occur in
a new matrix with species as rows and sites as columns.
The main matrix is "mat":
mat <- as.data.frame(cbind(sp1=c(rep(0, 5), rep(1, 5)),sp2=sample(c(rep(0,
6),rep(1, 4))),
fac=c(rep("a", 3), rep("b", 3),rep("c
I am installing R 2.15.1 onto RHEL 6, using gcc 4.7.0 with Intel MKL
10.3.7 and the following environment:
export BLAS_LIBS="-Wl,--start-group
/usr/caen/intel-12.1/mkl/lib/intel64/libmkl_gf_lp64.a
/usr/caen/intel-12.1/mkl/lib/intel64/libmkl_sequential.a
/usr/caen/intel-12.1/mkl/lib/intel64/libmkl_
This is a LaTeX question. I'll be surprised if \caption{} allows
arbitrary LaTeX environments inside it.
Perhaps it is irrelevant, but you can make it more compact with the
fig.cap option in knitr; see an example at
https://github.com/yihui/knitr-examples/blob/master/019-fig-caption.Rnw
\document
You're misreading the docs: from grep,
value: if ‘FALSE’, a vector containing the (‘integer’) indices of
the matches determined by ‘grep’ is returned, and if ‘TRUE’,
a vector containing the matching elements themselves is
returned.
Since there's a match somewhere
'grep' does not change strings. Use 'gsub' or 'regmatches':
# gsub
Front <- gsub("^.*?([1-9][0-9]*\\.).*?$", "\\1", a)
End <- gsub("^.*?(\\.[0-9]*[1-9]).*?$", "\\1", a)
# regexpr and regmatches (R >= 2.14.0)
Front <- regmatches(a, regexpr("[1-9][0-9]*\\.", a))
End <- regmatches(a, regexpr("\\.[0-9
grep() returns the matches. You want regexpr() and regmatches()
-- Bert
On Tue, Aug 21, 2012 at 12:24 PM, Dr. Holger van Lishaut
wrote:
> Dear r-help members,
>
> I have a number in the form of a string, say:
>
> a<-"-01020.909200"
>
> I'd like to extract "1020." as well as ".9092"
>
> Front<-gr
On Aug 21, 2012, at 2:03 PM, Alexander Shenkin wrote:
> Hi Folks,
>
> I'm surprised, but I didn't find this question addressed anywhere. I'd
> like to generate a LaTeX caption with R code. I've tried the code
> below, but I get the following TeX error:
>
>! Argument of \@caption has an ex
Dear r-help members,
I have a number in the form of a string, say:
a<-"-01020.909200"
I'd like to extract "1020." as well as ".9092"
Front<-grep(pattern="[1-9]+[0-9]*\\.", value=TRUE, x=a, fixed=FALSE)
End<-grep(pattern="\\.[0-9]*[1-9]+", value=TRUE, x=a, fixed=FALSE)
However, both strings gi
Thanks a lot, it works very well!
Inti
2012/8/21 Rui Barradas
> Hello,
>
> Try the following
>
>
> d <- read.table(text="
>
> id VI
> 1 -10
> 2 -4
> 3 5
> 4 -2
> 5 -5
> 6 -3
> 7 -2
> 8 -1
> 9 4
> 10 8
> ", header = TRUE)
>
>
> fun <- function(n, thres){
> r <- rle(!d$VI < thres)
> in
Hi Folks,
I'm surprised, but I didn't find this question addressed anywhere. I'd
like to generate a LaTeX caption with R code. I've tried the code
below, but I get the following TeX error:
! Argument of \@caption has an extra }.
\par
l.21 }
Any thoughts? Perhaps I'll have to
Hi,
In the example below, I am modeling the dependent variable Y as a function
of X when the errors are autoregressive at the first lag. There is a number
of ways/functions with which to model this. arima (tseries), gls (nlme), and
gamm should produce similar results in the simulated example belo
On 2012-08-21 11:02, m p wrote:
Hello,
I have a problem writing a variable to an existing file.
Below is a part of my script and how it fails.
I can't find "create.var.ncdf" in help
You probably want 'create.ncdf'.
Peter Ehlers
Thanks for any help.
Mark
nc <- open.ncdf(ncname, readunlim=FAL
Rui:
It's much simpler than you propose, which is why I left it to the OP.
Just use the results of rle to create a logical vector to index id.
For example, the solution for the OP's example becomes : (d is the
data frame containing id and VI)
> with(d,{
+ z <- rle(VI < 1)
+ vals <- with(z,values
Hello,
Try the following
d <- read.table(text="
id VI
1 -10
2 -4
3 5
4 -2
5 -5
6 -3
7 -2
8 -1
9 4
10 8
", header = TRUE)
fun <- function(n, thres){
r <- rle(!d$VI < thres)
inx <- which(!r$values & r$lengths >= n)
csum <- cumsum(r$lengths)
from <- ifelse(inx == 1, 1, csum[in
z <- rle(VI<1)
gets you most of the way there. ?rle will tell you what z is so that
you can then use it to go the rest of the way.
-- Bert
On Tue, Aug 21, 2012 at 10:54 AM, inti luna wrote:
> Hello,
>
> I have 2 variable: one is an "id" sequence from 1:1000 and the other is
> variable with real
Hello,
I have a problem writing a variable to an existing file.
Below is a part of my script and how it fails.
I can't find "create.var.ncdf" in help
Thanks for any help.
Mark
nc <- open.ncdf(ncname, readunlim=FALSE, write=TRUE )
missing <- 1.e+30
xdim <- nc$dim[["west_east"]]
ydim <- nc$dim[["s
Hello,
I have 2 variable: one is an "id" sequence from 1:1000 and the other is
variable with real values "VI" from -15.0 to 20.0 and I want to detect id
values that have indicator values less than a certain threshold, for
example (x=1) BUT that are in sequence equal or longer than 5.
For instance
On 2012-08-21 08:28, Brian Kriegler wrote:
All of these suggestions are great. Let me add one other dimension.
Suppose I want to add curly quotations, italicize, underline, or bold
some of the words. I can get this to work:
plot(-15:15,15:-15, pch=NA)
text(x=0, y=5,
expression(bold(This)~is~
On 12-08-19 3:47 PM, Marius Hofert wrote:
Dear Duncan,
I recently asked a question concerning patchDVI on r-help, see
,
| https://stat.ethz.ch/pipermail/r-help/2012-August/321780.html
`
Unfortunately, no one could help. I was wondering if you know a solution to the
above problem. Any h
... or using the handy with()
with(x,paste(n,l,sep="_"))
Slightly greater clarity and robustness, perhaps...
-- Bert
On Tue, Aug 21, 2012 at 9:24 AM, R. Michael Weylandt
wrote:
> You don't need loops or apply with paste since it's vectorized:
>
> x <- data.frame(n = 1:5, l = letters[1:5], stri
You don't need loops or apply with paste since it's vectorized:
x <- data.frame(n = 1:5, l = letters[1:5], stringsAsFactors = FALSE)
paste(x[,1], x[,2], sep = "_")
Cheers,
Michael
On Tue, Aug 21, 2012 at 10:57 AM, Chet Seligman wrote:
> This works, where zz is a dataframe:
>
> for(i in 1:nrow(
What is wrong with utilizing the vectorized nature of paste?
zz[[1]] <- paste(zz[[1]],zz[[2]],sep="_")
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Liv
Sorry - it is working as suggested by Joshua.
Thanks a lot and sorry for the horrible confusion and examples,
Rainer
On 21/08/12 18:12, Rainer M Krug wrote:
On 21/08/12 17:54, Joshua Wiley wrote:
What is wrong with what I suggested initially?
subst <- expression(trt > 0)
lm(weight ~ group,
Josh's solution should be fine.
just one more note,
trt>0 may not work as intended since trt is a factor in your example.
Just check
subset(dat,trt>0)
wich is just 'dat'.
Am 21.08.2012 17:41, schrieb Rainer M Krug:
> On 21/08/12 17:35, Eik Vettorazzi wrote:
>> Hi Rainer,
>> I got an error while
On 21/08/12 17:54, Joshua Wiley wrote:
What is wrong with what I suggested initially?
subst <- expression(trt > 0)
lm(weight ~ group, data=dat, subset=eval(subst))
That it does not work?
ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89
This works, where zz is a dataframe:
for(i in 1:nrow(zz)) {
zzz[i,1]<-paste(zz[i,1],zz[i,2],sep="_")
}
I would like to use "apply" to concatentate two columns of text along with
a separator.
How?
Chet
[[alternative HTML version deleted]]
__
What is wrong with what I suggested initially?
subst <- expression(trt > 0)
lm(weight ~ group, data=dat, subset=eval(subst))
??
On Tue, Aug 21, 2012 at 8:41 AM, Rainer M Krug wrote:
> On 21/08/12 17:35, Eik Vettorazzi wrote:
>>
>> Hi Rainer,
>> I got an error while replicating your data.frame
That simple code works for me as well, but for some reason my actual code
still isn't working. Here is exactly what I'm doing:
- The code for importing my data is:
/Bridges = read.csv(file = "C:/R/Bridges.RData.csv", head=TRUE, sep=",")/
- Within my group of highway bridges, I have defined a su
Did you download the zip file to your computer? Next install from zip
file.
On Aug 21, 2012 9:06 AM, wrote:
> Hy everyone,
>
> I am new in R so I will pose a simple question that I cannot find out
> reading tutorials.
> I would like to install the R-package "FunctSNP" by file.zip because it´s
>
On 21/08/12 17:35, Eik Vettorazzi wrote:
Hi Rainer,
I got an error while replicating your data.frame construction.
But this worked for me
ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
dat <- data.frame( group = gl(2,10,20,
Wow, sometimes writing it all out does really help to see the issue. I now
see that I need to run the "log" command before splitting my data into a
subset, or it won't see the newly created variable.
My code now runs without issue. Thanks for everyone who took the time to
read this and/or help. I
- Messaggio inoltrato da ltro...@mappi.helsinki.fi -
Data: Tue, 21 Aug 2012 17:13:34 +0300
Da: ltro...@mappi.helsinki.fi
Oggetto: Re: [R] R-package "FunctSNP"
A: "Steve Friedman"
Cc: r-help@r-project.org
I did it, but It can´t find the library.
Def. Quota "Steve
All of these suggestions are great. Let me add one other dimension. Suppose I
want to add curly quotations, italicize, underline, or bold some of the words.
I can get this to work:
plot(-15:15,15:-15, pch=NA)
text(x=0, y=5,
expression(bold(This)~is~a~underline(test)^1~of~the~italic(
Hi Rainer,
I got an error while replicating your data.frame construction.
But this worked for me
ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
dat <- data.frame( group = gl(2,10,20, labels=c("Ctl","Trt")),
weight = c
On 21/08/12 16:57, Bert Gunter wrote:
?? I do not groc what you mean. ... subset == subs would work fine in
your lm call. So unless someone else does get it, you may need to
elaborate.
OK - here is an example:
dat <- data.frame(
ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.1
Hi,
I am running a large number of jobs (thousands) in parallel (linux OS
64bit), R version 2.14.1 (2011-12-22), Platform: x86_64-redhat-linux-gnu
(64-bit). Up to yesterday everything ran fine with jobs in several
blocks (block1, block2 etc) of submission. They are sent to an LSF
platform to
?? I do not groc what you mean. ... subset == subs would work fine in
your lm call. So unless someone else does get it, you may need to
elaborate.
In general, ?substitute, ?bquote, and ?quote are useful to avoid
immediate evaluation of calls, but I don't know if that's relevant to
what you want he
Hi Rainer,
You could try:
subs <- expression(dead==FALSE & recTreat==FALSE)
lme(formula, subset = eval(subs))
Not tested, but something along those lines should work.
Cheers,
Josh
On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug wrote:
> Hi
>
> I want to do a series of linear models, and woul
Hi
I want to do a series of linear models, and would like to define the input arguments for lm() as
variables. I managed easily to define the formula arguments in a variable, but I also would like to
have the "subset" in a variable. My reasoning is, that I have the subset in the results object.
I've no idea why, but both seemed to work perfectly. I'll work through
the help files to see if I can figure it out.
Thank you!
Thomas
On 21 Aug 2012, at 14:03, Noia Raindrops wrote:
Try this:
table(nDeaths = rep(0:4, c(109, 65, 22, 3, 1)))
## or
as.table(array(c(109L, 65L, 22L, 3L, 1L), d
Thanks
I new that gsub shall be involved somehow, but the key is to use backreference
"-\\1".
Regards
Petr
> -Original Message-
> From: arun [mailto:smartpink...@yahoo.com]
> Sent: Tuesday, August 21, 2012 2:51 PM
> To: PIKAL Petr
> Cc: R help
> Subject: Re: [R] reexpr transform nonume
Hi: I'm not sure how to do 1, but I also tried strwrap() and that worked OK.
Although it's not pretty. But it'll do.
Simon
On 2012-08-20, at 5:52 PM, Yihui Xie wrote:
> Two possible solutions:
>
> 1. Redefine the LaTeX environment so it allows wrapping (see listings
> for example);
> 2. Manually
This is somewhat technical and I'd advise you post to R-Sig-Finance and cc the
package maintainer, Dr Pfaff, directly. Use maintainer("urca") to get his
email.
Cheers,
Michael
On Aug 21, 2012, at 4:48 AM, Rmillan wrote:
> Hi all R users,
>
> I'm trying to replicate the same results that are
Try this:
table(nDeaths = rep(0:4, c(109, 65, 22, 3, 1)))
## or
as.table(array(c(109L, 65L, 22L, 3L, 1L), dimnames = list(nDeaths = 0:4)))
--
Noia Raindrops
noia.raindr...@gmail.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/lis
Hi,
Slight modification to my earlier code to make it more simple.
dat1 <- readLines(textConnection("Granitic Hills 16-20 PZ
Loamy Upland 16-20 PZ
Sandy Loam Upland 12-16 PZ / Sandy Loam, Deep 12-16 PZ
Loamy Upland 12-16 PZ / Sandy Loam Upland 12-16 PZ
Loamy Upland 16-20 PZ"))
dat3<-strspli
Hy everyone,
I am new in R so I will pose a simple question that I cannot find out
reading tutorials.
I would like to install the R-package "FunctSNP" by file.zip because
it´s in the archive
(http://ftp.sunet.se/pub/lang/CRAN/src/contrib/Archive/FunctSNP/)
but It doesn´t work, I think some
HI,
Try this:
as.numeric(gsub("^([0-9].[0-9]{1,4})-$","-\\1",temp))
#or
as.numeric(gsub("^([[:digit:]].[[:digit:]]{1,4})-$","-\\1",temp))
#or
as.numeric(gsub("^([[:digit:]]..*{1,4})-$","-\\1",temp))
#or
as.numeric(gsub("^(.*.[[:digit:]]{1,4})-$","-\\1",temp))
A.K.
- Original Message -
F
Hi,
Try this:
dat1 <- readLines(textConnection("Granitic Hills 16-20 PZ
Loamy Upland 16-20 PZ
Sandy Loam Upland 12-16 PZ / Sandy Loam, Deep 12-16 PZ
Loamy Upland 12-16 PZ / Sandy Loam Upland 12-16 PZ
Loamy Upland 16-20 PZ"))
dat3<-strsplit(dat1,"/")
data.frame(col1=do.call(rbind,lapply(dat3,
Hi all R users,
I'm trying to replicate the same results that are given in a published
article after been granted
the same data that the authors use.
I'm having problems to determine the cointegration rank of my data set using
the Johnasen's trace test.
This trace test is already programmed in the
I have run this model for my data set on insect survival time for
different temperature treatments in R. I understand the Wald statistic
reports on whether the variable
makes a difference to the model but there is also a p-value so i am
not sure whther to report the Wald statistic. I have pasted a
Dear all,
I am trying to iterate an elastic net regression method on a matrix
1) Each column of the matrix ( nxp) will act as a response (y) and rest of
the variables (columns) which means p-1 will act as predictor set
2) i want to store selected variables as matrix
I wrote a code. Please help
Hi Nooshin,
Have you tried the RColorBrewer package?
After loading the package, the following gives you the whole array of available
colour palletes.
> display.brewer.all(n=NULL, type="all", select=NULL, exact.n=TRUE)
Then, you can add to your code something like this (I'm using palette "Set3"
I'm trying to enter a frequency table manually so that I can run a
goodness of fit test (I only have the frequencies, I don't have the
raw data).
So for example, let's say I want to re-create the HorseKicks table:
library(vcd)
data(HorseKicks)
str(HorseKicks)
'table' int [1:5(1d)] 109 65
Mitra,
Try this ...
# provide the data you want to plot, and the desired range
mydat <- c(3.34, 3.28, 1.37, 1.12, 3.52, 4.07, 3.66)
myrange <- c(0, 5)
# create a data frame with the max and min as the first two rows
mydf <- data.frame(rbind(max=myrange[2], min=myrange[1], mydat))
# create a rad
cbe...@tajo.ucsd.edu escreveu:
>"This rather unlikely linguistic cocktail would probably never have
>been prepared by computer scientists, yet the language has become
>surprisingly popular."
A linguist that is a good friend once observed that language is both expression
and a modeller of how we
Thank you
I knew that somebody can use power of gsub better than myself.
Petr
> -Original Message-
> From: Noia Raindrops [mailto:noia.raindr...@gmail.com]
> Sent: Tuesday, August 21, 2012 12:33 PM
> To: PIKAL Petr
> Cc: r-help
> Subject: Re: [R] reexpr transform nonumeric values to nume
Use backreference:
as.numeric(gsub("^(.*)-$", "-\\1", as.character(temp)))
--
Noia Raindrops
noia.raindr...@gmail.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-proje
How about this?
read.table(text = "Granitic Hills 16-20 PZ
Loamy Upland 16-20 PZ
Sandy Loam Upland 12-16 PZ / Sandy Loam, Deep 12-16 PZ
Loamy Upland 12-16 PZ / Sandy Loam Upland 12-16 PZ
Loamy Upland 16-20 PZ", header = FALSE, sep = "/", fill = TRUE, na.strings =
"", strip.white = TRUE)
##
Dear all
I have got this kind of data
temp <- structure(c(1L, 1L, 1L, 1L, 1L, 16L, 6L, 6L, 16L, 16L, 16L, 16L,
16L, 16L, 16L, 16L, 16L, 16L, 16L, 16L, 16L, 11L, 10L, 16L, 16L,
16L, 21L, 16L, 16L, 16L, 1L, 1L, 1L, 16L, 16L, 16L, 16L, 1L,
1L, 7L, 3L, 1L, 1L, 1L, 1L, 3L, 3L, 1L, 1L, 1L, 4L, 10L,
Hi Simon
try strwrap(test)
regards,
Enrico
Am 20.08.2012 23:03, schrieb Simon Kiss:
Hi there: I'm preparing a report in RStudio 0.96.330 on a Mac OS. I'm running R
2.15.0
I understand from Ross Ihaka's document
(http://www.stat.auckland.ac.nz/~stat782/downloads/Sweave-customisation.pdf)
t
try this:
> x <- scan(text = "Granitic Hills 16-20 PZ
+ Loamy Upland 16-20 PZ
+ Sandy Loam Upland 12-16 PZ / Sandy Loam, Deep 12-16 PZ
+ Loamy Upland 12-16 PZ / Sandy Loam Upland 12-16 PZ
+ Loamy Upland 16-20 PZ", what = '', sep = '\n')
Read 5 items
> x
[1] "Granitic Hills 16-20 PZ"
[2] "Loamy Up
filled.contour(Nooshin, main="Permeability",color.palette=rainbow)
filled.contour(Nooshin, main="Permeability",color.palette=terrain.colors)
filled.contour(Nooshin, main="Permeability",color.palette=topo.colors)
not quite sure what to do with the heat map, but it has a "col" attribute.
On 21.08.2
>
> > Hello guys,
> >
> > I want to increase contrast of colors in my matrix plot.
>
> Nooshin<-
as.matrix(matrix(sample(c(0.02,0.01,0.003,0.001,0.0001,0.0005,0.0004,0.00013),2401,replace=TRUE),ncol=49),
row.names=1, header=TRUE)
> filled.contour(Nooshin, main="Permeability")
> Nooshin_matrix
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of David A. Kim
> Sent: Monday, August 20, 2012 9:17 PM
> To: r-help@r-project.org
> Subject: [R] Iterative sampling with restrictions
>
> Hi all,
>
> I'm working on a seemingly tri
Dear Peter,
Your explanation makes perfect sense, thank you.
With regards,
Tal
Contact
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Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
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Thank you, Michael and Rui. I think I will adjust namespaces file.
Cheers,
Eva
--- El mar, 21/8/12, R. Michael Weylandt escribió:
De: R. Michael Weylandt
Asunto: Re: [R] Doubt in relation with packaging
Para: "Rui Barradas"
CC: "Eva Prieto Castro" , "r-help"
Fecha: martes, 21 de agosto, 2012
Hi
why not shuffle your values by
x <- 1:60
x <- sample(x, 60)
and then add groups
groups <- rep(paste("g",1:3, sep=""), each=20)
If you want you can suffle also groups.
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On
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