Hi,
I have been trying to build R with optimized BLAS library.
I am using a Ubuntu 13.10 x86_64 desktop, on which I am able to build R
with openblas without any problem:
#BEGIN_SRC sh
./configure --enable-BLAS-shlib --enable-R-shlib LIBnn=lib --disable-nls
--with-blas=-L/usr/lib/openblas-base/
There is no R code following
On 01 Nov 2013, at 05:34, Li Bowen bowenl...@gmail.com wrote:
Hi,
I have been trying to build R with optimized BLAS library.
I am using a Ubuntu 13.10 x86_64 desktop, on which I am able to build R
with openblas without any problem:
#BEGIN_SRC sh
Hi,
I have a data frame with one column and several rows of the form.
Peak Usage: init:2359296, used:15859328, committed:15892480,
max:50331648Current Usage : init:2359296, used:15857920,
committed:15892480, max:50331648|---|
I tested the regex
Current.*?[\|]
I have data that looks like this:
Friend1, Friend2
A, B
A, C
B, A
C, D
And I'd like to generate some more rows and another column. In the new
column I'd like to add a 1 beside all the existing rows. That bit's
easy enough.
Then I'd like to add rows for all the possible directed
Hi
Another option is sapply/split/sum construction
with(data, sapply(split(x, ID), function(x) sum(x==0)))
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Carlos Nasher
Sent: Thursday, October 31, 2013 6:46 PM
You could use the data.table package
require(data.table)
DT - data.table(Friend1 = sample(LETTERS, 10, replace = TRUE), Friend2 =
sample(LETTERS, 10, replace = TRUE), Indicator = 1)
ALL - data.table(unique(expand.grid(DT)))
setkey(ALL)
OTHERS - ALL[!DT]
OTHERS[, Indicator := 0]
RESULT -
Hi Thomas,
It depends whether you'd like to include all levels of each column in every
column. For including all values you could try something like this:
isAllDifferent - function(z) !any(duplicated(z))
myData - data.frame(Friend1=c(a, a, b, c), Friend2=c(b, c, a,
d),
You can override the legend aesthetics, e.g.,
ggplot(df,aes(x=Importance,y=Performance,fill=PBF,size=gapsize))+
geom_point(shape=21,colour=black)+
scale_size_area(max_size=pointsizefactor) +
scale_fill_discrete(guide = guide_legend(override.aes = list(size = 4)))
Best,
Ista
On Thu,
Thanks, Bill Duncan. Actually I tried values which are inside the defined
region. please find below the extracted script
xnew-rlnorm(seq(0,400,1), meanlog=9.7280055, sdlog=2.0443945)
f - ecdf(xnew)
y - f(x)
y1-f(200)## finding y for a given xnew value of
200
try this:
x - rbind(Peak Usage: init:2359296, used:15859328,
committed:15892480,max:50331648Current Usage : init:2359296,
used:15857920,committed:15892480, max:50331648|---|)
apply(x, 1, function(a) sub((Current.*?[/|]).*, \\1, a))
[1] Peak Usage: init:2359296,
Thanks.
I converted my data structure( that is most of the confusion in my case )
into a data frame and then applied this function
y - apply( y, 1, function(z) str_extract(z,Current.*?[/|]))
to get
Current Usage : init:2359296, used:15857920, committed:15892480,
max:50331648|
Mohan
From:
Thank you very much, guys - it worked beautifully.
On Thu, Oct 31, 2013 at 7:55 AM, John Kane jrkrid...@inbox.com wrote:
At a guess, don't use colour.
John Kane
Kingston ON Canada
-Original Message-
From: dimitri.liakhovit...@gmail.com
Sent: Wed, 30 Oct 2013 14:11:37 -0400
below is a code to compute hessian matrix , which i need to generate 29 number
of different matrices for example first element in x1 and x2 is use to generate
let say matrix (M1) and second element in x1 and x2 give matrix (M2) upto
matrix (M29) corresponding to the total number of
I have R version 2.15.3 When i try to load it:
library (tawny)
i receive this response:
package ‘parser’ could not be loaded
The package Parser in not on Cran anymore, it seems a dead project!
http://cran.r-project.org/web/packages/parser/index.html
If i try to manual install
Hello,
I´m using function aggregate in R 3.0.2. If I run the instruction
x-aggregate(cbind(mpg,hp)~cyl+gear,data=mtcars,quantile) I get the
result the following data.frame:
cyl
gear
mpg.0%
mpg.25%
mpg.50%
mpg.75%
mpg.100%
hp.0%
hp.25%
hp.50%
hp.75%
hp.100%
4
3
21.5
21.5
Hi,
Try this:
Lines1 - readLines(textConnection(Peak Usage : init:2359296,
used:15859328, committed:15892480,max:50331648Current Usage : init:2359296
used:15857920,committed:15892480,max:50331648|---|
Peak Usage : init:2359296, used:15859328,
I think this gives a different result than the one OP asked for:
df1 - structure(list(ID = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L,
2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), x = c(1, 0,
0, 1, 0, 0, 0, 1, 2, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0)), .Names = c(ID,
x), row.names = c(NA,
Lorenzo,
I may be able to help you get started. You can use the XML package to grab
the information off the internet.
library(XML)
mylines - readLines(url(http://bit.ly/1coCohq;))
closeAllConnections()
mylist - readHTMLTable(mylines, asText=TRUE)
mytable - mylist1$xTable
However, when I look
Daniel,
You can see better what is going on if you look at
as.list(x)
There you can see that cyl and gear are vectors but mpg and hp are matrices.
You can rearrange them using the do.call() function
x2 - do.call(cbind, x)
dim(x2)
Jean
On Fri, Nov 1, 2013 at 7:08 AM, Daniel Fernandes
Pretty much what the subject says:
I used an env as the basis for a Hashtable in R, based on information
that this is in fact the way environments are implemented under the hood.
I've been experimenting with doubling the number of entries, and so far
it has seemed to be scaling more or less
Hi,
Try:
do.call(data.frame,c(x,check.names=FALSE))
A.K.
Hello,
I´m using function aggregate in R 3.0.2. If I run the instruction
x-aggregate(cbind(mpg,hp)~cyl+gear,data=mtcars,quantile) I get the
result the following data.frame:
cyl
gear
mpg.0%
mpg.25%
mpg.50%
mpg.75%
It would be nice if you followed the posting guidelines and at least
showed the script that was creating your entries now so that we
understand the problem you are trying to solve. A bit more
explanation of why you want this would be useful. This gets to the
second part of my tag line: Tell me
The release version of tawny has no such dependency and builds just fine on
CRAN. Try updating that instead.
Michael
On Nov 1, 2013, at 7:10, Tstudent tstud...@gmail.com wrote:
I have R version 2.15.3 When i try to load it:
library (tawny)
i receive this response:
package
Install a recent version of tawny that does not depend on the other package?
Best,
Uwe Ligges
On 01.11.2013 12:10, Tstudent wrote:
I have R version 2.15.3 When i try to load it:
library (tawny)
i receive this response:
package ‘parser’ could not be loaded
The package Parser in not on
You could also try:
library(plyr)
newdf - function(.data, ...) {
eval(substitute(data.frame(...)), .data, parent.frame())
}
x1 - ddply(mtcars,.(cyl,gear), newdf, mgp=t(quantile(mpg)),hp=t(quantile(hp)))
#(found in one of the google group discussions)
#or
library(data.table)
dt1 -
All,
I've used the excellent package, spsurvey, to create spatially balanced samples
many times in the past. I'm now attempting to use the analysis portion of the
package, which compares CDFs among sub-populations to test for differences in
sub-population metrics.
- My data (count data) have
You are not using the inv_ecdf function that Rui sent. His was
inv_ecdf_orig -
function (f)
{
x - environment(f)$x
y - environment(f)$y
approxfun(y, x)
}
(There is no 'xnew' in the environment of f.)
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
From:
I am building a horizontal bar plot using ggplot2 - see the code below.
A couple of questions:
1. On the right side of the graph the value labels are inside the bars. How
could I move them to be outside the bars - the way they are on the left
side?
2. How can I make sure that the scale on my X
Sure,
I was attempting to be concise and boiling it down to what I saw as the
root issue, but you are right, I could have taken it a step further. So
here goes.
I have a set of around around 20M string pairs. A given string (say, A)
can either be equivalent to another string (B) or not. If
Good day all.
I am hoping you can help me (and I did this right). I've been working in
R for a week now, and have encountered a problem with forecast.lm().
I have a list of 12 variables, all type = double, with 15 data entries.
(I imported them from tab delimited text files, and then
Uwe Ligges ligges at statistik.tu-dortmund.de writes:
Install a recent version of tawny that does not depend on the other package?
The most recent version is this:
http://cran.r-project.org/web/packages/tawny/index.html
I can install, but can't load without parser package.
It seems true
Thanks a lot Achim!
This helped a lot. I do not have exactly what I want yet, but I now have
promising ideas to gather my data and find what I'm looking for (especially
as.numeric(x,
units = hours)).
Regards,
Sartene Bel
Message du 31/10/13 à 08h48
De : Achim Zeileis
A :
(Inline)
On Fri, Nov 1, 2013 at 7:33 AM, Tstudent tstud...@gmail.com wrote:
Uwe Ligges ligges at statistik.tu-dortmund.de writes:
Install a recent version of tawny that does not depend on the other package?
The most recent version is this:
Hi
Yes you are right. This gives number of zeroes not max number of consecutive
zeroes.
Regards
Petr
-Original Message-
From: arun [mailto:smartpink...@yahoo.com]
Sent: Friday, November 01, 2013 2:17 PM
To: R help
Cc: PIKAL Petr; Carlos Nasher
Subject: Re: [R] Count number of
Hi Duncan,
Thanks for that template. Not quite the solution I was hoping for, but
that works!
Richard
On Thu, Oct 31, 2013 at 3:47 PM, Duncan Mackay dulca...@bigpond.com wrote:
Hi Richard
If you cannot get a better suggestion this example from Deepayan Sarkar may
help.
It is way back in
Have you looked into the 'igraph' package?
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
Behalf
Of Magnus Thor Torfason
Sent: Friday, November 01, 2013 8:23 AM
To:
Dear R users,
I wonder how I can use R to identify the max value of each row, the column
number column name:
For example:
a - data.frame(x = rnorm(4), y = rnorm(4), z = rnorm(4))
a
x y z
1 -0.7289964 0.2194702 -2.4674780
2 1.0889353 0.3167629 -0.9208548
3
On Nov 1, 2013, at 6:50 AM, Ryan wrote:
Good day all.
I am hoping you can help me (and I did this right). I've been working in R
for a week now, and have encountered a problem with forecast.lm().
I have a list of 12 variables, all type = double, with 15 data entries.
(I imported them
?which.max should start you down the right path
Clint BowmanINTERNET: cl...@ecy.wa.gov
Air Quality Modeler INTERNET: cl...@math.utah.edu
Department of Ecology VOICE: (360) 407-6815
PO Box 47600FAX:
Hi,
Try:
cbind(a,do.call(rbind,apply(a,1,function(x) {data.frame(max=max(x),
max.col.num=which.max(x),
max.col.name=names(a)[which.max(x)],stringsAsFactors=FALSE)}))) ##assuming that
unique max for each row.
A.K.
On Friday, November 1, 2013 1:05 PM, Gary Dong pdxgary...@gmail.com wrote:
Yeah, now it works. Thanks a lot, William and everyone who helped me. This
forum is really helpful for beginners like me. :)
Mano.
On Fri, Nov 1, 2013 at 3:54 PM, William Dunlap wdun...@tibco.com wrote:
You are not using the inv_ecdf function that Rui sent. His was
inv_ecdf_orig
On Nov 1, 2013, at 10:03 AM, Gary Dong wrote:
Dear R users,
I wonder how I can use R to identify the max value of each row, the column
number column name:
For example:
a - data.frame(x = rnorm(4), y = rnorm(4), z = rnorm(4))
a
x y z
1 -0.7289964
Dear R users,
I wonder if R has a default function that I can use to do extraction of
roots.
Here is an example:
X N
2.5 5
3.4 7
8.9 9
6.4 1
2.1 0
1.1 2
I want to calculate Y = root(X)^N, where N
If you just want the nth root of X, use X^(1/n)
x - 256
x^(1/8)
[1] 2
x - -256
x^(1/8)
[1] NaN
It appears that you get the positive real root.
Is this all you wanted?
On 1-Nov-13, at 11:11 AM, Gary Dong wrote:
Dear R users,
I wonder if R has a default function that I can use to do
If you want complex roots, there is a post by Ravi Varadhan from
2010, a reprint of which I found quickly by a google search at
http://r.789695.n4.nabble.com/finding-complex-roots-in-R-td2541514.html
On 1-Nov-13, at 11:20 AM, Don McKenzie wrote:
If you just want the nth root of X, use
I use the spsurvey package a decent amount. The cont.cdftest function bins the
cdf in order to perform the test which I think is the root of the problem.
Unfortunately, the default is 3 which is the minimum number of bins.
I would contact Tom Kincaid or Tony Olsen at NHEERL WED directly to
Jason,
Thank you for your reply. Interesting ... so you think the 'classes' in the
error message The combined number of values in at least one class... is
referring to the CDF bins rather than the sub-population classes that I
defined.
That makes sense as I only defined two classes (!). I
Claudia,
I have not worked through the example myself. Since you seem to be getting
errors, perhaps a different example would help. Here are some more
choropleth maps (although these use US states rather than European
countries).
... but you may be interested in this:
http://andywoodruff.com/blog/why-are-choropleth-mercator-maps-bad-because-we-said-so/
Cheers,
Bert
On Fri, Nov 1, 2013 at 12:18 PM, Adams, Jean jvad...@usgs.gov wrote:
Claudia,
I have not worked through the example myself. Since you seem to be getting
A sample of my data looks like this.
Header: Time Sender Receiver
11 2
11 3
22 1
22 1
31 2
31 2
There are 3 time periods (sessions) and the edgelists between nodes.
Dear all,
We are conducting a study in with a set of covariates and a time to event
outcome.
Covariates b1 and b3 violate proportionality. We applied a coxph with a tt
term to evaluate the nature of time dependence.
Call:
coxph(formula = Surv(start - 1, stop, outcome) ~ tt(b1) +
b5
Hi Jean,
thanks again for your response. As I told you I did the downloads and
double checked if I selected the right directory.
But I noticed right now what happend:
The command in the example is :
eurMap - readShapePoly(fn=NUTS_2010_60M_SH/Shape/data/NUTS_RG_60M_2010)
But it should be :
Hi Jim,
that works nice.
Thanks again!
Have a nice weekend, best regards
Claudia
Zitat von Jim Lemon j...@bitwrit.com.au:
On 10/31/2013 03:04 AM, palad...@trustindata.de wrote:
Hi Jim,
thats the second time that you helped me in a short while so thanks a lot!
But it seems to me quite
Hello,
Dr. Simon Wood told me how to force a cubic spline passing through a
point. The code is as following. Anyone who knows how I can change the code
to force the first derivative to be certain value. For example, the first
derivative of the constrained cubic spline equals 2 at point (0,
I have no specific expertise here, but I just wanted to point out that
this sounds like a losing strategy long term: As new packages and
newer versions of packages come out that fix bugs and add features,
you'll be unable to use them because you'll be stuck with 2.15.3 . I
suggest you bite
Hi Jean,
nevertheless this page R-bloggers looks realy interesting so I'll
work through the tutorial.
Thanks again for recommanding this web-site.
Best regards
Claudia
Zitat von Adams, Jean jvad...@usgs.gov:
Claudia,
I have not worked through the example myself. Since you seem to be
On Nov 1, 2013, at 11:16 AM, cesar garcia perez de leon wrote:
Dear all,
We are conducting a study in with a set of covariates and a time to event
outcome.
Covariates b1 and b3 violate proportionality.
Can you describe the basis for that statement?
We applied a coxph with a „tt‰ term
Hi,
Check whether this works:
vec1 - c( 'eric', 'JOHN', 'eric', 'JOHN', 'steve', 'scott', 'steve', 'scott',
'JOHN', 'eric')
vec2 - c( 'eric', 'JOHN', 'eric', 'eric', 'JOHN', 'JOHN', 'steve', 'steve',
'scott', 'scott')
vec3 - c( 'eric', 'eric', 'JOHN', 'eric', 'JOHN', 'JOHN', 'steve',
Hi David,
thanks for your quick answer!
David Winsemius dwinsem...@comcast.net writes:
On Oct 31, 2013, at 1:27 PM, Andreas Leha wrote:
Hi all,
what is the recommended way to quickly (and without much burden on the
memory) extract the response from a formula?
If you want its expression
You can bullet-proof it a bit by making sure that length(formula)==3
before assuming that formula[[2]] is the response. If length(formula)==2
then there is no response term, only predictor terms. E.g., replace
resp - frm[[2]]
with
resp - if (length(frm)==3) frm[[2]] else NULL
(or call
On 11/01/2013 08:22 AM, Magnus Thor Torfason wrote:
Sure,
I was attempting to be concise and boiling it down to what I saw as the root
issue, but you are right, I could have taken it a step further. So here goes.
I have a set of around around 20M string pairs. A given string (say, A) can
Hi Richard
Untested Perhaps adding some dummy factors with NA and then have their
labels as and color of lines as 0 or transparent.
I think that I used it partly for the same reason and in addition I was
combining 2 purposes with the groups and wanted to split them
Duncan
-Original
Hi,
You may try:
dat1 - read.table(text=
Friend1,Friend2
A,B
A,C
B,A
C,D,sep=,,header=TRUE,stringsAsFactors=FALSE)
indx - as.vector(outer(unique(dat1[,1]),unique(dat1[,2]),paste))
res -
cbind(setNames(read.table(text=indx,sep=,header=FALSE,stringsAsFactors=FALSE),paste0(Friend,1:2)),
Dear all,
I am trying to make a series of waffle plot-like figures for my data to
visualize the ratios of amino acid residues at each position. For each one
of 37 positions, there may be one to four different amino acid residues. So
the data consist of the positions, what residues are there, and
William Dunlap wdun...@tibco.com writes:
You can bullet-proof it a bit by making sure that length(formula)==3
before assuming that formula[[2]] is the response. If length(formula)==2
then there is no response term, only predictor terms. E.g., replace
resp - frm[[2]]
with
resp - if
Dear R users,
I wonder if there is a way that I can plot a time series data which is in a
wide format like this:
CITY_NAME 2000Q12000Q2 2000Q32000Q4 2001Q1
2001Q2 2001Q3 2001Q4 2002Q1 2002Q2
CITY1100.5210 101.9667 103.24933
On 1 November 2013 11:06, IZHAK shabsogh ishaqb...@yahoo.com wrote:
below is a code to compute hessian matrix , which i need to generate 29
number of different matrices for example first
You may consider using Numerical Derivatives package for that instead, see:
wudadan wrote
Dear R users,
I wonder if there is a way that I can plot a time series data which is in
a
wide format like this:
CITY_NAME 2000Q12000Q2 2000Q32000Q4 2001Q1
2001Q2 2001Q3 2001Q4 2002Q1 2002Q2
CITY1100.5210
I'm trying to install.packages( RCurl ) as root but get
ERROR: 'configure' exists but is not executable
I remember having had something like that before on another machine and tried
in bash what is described here
http://mazamascience.com/WorkingWithData/?p=1185
and helped me before:
# mkdir
The error message doesn't seem to refer to the tmp directory. What do you
get from:
ls -l `which curl-config`
-- Mike
On Fri, Nov 1, 2013 at 7:43 PM, Rainer Schuermann rainer.schuerm...@gmx.net
wrote:
I'm trying to install.packages( RCurl ) as root but get
ERROR: 'configure' exists
# ls -l `which curl-config`
-rwxr-xr-x 1 root root 6327 Oct 20 15:25 /usr/bin/curl-config
On Friday 01 November 2013 20:21:36 Michael Hannon wrote:
The error message doesn't seem to refer to the tmp directory. What do you
get from:
ls -l `which curl-config`
-- Mike
On Fri,
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