Re: [R] compile error with C code and standalone R math C library
This is a known bug already fixed (we believe) in R-patched and R-devel.
On Sun, 14 Dec 2003, Dirk Eddelbuettel wrote:
On Sun, Dec 14, 2003 at 12:51:09AM -0500, Faheem Mitha wrote:
On Sat, 13 Dec 2003, Dirk Eddelbuettel wrote:
On Sat, Dec 13, 2003 at 07:44:46PM -0500, Faheem Mitha wrote:
I just went back to an old piece of C code. On trying to compile it with
the R math standalone C library I got the following error. Can anyone
enlighten me what I am doing wrong, if anything? C file (rr-sa.c) follows.
I'm on Debian sarge. I'm running R version 1.8.1. Gcc is version
3.3.1.
[...]
faheem ~/co/rr/trunkgcc -o rr rr-sa.c -lRmath -lm
/usr/lib/gcc-lib/i486-linux/3.3.2/../../../libRmath.so: undefined
reference to `Rlog1p'
collect2: ld returned 1 exit status
The linker tells you that it cannot find object code for a function Rlog1p.
So let's check:
[EMAIL PROTECTED]:~ grep Rlog1p /usr/include/Rmath.h
[EMAIL PROTECTED]:~ grep log1p /usr/include/Rmath.h
double log1p(double); /* = log(1+x) {care for small x} */
[EMAIL PROTECTED]:~
Turns out that there is none defined in Rmath.h, but log1p exists. This may
have gotten renamed since you first wrote your code.
Maybe I am being dense, but how is this my fault? I am not using either
Rlog1p or log1p in my code (as far as I can see).
Indeed -- it looks like we have a problem. Looking at the NEWS file, some
logic regarding (R)log1p was changed in the 1.8.* series, and it seems to be
going wrong here.
As a stop gap-measure, just define an empty Rlog1p() to complete linking:
[EMAIL PROTECTED]:/tmp grep Rlog1p rr-sa.c
void Rlog1p(void) {;}
[EMAIL PROTECTED]:/tmp gcc -Wall -o rr rr-sa.c -lRmath -lm
[EMAIL PROTECTED]:/tmp ls -l rr
-rwxr-xr-x1 edd edd 13889 Dec 14 00:12 rr
[EMAIL PROTECTED]:/tmp
For the record, on my build system, a log1p function is found but deemed not
good enough:
[EMAIL PROTECTED]:/tmp grep log1p ~/src/debian/build-logs/r-base_1.8.1-1.log |
head -2checking for log1p... yes
checking for working log1p... no
Dirk
--
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Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
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RE: [R] density plot for very large dataset
You might want to try hexbin (hexagonal binning) in the BioConductor suite
(see www.bioconductor.org).
HTH,
Andy
From: Obi Griffith
I'm new to R and am trying to perform a simple, yet
problematic task. I
have two variables for which I would like to measure the
correlation and
plot versus each other. However, I have ~30 million data points
measurements of each variable. I can read this into R from file and
produce a plot with plot(x0, x1) but as you would expect, its
not pretty
to look at and produces a postscript file of about 700MB. A google
search found a few mentions of doing density plots but they seemed to
assume you already have the density matrix. Can anyone point
me in the
right direction, keeping in mind that I am a complete R newbie.
Obi
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[R] Problem with data conversion
Hi All: I came across the following problem while working with a dataset, and wondered if there could be a solution I sought here. My dataset consists of information on 402 individuals with the followng five variables (age,sex, status = a binary variable with levels case or control, mma, dma). During data check, I found that in the raw data, the data entry operator had mistakenly put a 0 for one participant, so now, the levels show levels(status) [1] 0 control case The variables mma, and dma are actually numerical variables but in the dataframe, they are represented as characters. I tried to change the type of the variables (from character to numeric) using the edit function (and bringing up the data grid where then I made changes), but the changes were not saved. I tried mma1 - as.numeric(mma) but I was not successful in converting mma from a character variable to a numeric variable. So, to edit and clean the data, I exported the dataset as a text file to Epi Info 2002 (version 2, Windows). I used the following code: mysubset - subset(workingdat, select = c(age,sex,status, mma, dma)) write.table(mysubset, file=mysubset.txt, sep=\t, col.names=NA) After I made changes in the variables using Epi Info (I created a new variable called statusrec containing values case and control), I exported the file as a .rec file (filename mydata.rec). I used the following code to read the file in R: require(foreign) myData - read.epiinfo(mydata.rec, read.deleted=NA) Now, the problem is this, when I want to run a logistic regression, R returns the following error message: glm(statusrec~mma, family=binomial(link=logit)) Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : invalid variable type I cannot figure out the solution. I want to run a logistic regression now with the variable statusrec (which is a binary variable containing values case and control), and another variable (say mma, which is now a numeric variable). What does the above error message mean and what could be a possible solution? Would greatly appreciate your insights and wisdom. -Arin Basu __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Problem with data conversion
The variables mma, and dma are actually numerical variables but in the dataframe, they are represented as characters. I tried to change the type of the variables (from character to numeric) using the edit function (and bringing up the data grid where then I made changes), but the changes were not saved. I tried mma1 - as.numeric(mma) i'm not sure understanding your problem correct, but is it possible that you forget the data.frame ,suppose your data.frame is df df$mma - as.numeric (mma) should work df$mma[df$mma == 0 ] - 1 #or any other value regards,christian __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Problem with data conversion
The message probably means that the variable is a character variable and not numerical (as you intended) nor factor. Although you said there was a trip to epiinfo, you never said where the data came from. Try dumping out the data, editing the file, and reading it with read.table. There are other ways, but one of your steps has a bug and we have no idea what the steps actually were. When you are finished, try sapply(mfdf, class) on your dataframe `mydf'. You should see only numeric or factor variables. On Sun, 14 Dec 2003 [EMAIL PROTECTED] wrote: Hi All: I came across the following problem while working with a dataset, and wondered if there could be a solution I sought here. My dataset consists of information on 402 individuals with the followng five variables (age,sex, status = a binary variable with levels case or control, mma, dma). During data check, I found that in the raw data, the data entry operator had mistakenly put a 0 for one participant, so now, the levels show levels(status) [1] 0 control case The variables mma, and dma are actually numerical variables but in the dataframe, they are represented as characters. I tried to change the type of the variables (from character to numeric) using the edit function (and bringing up the data grid where then I made changes), but the changes were not saved. I tried mma1 - as.numeric(mma) but I was not successful in converting mma from a character variable to a numeric variable. So, to edit and clean the data, I exported the dataset as a text file to Epi Info 2002 (version 2, Windows). I used the following code: mysubset - subset(workingdat, select = c(age,sex,status, mma, dma)) write.table(mysubset, file=mysubset.txt, sep=\t, col.names=NA) After I made changes in the variables using Epi Info (I created a new variable called statusrec containing values case and control), I exported the file as a .rec file (filename mydata.rec). I used the following code to read the file in R: require(foreign) myData - read.epiinfo(mydata.rec, read.deleted=NA) Now, the problem is this, when I want to run a logistic regression, R returns the following error message: glm(statusrec~mma, family=binomial(link=logit)) Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : invalid variable type I cannot figure out the solution. I want to run a logistic regression now with the variable statusrec (which is a binary variable containing values case and control), and another variable (say mma, which is now a numeric variable). What does the above error message mean and what could be a possible solution? Would greatly appreciate your insights and wisdom. -Arin Basu __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] half normal probability plot in R
On 13 Dec 2003 at 4:00, Allison Jones wrote: I have generated the effects in a factorial design and now want to put them in a half normal probability plot. Is there an easy way to do this in R??? I can't find the command. Thanks much - qqnorm.aov in package gregmisc (on CRAN) Kjetil Halvorsen Ali Jones __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] correlation and causality examples
Dear R-users, many thanks to all who replied to my request, approx. one month ago, about examples illustrating that correlation does imply causality. I have tried to compile your suggestions in a web-site, which URL is given in the screenshot in png format there: http://pbil.univ-lyon1.fr/members/lobry/z.png and in jpeg format there: http://pbil.univ-lyon1.fr/members/lobry/z.jpg It's far for being perfect because of an over-teaching period, but I hope to improve it in the future, so that your suggestions and comments are always welcome. All the best, Jean -- Jean R. Lobry Laboratoire BBE-CNRS-UMR-5558, Univ. C. Bernard - LYON I, 43 Bd 11/11/1918, F-69622 VILLEURBANNE CEDEX, FRANCE allo : +33 472 43 12 87 fax: +33 472 43 13 88 http://pbil.univ-lyon1.fr/members/lobry/ __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] contour() should handle the asp parameter
Hi all,
To my knowledge, the current version of contour.default() does not handle the 'asp'
parameter. This can be embarassing when displaying eg geographical maps, etc...
Submitted to the opinion of more experienced R programmers, contour.defaut() function
should be changed according to the followings:
line 7: add = FALSE,asp=NA,...)
line 33: plot.window(xlim, ylim, asp=asp,)
The new script would be:
contour.default-
function (x = seq(0, 1, len = nrow(z)), y = seq(0, 1,
len = ncol(z)),z, nlevels = 10, levels = pretty(zlim, nlevels), labels = NULL, xlim =
range(x, finite = TRUE), ylim = range(y, finite = TRUE),
zlim = range(z, finite = TRUE), labcex = 0.6, drawlabels = TRUE,
method = flattest, vfont = c(sans serif, plain), axes = TRUE,
frame.plot = axes, col = par(fg), lty = par(lty), lwd = par(lwd),
add = FALSE,asp=NA,...)
{
if (missing(z)) {
if (!missing(x)) {
if (is.list(x)) {
z - x$z
y - x$y
x - x$x
}
else {
z - x
x - seq(0, 1, len = nrow(z))
}
}
else stop(no `z' matrix specified)
}
else if (is.list(x)) {
y - x$y
x - x$x
}
if (any(diff(x) = 0) || any(diff(y) = 0))
stop(increasing x and y values expected)
if (!is.matrix(z) || nrow(z) = 1 || ncol(z) = 1)
stop(no proper `z' matrix specified)
if (!add) {
plot.new()
plot.window(xlim, ylim, asp=asp,)
title(...)
}
if (!is.double(z))
storage.mode(z) - double
method - pmatch(method[1], c(simple, edge, flattest))
if (!is.null(vfont))
vfont - c(typeface = pmatch(vfont[1], Hershey$typeface) -
1, fontindex = pmatch(vfont[2], Hershey$fontindex))
if (!is.null(labels))
labels - as.character(labels)
.Internal(contour(as.double(x), as.double(y), z, as.double(levels),
labels, labcex, drawlabels, method, vfont, col = col,
lty = lty, lwd = lwd))
if (!add) {
if (axes) {
axis(1)
axis(2)
}
if (frame.plot)
box()
}
invisible()
}
environment: namespace:base
Best regards,
Patrick
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Re: [R] contour() should handle the asp parameter
Given contour() has an add= argument whose use is demonstrated for this exact reason in `all good books on S', why complicate the contour function? Contrary to popular belief, the `asp parameter' is not a parameter, but an argument of plot's default method. On Sun, 14 Dec 2003, Patrick Giraudoux wrote: To my knowledge, the current version of contour.default() does not handle the 'asp' parameter. This can be embarassing when displaying eg geographical maps, etc... Submitted to the opinion of more experienced R programmers, contour.defaut() function should be changed according to the followings: line 7: add = FALSE,asp=NA,...) line 33: plot.window(xlim, ylim, asp=asp,) And of course alter the documentation. I suspect you could get away with passing ... to the plot.window call, which for clarity should be plot.window(xlim, ylim, , asp = asp) -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] A faster plotOHLC() for the tseries package
The plotOHLC function in the tseries package is useful to plot timeseries of
various financial assets with open/high/low/close data. I had often
wondered if it could be made to run a little faster. It turns out that the
following patch does
--- plotOHLC.R.orig 2003-12-14 12:02:20.0 -0600
+++ plotOHLC.R 2003-12-14 12:03:42.0 -0600
@@ -21,14 +21,9 @@
ylim - range(x[is.finite(x)])
plot.new()
plot.window(xlim, ylim, ...)
-for (i in 1:NROW(x)) {
-segments(time.x[i], x[i, High], time.x[i], x[i, Low],
-col = col[1], bg = bg)
-segments(time.x[i] - dt, x[i, Open], time.x[i], x[i,
-Open], col = col[1], bg = bg)
-segments(time.x[i], x[i, Close], time.x[i] + dt, x[i,
-Close], col = col[1], bg = bg)
-}
+segments(time.x, x[, High], time.x, x[, Low], col = col[1], bg = bg)
+segments(time.x - dt, x[, Open], time.x, x[, Open], col = col[1], bg = bg)
+segments(time.x, x[, Close], time.x + dt, x[, Close], col = col[1], bg = bg)
if (ann)
title(main = main, xlab = xlab, ylab = ylab, ...)
if (axes) {
decrease the time spent on a series of ~500 points by a factor of sixty:
IBM-get.hist.quote(IBM, 2001-12-14)
trying URL
http://chart.yahoo.com/table.csv?s=IBMa=11b=13c=2001d=11e=12f=2003g=dq=qy=0z=IBMx=.csv'
Content type application/octet-stream' length unknown
opened URL
.. .. ...
downloaded 23Kb
time series starts 2001-12-12
time series ends 2003-12-11
system.time(plotOHLC(IBM))# original
[1] 1.56 0.26 5.11 0.00 0.00
system.time(fastplotOHLC(IBM))# patched
[1] 0.02 0.00 0.05 0.00 0.00
Regards, Dirk
--
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Re: [R] A faster plotOHLC() for the tseries package
Hi Gabor,
On Sun, Dec 14, 2003 at 02:35:40PM -0500, Gabor Grothendieck wrote:
Dirk,
Could you please explain to me how to interpret the lines
that you posted (which I gather are intended to be used with
some program that combines them with the original source)?
That is the usual paradigm of using output of diff(1)
$ diff -u old new diff.txt
as input to the patch(1) program as e.g. in
$ patch diff.txt# try patch --dry-run diff.txt first
On win2k, you can get them for sure with Cygwin, probably also with
mingw/msys and likely also with BDR's set of tools to build R from source.
Patch, written by Larry Wall of Perl fame, reads an entry such as
--- plotOHLC.R.orig 2003-12-14 12:02:20.0 -0600
+++ plotOHLC.R 2003-12-14 12:03:42.0 -0600
@@ -21,14 +21,9 @@
and knows to replace lines marked with '-' (taken be old the old file) with
those marked '+'. In this example, it is trivial as there is only one
segment in which the code
for (i in 1:NROW(x)) {
segments(time.x[i], x[i, High], time.x[i], x[i, Low],
col = col[1], bg = bg)
segments(time.x[i] - dt, x[i, Open], time.x[i], x[i,
Open], col = col[1], bg = bg)
segments(time.x[i], x[i, Close], time.x[i] + dt, x[i,
Close], col = col[1], bg = bg)
}
with
segments(time.x, x[, High], time.x, x[, Low], col = col[1], bg = bg)
segments(time.x - dt, x[, Open], time.x, x[, Open], col = col[1], bg = bg)
segments(time.x, x[, Close], time.x + dt, x[, Close], col = col[1], bg = bg)
Alternately, could you just send the revised OHLC source?
Well, the source is different from what you find in $R_HOME/library/tseries/R
so you may as well edit there by hand. I don't have access to a windows box
right now, but if the above doesn't help email off-line and I start up the
laptop from work.
Hope this helps, Dirk
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[R] reverse lexicographic order
Hi all, I have some email addresses that I would like to sort in reverse lexicographic order so that addresses from the same domain will be grouped together. How might that be done? Murray -- Dr Murray Jorgensen http://www.stats.waikato.ac.nz/Staff/maj.html Department of Statistics, University of Waikato, Hamilton, New Zealand Email: [EMAIL PROTECTED]Fax 7 838 4155 Phone +64 7 838 4773 wk+64 7 849 6486 homeMobile 021 1395 862 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] pca
Dear listmates, i've done a pca analisys in R (1.8 v.) with the command pca(Matrix, cent=FALSE, scle=FALSE) I have obtained a v matrix very different from the component matrix resulted by a factor analysis in SPSS, unrotated and with a extraction from a correlation matrix. Any clues about this diference? Thanks in advance, Alex. __ Acabe com aquelas janelinhas que pulam na sua tela. AntiPop-up UOL - É grátis! http://antipopup.uol.com.br/ __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] reverse lexicographic order
Thomas W Blackwell [EMAIL PROTECTED] writes: Murray - If you could guarantee that all of the email addresses have exactly one occurrence of the @ character in them, then something like snip Otherwise, try something like this (I don't think we have a string reversal function anywhere, do we?): mychar - scan(what=) 1: I have some email addresses that I would like to sort in reverse 14: lexicographic order so that addresses from the same domain will be 25: grouped together. How might that be done? 32: Read 31 items mychar[order(sapply(lapply(strsplit(mychar,),rev),paste,collapse=))] [1] together. done? I I [5] lexicographic grouped would be [9] bethe like same [13] some reverse have email [17] will from indomain [21] sotoorder addresses [25] addresses that that that [29] might sort How Hi all, I have some email addresses that I would like to sort in reverse lexicographic order so that addresses from the same domain will be grouped together. How might that be done? Murray -- Dr Murray Jorgensen http://www.stats.waikato.ac.nz/Staff/maj.html Department of Statistics, University of Waikato, Hamilton, New Zealand Email: [EMAIL PROTECTED]Fax 7 838 4155 Phone +64 7 838 4773 wk+64 7 849 6486 homeMobile 021 1395 862 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] reverse lexicographic order
Hi Duncan, Hi Peter,
thanks for those ideas! I'm sure I will learn a lot be fooling around
with them.
Cheers,
Murray
Duncan Murdoch wrote:
On Mon, 15 Dec 2003 10:08:31 +1300, you wrote:
Hi all,
I have some email addresses that I would like to sort in reverse
lexicographic order so that addresses from the same domain will be
grouped together. How might that be done?
I'm not sure this is what you want, but this function sorts a
character vector by last letters, then 2nd last, 3rd last, and so on:
revsort - function(x,...) {
x[order(unlist(lapply(strsplit(x,''),
function(x) paste(rev(x),collapse=''))),...)]
}
revsort(as.character(1:20))
[1] 10 20 1 11 2 12 3 13 4 14 5 15 6
16 7
[16] 17 8 18 9 19
The ... args are given to order(), so na.last=FALSE and
decreasing=TRUE are possibilities.
Duncan Murdoch
--
Dr Murray Jorgensen http://www.stats.waikato.ac.nz/Staff/maj.html
Department of Statistics, University of Waikato, Hamilton, New Zealand
Email: [EMAIL PROTECTED]Fax 7 838 4155
Phone +64 7 838 4773 wk+64 7 849 6486 homeMobile 021 1395 862
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Re: [R] extensive grid docs?
Hi Most of the documentation that exists for grid is currently linked off my grid web site (http://www.stat.auckland.ac.nz/~paul/grid/grid.html). Paul [EMAIL PROTECTED] wrote: [EMAIL PROTECTED] wrote on 10/12/2003 22:23:43: I'm looking for extensive docs on using grid (for the somewhat newbie). I'm attempting to create a set of graphics that look similar to the attached image (I hope this doesn't get bounced) and have only come across the R newsletters and it appears that grid was new as of 1.8.0? I think the best way to proceed is to create the plot, clip it using a polygon, then manually add the annotation. Is that correct? I couldn't find much on the FAQ regarding creating really goofy plots with grid and any hints would be greatly appreciated. Here's an R graphics tutorial by Paul Murrell http://www.ci.tuwien.ac.at/Conferences/DSC-2003/tutorials.html HTH, Tobias __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help -- Dr Paul Murrell Department of Statistics The University of Auckland Private Bag 92019 Auckland New Zealand 64 9 3737599 x85392 [EMAIL PROTECTED] http://www.stat.auckland.ac.nz/~paul/ __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] read.spss question warning compression bias
So it would appear that if the above is correct, there is no user adjustment to the
bias value.
The only scenario that I can envision is if the user SAVE's the .sav file in an
uncompressed
format, where the bias value **might** be set to 0.
Perhaps a r-help reader with access to current SPSS manuals can confirm the above.
The windows version 11.5.0 appears the same (I assume the negative sign on -99 was
somehow dropped)
COMPRESSED and UNCOMPRESSED Subcommands
COMPRESSED saves the file in compressed form. UNCOMPRESSED saves the file in
uncom-pressed form.
In a compressed file, small integers (from 99 to 155) are stored in one byteinstead
of the
eight bytes used in an uncompressed file.
The only specification is the keyword COMPRESSED or UNCOMPRESSED. There are
noadditional specifications.
Compressed data files occupy less disk space than do uncompressed data files.
Compressed data files take longer to read than do uncompressed data files.
The GET command, which reads SPSS-format data files, does not need to specify
whetherthe files it reads are compressed or uncompressed.
Only one of the subcommands COMPRESSED or UNCOMPRESSED can be specified perSAVE
command. COMPRESSED is usually the default, though UNCOMPRESSED may bethe default on
some systems.
Ciao, Tom
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-Original Message-
From: Marc Schwartz [mailto:[EMAIL PROTECTED]
Sent: Friday, 12 December 2003 3:56 AM
To: Thomas Lumley
Cc: [EMAIL PROTECTED]
Subject: Re: [R] read.spss question warning compression bias
On Thu, 2003-12-11 at 12:32, Thomas Lumley wrote:
On Thu, 11 Dec 2003, Marc Schwartz wrote:
An additional question might be, if the file is not compressed, what
is the default bias value set by SPSS? If it is 0, then the check is
meaningless. On the other hand, if the default value is 100, whether
or not the file is compressed, then the warning message would serve
a purpose in flagging the possibility of other issues. Reasonably,
that setting may be SPSS version specific.
I think the issue is that the format is not documented, so the author
of the code (Ben Pfaff) didn't know what a change in the value would
imply. If the file is apparently read correctly it seems that it
doesn't imply anything.
-thomas
Thanks for the clarification Thomas.
I did some searching of the PSPP site and found the following:
http://www.gnu.org/software/pspp/manual/pspp_18.html#SEC170
The compression bias is defined as:
flt64 bias;
Compression bias. Always set to 100. The significance of this
value is that only numbers between (1 - bias) and (251 - bias)
can be compressed.
So it would seem to potentially impact aspects of the file compression data structure,
when compression is used.
I am not sure if the Always set to 100 is unique to PSPP in how Ben elected to do
things. Presumably if that is always the case, even with SPSS, one might reasonably
wonder: why have it, if it does not vary?
It leaves things unclear as to under what circumstances this value would change.
I did some Googling and found the following text snippet from a presumably dated SPSS
manual for the syntax of the SAVE command:
SAVE OUTFILE=file
[/VERSION={3**}] {2 }
[/UNSELECTED=[{RETAIN}] {DELETE}
[/KEEP={ALL** }] [/DROP=varlist] {varlist}
[/RENAME=(old varlist=new varlist)...]
[/MAP]
[/{COMPRESSED }] {UNCOMPRESSED}
**Default if the subcommand is omitted.
COMPRESSED and UNCOMPRESSED Subcommands
COMPRESSED saves the file in compressed form. UNCOMPRESSED saves the file in
uncompressed form. In a compressed file, small integers (from
99 to 155) are stored in one byte instead of the eight bytes used in an uncompressed
file.
The only specification is the keyword COMPRESSED or UNCOMPRESSED. There are no
additional specifications.
Compressed data files occupy less disk space than do uncompressed data files.
Compressed data files take longer to read than do uncompressed data files.
The GET command, which reads SPSS-format data files, does not need to specify whether
the files it reads are compressed or uncompressed.
Only one of the subcommands COMPRESSED or UNCOMPRESSED can be specified per SAVE
command. COMPRESSED is usually the default, though UNCOMPRESSED may be the default on
some systems.
So it would appear that if the above is correct, there is no user adjustment to the
bias value. The only scenario that I can envision is if
Re: [R] Problem with data conversion
I sympathize with your trouble bringing in data, but you need to catch your breath and figure out what you really have. I think when you get a bit more R practice, you will be able to manage what you bring in without going back to that editor so much. I feel certain your data is not what you think it is. Here's an example where a factor DOES work on the lhs of a glm: y - factor(c(S,N,S,N,S,N,S,N)) x - rnorm(8) glm(y~x,family=binomial(link=logit)) Look here: the system knows y is a factor: attributes(y) $levels [1] N S $class [1] factor My guess is that your variables are not really factors, but rather character vectors. You have to convert them into factors. Watch the error I get is the same that you got. y - c(S,N,S,N,S,N,S,N) glm(y~x,family=binomial(link=logit)) Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : invalid variable type Note the system doesn't know y is supposed to be a factor. It just sees characters. y [1] S N S N S N S N levels(y) NULL attributes(y) NULL but look: glm(as.factor(y)~x,family=binomial(link=logit)) [EMAIL PROTECTED] wrote: Hi All: I came across the following problem while working with a dataset, and wondered if there could be a solution I sought here. My dataset consists of information on 402 individuals with the followng five variables (age,sex, status = a binary variable with levels case or control, mma, dma). During data check, I found that in the raw data, the data entry operator had mistakenly put a 0 for one participant, so now, the levels show levels(status) [1] 0 control case The variables mma, and dma are actually numerical variables but in the dataframe, they are represented as characters. I tried to change the type of the variables (from character to numeric) using the edit function (and bringing up the data grid where then I made changes), but the changes were not saved. I tried mma1 - as.numeric(mma) but I was not successful in converting mma from a character variable to a numeric variable. So, to edit and clean the data, I exported the dataset as a text file to Epi Info 2002 (version 2, Windows). I used the following code: mysubset - subset(workingdat, select = c(age,sex,status, mma, dma)) write.table(mysubset, file=mysubset.txt, sep=\t, col.names=NA) After I made changes in the variables using Epi Info (I created a new variable called statusrec containing values case and control), I exported the file as a .rec file (filename mydata.rec). I used the following code to read the file in R: require(foreign) myData - read.epiinfo(mydata.rec, read.deleted=NA) Now, the problem is this, when I want to run a logistic regression, R returns the following error message: glm(statusrec~mma, family=binomial(link=logit)) Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : invalid variable type I cannot figure out the solution. I want to run a logistic regression now with the variable statusrec (which is a binary variable containing values case and control), and another variable (say mma, which is now a numeric variable). What does the above error message mean and what could be a possible solution? Would greatly appreciate your insights and wisdom. -Arin Basu __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help -- Paul E. Johnson email: [EMAIL PROTECTED] Dept. of Political Sciencehttp://lark.cc.ukans.edu/~pauljohn University of Kansas Office: (785) 864-9086 Lawrence, Kansas 66045FAX: (785) 864-5700 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] help in lme
To anyone who can help, I have two stupid questions, and one fairly intelligent question Stupid question (1): is there an R function to calculate a factorial of a number? That is...is there a function g(.) such that g(3) = 6, g(4) = 24, g(6) = 720, etc? Stupid question (2): how do you extract the estimated covariance matrix of the random effects in an lme object? Intelligent question (1) I keep on trying to fit a linear mixed model in R using 'lme(y~fxd.dsgn, data = data.mtrx, ~rnd.dsgn|group)' where fxd.dsgn and rnd.dsgn are the fixed and random design matrices, respectively. The function won't work, though. It keeps telling me that it can't find the object 'rnd.dsgn'.What's the matter here? Any help would be greatly appreciated. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] help with random numbers and Rmpi
On Mon, 1 Dec 2003, A.J. Rossini wrote: Faheem Mitha [EMAIL PROTECTED] writes: So, can this (parallelization at the C level) be done without running a bunch of C slaves along the lines I had previously written? Any examples would be helpful. How much heavy lifting happens before you spawn the slaves, and can that not be moved to R? Your best bet is to read the SNOW code for handling SPRNG/RSPRNG, otherwise. I've tried to use snow as suggested. I have a R function mg.randvec which generates a vector of random variates. This function calls a C routine via the .C call. This works fine if I call it like say... * mg.randvec(3,2,10,5) $val [1] -1.9967464 -1.8634205 -0.7459255 -1.7591047 -1.7811685 -1.9953316 [7] -1.7932502 -1.9823565 -1.7999789 -1.0501179 -1.9679886 0.1484859 [13] 0.5768898 1.9117889 1.9366872 -1.3847453 -1.5554107 -1.4933195 [19] -1.8508795 -1.6715850 -1.8951212 -1.8900167 -1.1630852 -1.3989748 [25] -1.9400337 -1.6774471 -1.8136065 -1.8685709 -1.9119879 -1.3378416 * However, with snow I get ** clusterCall(cl,mg.randvec,3,2,10,5) [[1]] [1] Error in .C(\rocftp\, as.integer(k), as.integer(len), as.double(theta), : \n\tC/Fortran function name not in load table\n attr(,class) [1] try-error [[2]] [1] Error in .C(\rocftp\, as.integer(k), as.integer(len), as.double(theta), : \n\tC/Fortran function name not in load table\n attr(,class) [1] try-error [[3]] [1] Error in .C(\rocftp\, as.integer(k), as.integer(len), as.double(theta), : \n\tC/Fortran function name not in load table\n attr(,class) [1] try-error In the cluster case it seems to have difficulty loading up the C routine. I think snow is working Ok, because basic examples like the following work. clusterCall(cl,runif,3) [[1]] [1] 0.1527429 0.1134621 0.8663094 [[2]] [1] 0.2256776 0.8981241 0.1120226 [[3]] [1] 0.2371450 0.5090693 0.2776081 ** Can anyone tell me what I am doing wrong? All data files is shared across all three machines I am using (AFS space). Thanks in advance. Faheem. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
