[R] Error when compiling R-2.5.1 / *** [d-p-q-r-tests.Rout] Fehler 1
Hi everyone, I am new to Linux and R - but I managed to build R-2.5.1 from source to use it in Genepattern. Genepattern does only support R-2.5.1 which I could not find anywhere for installation via apt-get or in the Ubuntu Software-Centre (I am using Ubuntu 14.04 (Trusty Tahr) 32-bit) But after doing make check I get comparing 'method-dispatch.Rout' to './method-dispatch.Rout.save' ... OK running code in 'd-p-q-r-tests.R' ...make[3]: *** [d-p-q-r-tests.Rout] Fehler 1 make[3]: Verzeichnis »/home/karin/Downloads/R-2.5.1/tests« wird verlassen make[2]: *** [test-Specific] Fehler 2 make[2]: Verzeichnis »/home/karin/Downloads/R-2.5.1/tests« wird verlassen make[1]: *** [test-all-basics] Fehler 1 make[1]: Verzeichnis »/home/karin/Downloads/R-2.5.1/tests« wird verlassen make: *** [check] Fehler 2 but I can make install and use R for simple plots etc. afterwards - still I am worried something is wrong, can you give some advice. A closer look at the error gives ## PR#7099 : pf() with large df1 or df2: nu - 2^seq(25,34, 0.5) y - 1e9*(pf(1,1,nu) - 0.68268949) stopifnot(All.eq(pf(1,1,Inf), 0.68268949213708596), + diff(y) 0, # i.e. pf(1,1, *) is monotone increasing + All.eq(y [1], -5.07420372386491), + All.eq(y[19], 2.12300110824515)) Error: All.eq(y[1], -5.07420372386491) is not TRUE Execution halted As I understand so far some errors are critical some are not - can you please give some advice on the error above? Can I still use R installed with that error? What do I need to solve the error? Thanks, Joerg [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Clarification on Simulation and Iteration
On Jul 31, 2015, at 8:41 PM, Christopher Kelvin wrote:
Thanks Dave.
What I actually want is to obtain say 10, different sets of (n=50) data for
every 10,000 iterations I run. You will realise that the current code
produces one set of data (n=50). I want 10 different sets of 50 observations
at one run. I hope this makes sense.
I would think that either `replicate(50, ...)` or `for(i in 1:50) {...}` would
suffice. Unless of course the phrase want 10 different sets of 50 observations
at one run` means something different than it appears to request.
Chris Guure
On Saturday, August 1, 2015 3:32 AM, David Winsemius dwinsem...@comcast.net
wrote:
On Jul 31, 2015, at 6:36 PM, Christopher Kelvin via R-help wrote:
Dear All,
I am performing some simulations for a new model. I run about 10,000
iterations with a sample of 50 datasets and this returns one set of 50
simulated data.
Now, what I need to obtain is 10 sets of the 50 simulated data out of the
10,000 iterations and not just only 1 set. The model is the Copas selection
model for publication bias in Mete-analysis. Any one who knows this model
has any suggestion for the improvement of my code is most welcome.
Below is my code.
Kind regards
Chris Guure
University of Ghana
install.packages(msm)
library(msm)
rho1=-0.3; tua=0.020; n=50; d=-0.2; rr=1; a=-1.3; b=0.06
si-rtnorm(n, mean=0, sd=1, lower=0, upper=0.2)# I used this to generate
standard errors for each study
set.seed(2) ## I have stored the data and the output in this seed
for( i in 1:rr){
mu-rnorm(n,d,tua^2) # prob. of each effect estimate
rho-si*rho1/sqrt(tua^2 + si^2) # estimate of the correlation coefficient
mu0- a + b/si # mean of the truncated normal model (Copas selection
model)
y1-rnorm(mu,si^2)# observed effects zise
z-rnorm(mu0,1) # selection model
rho2-cor(y1, z)
select-pnorm((mu0 + rho*(y1-mu)/sqrt(tua^2 + si^2))/sqrt(1-rho^2))
probselect-ifelse(selectz, y1, NA)# the prob that the study is selected
probselect
data-data.frame(probselect,si)# this contains both include and exclude
data
data
data1-data[complete.cases(data),] # Contains only the included data for
analysis
data1
}
OK. The code runs without error. So what exactly is the problem? (I have
no experience with the Copas selection model if in fact that is what is being
exemplified.)
--
David Winsemius
Alameda, CA, USA
David Winsemius
Alameda, CA, USA
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Re: [R] vectorized sub, gsub, grep, etc.
Adam,
You reopened an old thread noting its age, but did you begin at its beginning?
Subject: vectorized sub, gsub, grep, etc. Date: Oct 7, 2008
R pattern-matching and replacement functions are
vectorized: they can operate on vectors of targets.
However, they can only use one pattern and replacement.
Here is code to apply a different pattern and replacement
for every target. My question: can it be done better?
sub2 - function(pattern, replacement, x) {
len - length(x)
if (length(pattern) == 1)
pattern - rep(pattern, len)
if (length(replacement) == 1)
replacement - rep(replacement, len)
FUN - function(i, ...) {
sub(pattern[i], replacement[i], x[i], fixed = TRUE)
}
idx - 1:length(x)
sapply(idx, FUN)
}
#Example
X - c(ab, cd, ef)
patt - c(b, cd, a)
repl - c(B, CD, A)
sub2(patt, repl, X)
If you run that code, you'll see the correct answer is not CD , it is
[1] aB CD ef
And the same answer is given by the shorter (but slower) code suggested later
that day by Christos
mapply(function(p, r, x) sub(p, r, x, fixed = TRUE), p=patt, r=repl, x=X)
b cd a
aB CD ef
By talking instead about simple string matching, I'm afraid you've rather
hijacked the thread.
-John
-John
Adam wrote
I'm not sure I understand your question. Both functions return CD
because they
perform exact string matching. The first demonstrates how string or character
replacements
can be vectorized, while the second merely demonstrates how Rcpp can
accelerate this type of operation.
On Jul 30, 2015, at 21:09, John Thaden jjtha...@flash.net wrote:
| Can you show what is its solution for the original sample data? Why that
discrepancy for you original sub2() function?
| From:Adam Erickson adam.michael.erick...@gmail.com
Date:Thu, Jul 30, 2015 at 6:11 pm
Subject:Re: [R] vectorized sub, gsub, grep, etc.
Here is a Rcpp version for exact character matching (for example) written in
C++ that is substantially faster. Hence, I think this is the way to go where
loops may be unavoidable. However, the input vector length has to match the
length of the pattern and replacement vectors, as your original code did. That
can be changed though.
#include Rcpp.husing namespace Rcpp;
// [[Rcpp::export]]CharacterVector subCPP(CharacterVector pattern,
CharacterVector replacement, CharacterVector x) { int len = x.size();
CharacterVector y(len); int patlen = pattern.size(); int replen =
replacement.size(); if (patlen != replen) RcoutError: Pattern and
replacement length do not match; for(int i = 0; i patlen; ++i) { if
(*(char*)x[i] == *(char*)pattern[i]) y[x[i] == pattern[i]] =
replacement[i]; } return y;}
CD
system.time(for(i in 1:5) subCPP(patt, repl, X)) user system elapsed
0.16 0.00 0.16
Cheers,
Adam
On Wednesday, July 29, 2015 at 2:42:23 PM UTC-7, Adam Erickson wrote:
Further refining the vectorized (within a loop) exact string match function, I
get times below 0.9 seconds while maintaining error checking. This is
accomplished by removing which() and replacing 1:length() with seq_along().
sub2 - function(pattern, replacement, x) { len - length(x) y -
character(length=len) patlen - length(pattern) replen -
length(replacement) if(patlen != replen) stop('Error: Pattern and replacement
length do not match') for(i in seq_along(pattern)) { y[x==pattern[i]] -
replacement[i] } return(y) }
system.time(for(i in 1:5) sub2(patt, repl, X)) user system elapsed
0.86 0.00 0.86
Since the ordered vectors are perfectly aligned, might as well do an exact
string match. Hence, I think this is not off-topic.
Cheers,
Adam
On Wednesday, July 29, 2015 at 8:15:52 AM UTC-7, Bert Gunter wrote:
There is confusion here. apply() family functions are **NOT**
vectorization -- they ARE loops (at the interpreter level), just done
in functionalized form. Please read background material (John
Chambers's books, MASS, or numerous others) to improve your
understanding and avoid posting erroneous comments.
Cheers,
Bert
Bert Gunter
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
-- Clifford Stoll
On Tue, Jul 28, 2015 at 3:00 PM, John Thaden jjth...@flash.net wrote:
Adam, The method you propose gives a different result than the prior
methods for these example vectors
X - c(ab, cd, ef)
patt - c(b, cd, a)
repl - c(B, CD, A)
Old method 1
mapply(function(p, r, x) sub(p, r, x, fixed = TRUE), p=patt, r=repl, x=X)
gives
b cd a
aB CD ef
Old method 2
sub2 - function(pattern, replacement, x) {
len - length(x)
if (length(pattern) == 1)
pattern - rep(pattern, len)
if (length(replacement) == 1)
replacement - rep(replacement, len)
FUN - function(i, ...) {
sub(pattern[i], replacement[i], x[i], fixed = TRUE)
}
Re: [R] R and AWS
How about this: Setting Rstudio server using Amazon Web Services (AWS) – a step by step (screenshots) tutorial http://www.r-statistics.com/2015/06/setting-rstudio-server-using-amazon-web-services-aws-a-step-by-step-screenshots-tutorial/ Contact Details:--- Contact me: tal.gal...@gmail.com | Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- On Thu, Jul 30, 2015 at 3:41 PM, My List mylistt...@gmail.com wrote: Hello All, I wanted to know if there is a quick tutorial which I could be pointed to, for the understanding of setting R and R studio on Amazon web services. Thanks in Advance, Harmeet [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] missing data in R
Hello Mr. FeldesmanI am a master student in biostatistic my thesis about missing values in microarray data, but � can't create any values. � want to create %10, %20,...%90 missing values for all colums in microarray data set . Can you help me any code? thank you for your attention. Asena Ay�a �zdemirMersin University Biostatistics [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Clarification on Simulation and Iteration
I am not sure how you are doing this but there is a package on CRAN
which implements the Copas model (metasens). I am not sure whether that
would help in your modelling.
On 01/08/2015 02:36, Christopher Kelvin via R-help wrote:
Dear All,
I am performing some simulations for a new model. I run about 10,000 iterations
with a sample of 50 datasets and this returns one set of 50 simulated data.
Now, what I need to obtain is 10 sets of the 50 simulated data out of the
10,000 iterations and not just only 1 set. The model is the Copas selection
model for publication bias in Mete-analysis. Any one who knows this model has
any suggestion for the improvement of my code is most welcome.
Below is my code.
Kind regards
Chris Guure
University of Ghana
install.packages(msm)
library(msm)
rho1=-0.3; tua=0.020; n=50; d=-0.2; rr=1; a=-1.3; b=0.06
si-rtnorm(n, mean=0, sd=1, lower=0, upper=0.2)# I used this to generate
standard errors for each study
set.seed(2) ## I have stored the data and the output in this seed
for( i in 1:rr){
mu-rnorm(n,d,tua^2) # prob. of each effect estimate
rho-si*rho1/sqrt(tua^2 + si^2) # estimate of the correlation coefficient
mu0- a + b/si # mean of the truncated normal model (Copas selection
model)
y1-rnorm(mu,si^2)# observed effects zise
z-rnorm(mu0,1) # selection model
rho2-cor(y1, z)
select-pnorm((mu0 + rho*(y1-mu)/sqrt(tua^2 + si^2))/sqrt(1-rho^2))
probselect-ifelse(selectz, y1, NA)# the prob that the study is selected
probselect
data-data.frame(probselect,si)# this contains both include and exclude data
data
data1-data[complete.cases(data),] # Contains only the included data for
analysis
data1
}
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Michael
http://www.dewey.myzen.co.uk/home.html
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] vectorized sub, gsub, grep, etc.
Hi John,
So, you think looping over the sub() function with regular expressions
disabled is somehow more nuanced? Perhaps you should have specified that
you were only interested in sub() function results. Regardless, the
original function failed to match the 'a' in 'ab,' which should have
returned 'AB' at that vector position. Hence, the 'correct' results are not
really correct. It only returned the first match. The qdap library provides
a function for this:
library(qdap)
mgsub(patt,repl,X)
[1] AB CD ef
Based on my previous pattern matching and replacement code, there is
clearly an opportunity to accelerate these function with Rcpp. I think that
is the main takeaway from all of this. Unfortunately, I cannot dedicade my
own time to this at the moment. Until then, exact string matching provides
an elegant and efficient solution. A little data preparation, which is also
quite fast, is all that is needed.
Cheers,
Adam
On Fri, Jul 31, 2015 at 11:15 PM, John Thaden jjtha...@flash.net wrote:
Adam,
You reopened an old thread noting its age, but did you begin at its
beginning?
Subject: vectorized sub, gsub, grep, etc.
Date: Oct 7, 2008
R pattern-matching and replacement functions are
vectorized: they can operate on vectors of targets.
However, they can only use one pattern and replacement.
Here is code to apply a different pattern and replacement
for every target. My question: can it be done better?
sub2 - function(pattern, replacement, x) {
len - length(x)
if (length(pattern) == 1)
pattern - rep(pattern, len)
if (length(replacement) == 1)
replacement - rep(replacement, len)
FUN - function(i, ...) {
sub(pattern[i], replacement[i], x[i], fixed = TRUE)
}
idx - 1:length(x)
sapply(idx, FUN)
}
#Example
X - c(ab, cd, ef)
patt - c(b, cd, a)
repl - c(B, CD, A)
sub2(patt, repl, X)
If you run that code, you'll see the correct answer is not CD , it is
[1] aB CD ef
And the same answer is given by the shorter (but slower) code suggested later
that day by Christos
mapply(function(p, r, x) sub(p, r, x, fixed = TRUE), p=patt, r=repl, x=X)
b cda
aB CD ef
By talking instead about simple string matching, I'm afraid you've rather
hijacked the thread.
-John
-John
Adam wrote
I'm not sure I understand your question. Both functions return CD
because they
perform exact string matching. The first demonstrates how string or
character replacements
can be vectorized, while the second merely demonstrates how Rcpp can
accelerate this type of operation.
On Jul 30, 2015, at 21:09, John Thaden jjtha...@flash.net wrote:
Can you show what is its solution for the original sample data? Why that
discrepancy for you original sub2() function?
From:Adam Erickson adam.michael.erick...@gmail.com
Date:Thu, Jul 30, 2015 at 6:11 pm
Subject:Re: [R] vectorized sub, gsub, grep, etc.
Here is a Rcpp version for exact character matching (for example) written
in C++ that is substantially faster. Hence, I think this is the way to go
where loops may be unavoidable. However, the input vector length has to
match the length of the pattern and replacement vectors, as your original
code did. That can be changed though.
#include Rcpp.h
using namespace Rcpp;
// [[Rcpp::export]]
CharacterVector subCPP(CharacterVector pattern, CharacterVector
replacement, CharacterVector x) {
int len = x.size();
CharacterVector y(len);
int patlen = pattern.size();
int replen = replacement.size();
if (patlen != replen)
RcoutError: Pattern and replacement length do not match;
for(int i = 0; i patlen; ++i) {
if (*(char*)x[i] == *(char*)pattern[i])
y[x[i] == pattern[i]] = replacement[i];
}
return y;
}
CD
system.time(for(i in 1:5) subCPP(patt, repl, X))
user system elapsed
0.160.000.16
Cheers,
Adam
On Wednesday, July 29, 2015 at 2:42:23 PM UTC-7, Adam Erickson wrote:
Further refining the vectorized (within a loop) exact string match
function, I get times below 0.9 seconds while maintaining error checking.
This is accomplished by removing which() and replacing 1:length() with
seq_along().
sub2 - function(pattern, replacement, x) {
len- length(x)
y - character(length=len)
patlen - length(pattern)
replen - length(replacement)
if(patlen != replen) stop('Error: Pattern and replacement length do not
match')
for(i in seq_along(pattern)) {
y[x==pattern[i]] - replacement[i]
}
return(y)
}
system.time(for(i in 1:5) sub2(patt, repl, X))
user system elapsed
0.860.000.86
Since the ordered vectors are perfectly aligned, might as well do an exact
string match. Hence, I think this is not off-topic.
Cheers,
Adam
On Wednesday, July 29, 2015 at 8:15:52 AM UTC-7, Bert Gunter wrote:
There is confusion here. apply()
[R] Using R to fit a curve to a dataset using a specific equation
Hi there I would like to use a specific equation to fit a curve to one of my data sets (attached) dput(data) structure(list(Gossypol = c(1036.331811, 4171.427741, 6039.995102, 5909.068158, 4140.242559, 4854.985845, 6982.035521, 6132.876396, 948.2418407, 3618.448997, 3130.376482, 5113.942098, 1180.171957, 1500.863038, 4576.787021, 5629.979049, 3378.151945, 3589.187889, 2508.417927, 1989.576826, 5972.926124, 2867.610671, 450.7205451, 1120.955, 3470.09352, 3575.043632, 2952.931863, 349.0864019, 1013.807628, 910.8879471, 3743.331903, 3350.203452, 592.3403778, 1517.045807, 1504.491931, 3736.144027, 2818.419785, 723.885643, 1782.864308, 1414.161257, 3723.629772, 3747.076592, 2005.919344, 4198.569251, 2228.522959, 3322.115942, 4274.324792, 720.9785449, 2874.651764, 2287.228752, 5654.858696, 1247.806111, 1247.806111, 2547.326207, 2608.716056, 1079.846532), Treatment = structure(c(2L, 3L, 4L, 5L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 1L), .Label = c(C, 1c_2d, 3c_2d, 9c_2d, 1c_7d), class = factor), Damage_cm = c(0.4955, 1.516, 4.409, 3.2665, 0.491, 2.3035, 3.51, 1.8115, 0, 0.4435, 1.573, 1.8595, 0, 0.142, 2.171, 4.023, 4.9835, 0, 0.6925, 1.989, 5.683, 3.547, 0, 0.756, 2.129, 9.437, 3.211, 0, 0.578, 2.966, 4.7245, 1.8185, 0, 1.0475, 1.62, 5.568, 9.7455, 0, 0.8295, 2.411, 7.272, 4.516, 0, 0.4035, 2.974, 8.043, 4.809, 0, 0.6965, 1.313, 5.681, 3.474, 0, 0.5895, 2.559, 0)), .Names = c(Gossypol, Treatment, Damage_cm), row.names = c(NA, -56L), class = data.frame) The equation is: y~yo+a*(1-b^x) Where: y =Gossypol (from my data set) x= Damage_cm (from my data set) The other 3 parameters are unknown: yo=Intercept, a= assymptote ans b=slope In the end I would like to use the equation to plot a curve (with SE interval, I usually use ggplot2) Furthermore, I would like to know the R2 and p value. I would also be interested in the parameters yo , a and b I have never done this before and would be extremely grateful if anyone could help me? I suppose I have to use a non linear approach (glm(...)). I found out that the mosaic package might be helpful. thanks a lot, Mike [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error when compiling R-2.5.1 / *** [d-p-q-r-tests.Rout] Fehler 1
On 07/31/2015 10:48 PM, Joerg Kirschner wrote: Hi everyone, I am new to Linux and R - but I managed to build R-2.5.1 from source to use it in Genepattern. Genepattern does only support R-2.5.1 which I could not find anywhere for installation via apt-get or in the Ubuntu Software-Centre (I am using Ubuntu 14.04 (Trusty Tahr) 32-bit) Are you sure you want to do this? R 2.5.1 is from 2007, which is a very long time ago. It seems like GenePattern is not restricted to R-2.5.1, http://www.broadinstitute.org/cancer/software/genepattern/administrators-guide#using-different-versions-of-r and if their default distribution uses it, then I'm not sure I'd recommend using GenePattern for new analysis! (Maybe you're trying to re-do a previous analysis?) Since GenePattern modules that use R typically wrap individual CRAN or Bioconductor (http://bioconductor.org) packages, maybe you can take out the middleman ? Martin Morgan But after doing make check I get comparing 'method-dispatch.Rout' to './method-dispatch.Rout.save' ... OK running code in 'd-p-q-r-tests.R' ...make[3]: *** [d-p-q-r-tests.Rout] Fehler 1 make[3]: Verzeichnis »/home/karin/Downloads/R-2.5.1/tests« wird verlassen make[2]: *** [test-Specific] Fehler 2 make[2]: Verzeichnis »/home/karin/Downloads/R-2.5.1/tests« wird verlassen make[1]: *** [test-all-basics] Fehler 1 make[1]: Verzeichnis »/home/karin/Downloads/R-2.5.1/tests« wird verlassen make: *** [check] Fehler 2 but I can make install and use R for simple plots etc. afterwards - still I am worried something is wrong, can you give some advice. A closer look at the error gives ## PR#7099 : pf() with large df1 or df2: nu - 2^seq(25,34, 0.5) y - 1e9*(pf(1,1,nu) - 0.68268949) stopifnot(All.eq(pf(1,1,Inf), 0.68268949213708596), + diff(y) 0, # i.e. pf(1,1, *) is monotone increasing + All.eq(y [1], -5.07420372386491), + All.eq(y[19], 2.12300110824515)) Error: All.eq(y[1], -5.07420372386491) is not TRUE Execution halted As I understand so far some errors are critical some are not - can you please give some advice on the error above? Can I still use R installed with that error? What do I need to solve the error? Thanks, Joerg [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Computational Biology / Fred Hutchinson Cancer Research Center 1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109 Location: Arnold Building M1 B861 Phone: (206) 667-2793 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using R to fit a curve to a dataset using a specific equation
I can get you started, but you should really read up on non-linear least squares. Calling your data frame dta (since data is a function): plot(Gossypol~Damage_cm, dta) # Looking at the plot, 0 is a plausible estimate for y0: # a+y0 is the asymptote, so estimate about 4000; # b is between 0 and 1, so estimate .5 dta.nls - nls(Gossypol~y0+a*(1-b^Damage_cm), dta, start=list(y0=0, a=4000, b=.5)) xval - seq(0, 10, length.out=200) lines(xval, predict(dta.nls, data.frame(Damage_cm=xval))) profile(dta.nls, alpha= .05) === Number of iterations to convergence: 3 Achieved convergence tolerance: 1.750586e-06 attr(,summary) Formula: Gossypol ~ y0 + a * (1 - b^Damage_cm) Parameters: Estimate Std. Error t value Pr(|t|) y0 1303.4529432 386.1515684 3.37550 0.0013853 ** a 2796.0464520 530.4140959 5.27144 2.5359e-06 *** b 0.49391110.1809687 2.72926 0.0085950 ** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 1394.375 on 53 degrees of freedom Number of iterations to convergence: 3 Achieved convergence tolerance: 1.750586e-06 David Carlson Dept of Anthropology Texas AM College Station, TX 77843 From: R-help [r-help-boun...@r-project.org] on behalf of Michael Eisenring [michael.eisenr...@gmx.ch] Sent: Saturday, August 01, 2015 10:17 AM To: r-help@r-project.org Subject: [R] Using R to fit a curve to a dataset using a specific equation Hi there I would like to use a specific equation to fit a curve to one of my data sets (attached) dput(data) structure(list(Gossypol = c(1036.331811, 4171.427741, 6039.995102, 5909.068158, 4140.242559, 4854.985845, 6982.035521, 6132.876396, 948.2418407, 3618.448997, 3130.376482, 5113.942098, 1180.171957, 1500.863038, 4576.787021, 5629.979049, 3378.151945, 3589.187889, 2508.417927, 1989.576826, 5972.926124, 2867.610671, 450.7205451, 1120.955, 3470.09352, 3575.043632, 2952.931863, 349.0864019, 1013.807628, 910.8879471, 3743.331903, 3350.203452, 592.3403778, 1517.045807, 1504.491931, 3736.144027, 2818.419785, 723.885643, 1782.864308, 1414.161257, 3723.629772, 3747.076592, 2005.919344, 4198.569251, 2228.522959, 3322.115942, 4274.324792, 720.9785449, 2874.651764, 2287.228752, 5654.858696, 1247.806111, 1247.806111, 2547.326207, 2608.716056, 1079.846532), Treatment = structure(c(2L, 3L, 4L, 5L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 1L), .Label = c(C, 1c_2d, 3c_2d, 9c_2d, 1c_7d), class = factor), Damage_cm = c(0.4955, 1.516, 4.409, 3.2665, 0.491, 2.3035, 3.51, 1.8115, 0, 0.4435, 1.573, 1.8595, 0, 0.142, 2.171, 4.023, 4.9835, 0, 0.6925, 1.989, 5.683, 3.547, 0, 0.756, 2.129, 9.437, 3.211, 0, 0.578, 2.966, 4.7245, 1.8185, 0, 1.0475, 1.62, 5.568, 9.7455, 0, 0.8295, 2.411, 7.272, 4.516, 0, 0.4035, 2.974, 8.043, 4.809, 0, 0.6965, 1.313, 5.681, 3.474, 0, 0.5895, 2.559, 0)), .Names = c(Gossypol, Treatment, Damage_cm), row.names = c(NA, -56L), class = data.frame) The equation is: y~yo+a*(1-b^x) Where: y =Gossypol (from my data set) x= Damage_cm (from my data set) The other 3 parameters are unknown: yo=Intercept, a= assymptote ans b=slope In the end I would like to use the equation to plot a curve (with SE interval, I usually use ggplot2) Furthermore, I would like to know the R2 and p value. I would also be interested in the parameters yo , a and b I have never done this before and would be extremely grateful if anyone could help me? I suppose I have to use a non linear approach (glm(...)). I found out that the mosaic package might be helpful. thanks a lot, Mike [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] PythonInR. Python script in R with parameters required. Download satellite images from NASA
Magi, is there a reason that you need to run the script via R? If your plan is to download the data via python than then process with R, you might consider using the Rpy2 package to link them. This would allow you to call the downloading code from python and then have python feed the data to R. Collin. On Wed, Jul 29, 2015 at 12:29 PM, Magi Franquesa magifranqu...@gmail.com wrote: Hello, I'm trying to execute a python script within R (3.2.1 x 64) with the PythonInR package. I would like to download an order of satellite images from Nasa using a python script ( http://landsat.usgs.gov/documents/espa_bulk_downloader_v1.0.0.zip) but I have no success. I first run the pyExecfile command with the *feedparser.py* script and then the *download_espa_order.py* giving the required parameters (my mail acount and the order number), here is the code: setwd(C:/Python27) install.packages(PythonInR) library(PythonInR) pyConnect(pythonExePath=C:/Python27/python.exe) pyIsConnected() # autodetectPython(C:/Python27/python.exe) pyExecfile(C:/Landsat/feedparser.py) pyExecfile(C:/Landsat/download_espa_order.py -e magifranqu...@gmail.com -o magifranqu...@gmail.com-07222015-120911 -d C:/Landsat/ESPA) and I get this error: Error: unexpected string constant in pyExecfile(C:/Landsat/download_espa_order.py -e magifranqu...@gmail.com The code C:/Landsat/download_espa_order.py -e magifranqu...@gmail.com -o magifranqu...@gmail.com-07222015-120911 -d C:/Landsat/ESPA runs ok when I use it within system console. I appreciate if someone could help me to solve this problem. Thank you [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] missing data in R
Hi Asena, If you already have microarray data, you can simply change some of the existing values to NA (datum Not Available). Say you have a toy 10x10 array containing absolute (initial) values: array_values-matrix(sample(0:400,100,TRUE),nrow=10) # create a 10% missing array array_values_10-array_values array_values_10[sample(1:100,10)]-NA # next a 20% missing array array_values_20-array_values array_values_20[sample(1:100,20)]-NA # and so on This is possible because a matrix can be indexed as though it was a vector. Jim On Sat, Aug 1, 2015 at 10:21 PM, asena ayça özdemir a.aycaozde...@hotmail.com wrote: Hello Mr. FeldesmanI am a master student in biostatistic my thesis about missing values in microarray data, but ý can't create any values. ý want to create %10, %20,...%90 missing values for all colums in microarray data set . Can you help me any code? thank you for your attention. Asena Ayça ÖzdemirMersin University Biostatistics [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R parallel / foreach - aggregation of results
You can always just pull the last one off the list. When running things in
parallel, what does the last one mean? Do you want the last from each of
the parallel threads, or just the last one on the list? You might want to
put some flag on the data being returned so you can determine which one you
want to process.
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Sat, Aug 1, 2015 at 3:19 PM, Martin Spindler martin.spind...@gmx.de
wrote:
Dear Jim,
Thank you very much for your response. It seems to work now, but the
return value is not the required matrix but a list of matrices (one for
each repition j).
Any idea how it is possible to return only the last matrix and not all?
Thanks and best,
Martin
Gesendet: Freitag, 31. Juli 2015 um 18:22 Uhr
Von: jim holtman jholt...@gmail.com
An: Martin Spindler martin.spind...@gmx.de
Cc: r-help@r-project.org r-help@r-project.org
Betreff: Re: [R] R parallel / foreach - aggregation of results
Try this chance to actually return values:
library(doParallel)
Simpar3 - function(n1) {
L2distance - matrix(NA, ncol=n1, nrow=n1)
data - rnorm(n1)
diag(L2distance)=0
cl - makeCluster(4)
registerDoParallel(cl)
x - foreach(j=1:n1) %dopar% {
library(np)
datj - data[j]
for(k in j:n1) {
L2distance[j,k] - k*datj
}
L2distance # return the value
}
stopCluster(cl)
return(x)
}
Res - Simpar3(100)
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Fri, Jul 31, 2015 at 8:39 AM, Martin Spindler martin.spind...@gmx.de
wrote:Dear all,
when I am running the code attached below, it seems that no results are
returned, only the predefined NAs. What mistake do I make?
Any comments and help is highly appreciated.
Thanks and best,
Martin
Simpar3 - function(n1) {
L2distance - matrix(NA, ncol=n1, nrow=n1)
data - rnorm(n1)
diag(L2distance)=0
cl - makeCluster(4)
registerDoParallel(cl)
foreach(j=1:n1) %dopar% {
library(np)
datj - data[j]
for(k in j:n1) {
L2distance[j,k] - k*datj
}
}
stopCluster(cl)
return(L2distance)
}
Res - Simpar3(100)
__
R-help@r-project.org[R-help@r-project.org] mailing list -- To UNSUBSCRIBE
and more, see
https://stat.ethz.ch/mailman/listinfo/r-help[https://stat.ethz.ch/mailman/listinfo/r-help]
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html[http://www.R-project.org/posting-guide.html]
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] R parallel / foreach - aggregation of results
Dear Jim,
Thank you very much for your response. It seems to work now, but the return
value is not the required matrix but a list of matrices (one for each repition
j).
Any idea how it is possible to return only the last matrix and not all?
Thanks and best,
Martin
Gesendet: Freitag, 31. Juli 2015 um 18:22 Uhr
Von: jim holtman jholt...@gmail.com
An: Martin Spindler martin.spind...@gmx.de
Cc: r-help@r-project.org r-help@r-project.org
Betreff: Re: [R] R parallel / foreach - aggregation of results
Try this chance to actually return values:
library(doParallel)
Simpar3 - function(n1) {
L2distance - matrix(NA, ncol=n1, nrow=n1)
data - rnorm(n1)
diag(L2distance)=0
cl - makeCluster(4)
registerDoParallel(cl)
x - foreach(j=1:n1) %dopar% {
library(np)
datj - data[j]
for(k in j:n1) {
L2distance[j,k] - k*datj
}
L2distance # return the value
}
stopCluster(cl)
return(x)
}
Res - Simpar3(100)
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Fri, Jul 31, 2015 at 8:39 AM, Martin Spindler martin.spind...@gmx.de
wrote:Dear all,
when I am running the code attached below, it seems that no results are
returned, only the predefined NAs. What mistake do I make?
Any comments and help is highly appreciated.
Thanks and best,
Martin
Simpar3 - function(n1) {
L2distance - matrix(NA, ncol=n1, nrow=n1)
data - rnorm(n1)
diag(L2distance)=0
cl - makeCluster(4)
registerDoParallel(cl)
foreach(j=1:n1) %dopar% {
library(np)
datj - data[j]
for(k in j:n1) {
L2distance[j,k] - k*datj
}
}
stopCluster(cl)
return(L2distance)
}
Res - Simpar3(100)
__
R-help@r-project.org[R-help@r-project.org] mailing list -- To UNSUBSCRIBE and
more, see
https://stat.ethz.ch/mailman/listinfo/r-help[https://stat.ethz.ch/mailman/listinfo/r-help]
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html[http://www.R-project.org/posting-guide.html]
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] R parallel / foreach - aggregation of results
If you return just the row that the foreach procedure produces instead of
the entire matrix containing that row and use .combine=rbind then you will
end up with the matrix of interest. E.g.,
Simpar3a - function (n1)
{
L2distance - matrix(NA, ncol = n1, nrow = n1)
data - rnorm(n1)
diag(L2distance) = 0
cl - makeCluster(4)
registerDoParallel(cl)
x - foreach(j = 1:n1, .combine = rbind) %dopar% {
library(np)
datj - data[j]
rowJ - numeric(n1)
for (k in j:n1) {
rowJ[k] - k * datj
}
rowJ
}
stopCluster(cl)
x
}
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sat, Aug 1, 2015 at 12:19 PM, Martin Spindler martin.spind...@gmx.de
wrote:
Dear Jim,
Thank you very much for your response. It seems to work now, but the
return value is not the required matrix but a list of matrices (one for
each repition j).
Any idea how it is possible to return only the last matrix and not all?
Thanks and best,
Martin
Gesendet: Freitag, 31. Juli 2015 um 18:22 Uhr
Von: jim holtman jholt...@gmail.com
An: Martin Spindler martin.spind...@gmx.de
Cc: r-help@r-project.org r-help@r-project.org
Betreff: Re: [R] R parallel / foreach - aggregation of results
Try this chance to actually return values:
library(doParallel)
Simpar3 - function(n1) {
L2distance - matrix(NA, ncol=n1, nrow=n1)
data - rnorm(n1)
diag(L2distance)=0
cl - makeCluster(4)
registerDoParallel(cl)
x - foreach(j=1:n1) %dopar% {
library(np)
datj - data[j]
for(k in j:n1) {
L2distance[j,k] - k*datj
}
L2distance # return the value
}
stopCluster(cl)
return(x)
}
Res - Simpar3(100)
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Fri, Jul 31, 2015 at 8:39 AM, Martin Spindler martin.spind...@gmx.de
wrote:Dear all,
when I am running the code attached below, it seems that no results are
returned, only the predefined NAs. What mistake do I make?
Any comments and help is highly appreciated.
Thanks and best,
Martin
Simpar3 - function(n1) {
L2distance - matrix(NA, ncol=n1, nrow=n1)
data - rnorm(n1)
diag(L2distance)=0
cl - makeCluster(4)
registerDoParallel(cl)
foreach(j=1:n1) %dopar% {
library(np)
datj - data[j]
for(k in j:n1) {
L2distance[j,k] - k*datj
}
}
stopCluster(cl)
return(L2distance)
}
Res - Simpar3(100)
__
R-help@r-project.org[R-help@r-project.org] mailing list -- To UNSUBSCRIBE
and more, see
https://stat.ethz.ch/mailman/listinfo/r-help[https://stat.ethz.ch/mailman/listinfo/r-help]
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html[http://www.R-project.org/posting-guide.html]
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
