Re: [R] solving simultaneous Equations in R
On 11-12-2013, at 23:56, eliza botto eliza_bo...@hotmail.com wrote:
Dear Berend,
Thankyou very much indeed for you reply. By taking help from your previous
reply @ http://r.789695.n4.nabble.com/Simultaneous-equations-td2524645.html i
was able to generate the following loop for the calculation of x=x[1] and
y=x[2].
fun - function(x) {
f - numeric(length(x))
# read as:
f[1] -
1-0.514-(gamma(1/x[1])*gamma(2/x[1]-1/x[2]))/(gamma(2/x[1])*gamma(1/x[1]-1/x[2]))
f[2] - 0.57-
(gamma(1/x[1]-1/x[2])/gamma(1/x[1])-3*gamma(2/x[1]-1/x[2])/gamma(2/x[1])+2*gamma(3/x[1]-1/x[2])/gamma(3/x[1]))/(gamma(1/x[1]-1/x[2])/gamma(1
/x[1])-gamma(2/x[1]-1/x[2])/gamma(2/x[1]))
f
}
startx - c(0.1,0.15) # start the answer search here
answers-as.data.frame(nleqslv(startx,fun))
answers
What i cant understand is the concept involved for setting startx. my x[1]
should always be smaller than x[2] and they both should be less than 1. how
can i demonstrate it to startx command line?
I don’t quite understand what you you mean.
Your starting values obey the restrictions you specify (if that is what you
meant).
The solution vector has all elements 1. But it is a solution.
Do you mean that you want a solution satisfying the constraints you mention?
I cannot tell if that is possible.
Generally speaking a square system of equations is solved or not.
Sometimes you can vary the starting values to get a different solution that
obeys the specified constraints.
If these constraints are necessary you are not solving a system of equations
but trying to find a parameter set that satisfies certain criteria. Together
with a criterion (sum of squares of function values for example) you could use
an optimizing algorithm (optim, nlmin, constrOptim to name a few).
Berend
thanks for your help. I m grateful.
Eliza
Subject: Re: [R] solving simultaneous Equations in R
From: b...@xs4all.nl
Date: Wed, 11 Dec 2013 12:43:02 +0100
CC: r-help@r-project.org
To: eliza_bo...@hotmail.com
On 11-12-2013, at 12:16, eliza botto eliza_bo...@hotmail.com wrote:
Dear users of R,
I'm trying to solve the following 2 equations simultaneously in R for x
and y. I couldn't get through due to my limited knowledge of R.
3=1-[(x-1)!(2x-y-1)!/(2x-1)!(x-y-1)!]
6={[(x-y-1)!/(x-1)!]-[3(2x-y-1)!/(2x-1)!]+[2(3x-y-1)!/(3x-1)!]}/{[(x-y-1)!/(x-1)!]-[(2x-y-1)!/(2x-1)!]}
obviously, ! is factorial.
kindly help me out on it or at least suggest something.
I'll be extremely grateful.
There are several packages that solve a system of equations.
ktsolve, nleqslv, BB, which you can find in CRAN Task views: Numerical
Mathematics” and “Optimization”.
You will have to write your equations in standard R notation.
I can’t tell if your system is solvable.
Berend
Eliza
[[alternative HTML version deleted]]
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Re: [R] refline in forest() {metafor}
Regarding your question: Do you want *another* line or do you just want to move
the reference line to the value of the summary estimate? The latter can be done
by passing the value of the summary estimate to the 'refline' argument. If you
want another line, you could just use the abline function, or, for finer
control, the segments function. Something like:
segments(coef(res), 0, coef(res), res$k, lty=dashed)
where 'res' is the name of the fitted model object. You may have to play around
with the 0 and res$k values, so that the line has the 'correct' length.
Thanks for the positive feedback about the package.
Best,
Wolfgang
--
Wolfgang Viechtbauer, Ph.D., Statistician
Department of Psychiatry and Psychology
School for Mental Health and Neuroscience
Faculty of Health, Medicine, and Life Sciences
Maastricht University, P.O. Box 616 (VIJV1)
6200 MD Maastricht, The Netherlands
+31 (43) 388-4170 | http://www.wvbauer.com
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
johnwilli...@fas.harvard.edu [johnwilli...@fas.harvard.edu]
Sent: Thursday, December 12, 2013 6:49 AM
To: r-help@r-project.org
Subject: [R] refline in forest() {metafor}
Hello all,
I am using forest.rma to plot a random effects model meta-analysis. I noticed
that refline sets a vertical line indicating the null hypothesis.
Is there a way to draw another vertical line, possibly dashed, centered on the
summary estimate?
Prof. Viechtbauer, if you happen to read this, I'd like to thank you for making
an excellent package. I have been using the metafor package to do my first
meta-analysis, having never used R before. The documentation is thorough and
intuitive.
Thanks,
John
John Williams
ALB Candidate
Harvard University Extension School
johnwilli...@fas.harvard.edu
__
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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.
__
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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] refline in forest() {metafor}
I would like another line, and the solution below will work great. Thank you.
- John
Quoting Viechtbauer Wolfgang (STAT)
wolfgang.viechtba...@maastrichtuniversity.nl:
Regarding your question: Do you want *another* line or do you just want to
move the reference line to the value of the summary estimate? The latter can
be done by passing the value of the summary estimate to the 'refline'
argument. If you want another line, you could just use the abline function,
or, for finer control, the segments function. Something like:
segments(coef(res), 0, coef(res), res$k, lty=dashed)
where 'res' is the name of the fitted model object. You may have to play
around with the 0 and res$k values, so that the line has the 'correct'
length.
Thanks for the positive feedback about the package.
Best,
Wolfgang
--
Wolfgang Viechtbauer, Ph.D., Statistician
Department of Psychiatry and Psychology
School for Mental Health and Neuroscience
Faculty of Health, Medicine, and Life Sciences
Maastricht University, P.O. Box 616 (VIJV1)
6200 MD Maastricht, The Netherlands
+31 (43) 388-4170 | http://www.wvbauer.com
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf
Of johnwilli...@fas.harvard.edu [johnwilli...@fas.harvard.edu]
Sent: Thursday, December 12, 2013 6:49 AM
To: r-help@r-project.org
Subject: [R] refline in forest() {metafor}
Hello all,
I am using forest.rma to plot a random effects model meta-analysis. I noticed
that refline sets a vertical line indicating the null hypothesis.
Is there a way to draw another vertical line, possibly dashed, centered on
the
summary estimate?
Prof. Viechtbauer, if you happen to read this, I'd like to thank you for
making
an excellent package. I have been using the metafor package to do my first
meta-analysis, having never used R before. The documentation is thorough and
intuitive.
Thanks,
John
John Williams
ALB Candidate
Harvard University Extension School
johnwilli...@fas.harvard.edu
__
R-help@r-project.org mailing list
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
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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.
[R] Logistic Regression with 200K features in R?
Dear List,
I'm quite new to R and want to do logistic regression with a 200K
feature data set (around 150 training examples).
I'm aware that I should use Naive Bayes but I have a more general
question about the capability of R handling very high dimensional data.
Please consider the following R code where mygenestrain.tab is a 150
by 20 matrix:
traindata - read.table('mygenestrain.tab');
mylogit - glm(V1 ~ ., data = traindata, family = binomial);
When executing this code I get the following error:
Error in terms.formula(formula, data = data) :
allocMatrix: too many elements specified
Calls: glm ... model.frame - model.frame.default - terms - terms.formula
Execution halted
Is this because R can't handle 200K features or am I doing something
completely wrong here?
Thanks a lot for your help!
best Regards,
Romeo
__
R-help@r-project.org mailing list
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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] Logistic Regression with 200K features in R?
it is simply because you can't do a regression with more predictors than
observations.
Cheers.
Am 12.12.2013 09:00, schrieb Romeo Kienzler:
Dear List,
I'm quite new to R and want to do logistic regression with a 200K
feature data set (around 150 training examples).
I'm aware that I should use Naive Bayes but I have a more general
question about the capability of R handling very high dimensional data.
Please consider the following R code where mygenestrain.tab is a 150
by 20 matrix:
traindata - read.table('mygenestrain.tab');
mylogit - glm(V1 ~ ., data = traindata, family = binomial);
When executing this code I get the following error:
Error in terms.formula(formula, data = data) :
allocMatrix: too many elements specified
Calls: glm ... model.frame - model.frame.default - terms - terms.formula
Execution halted
Is this because R can't handle 200K features or am I doing something
completely wrong here?
Thanks a lot for your help!
best Regards,
Romeo
__
R-help@r-project.org mailing list
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.
--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Besuchen Sie uns auf: www.uke.de
_
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Christian Gerloff (Vertreter des Vorsitzenden),
Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Rainer Schoppik
_
SAVE PAPER - THINK BEFORE PRINTING
__
R-help@r-project.org mailing list
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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] solving simultaneous Equations in R
Dear Berend,Got it! Thankyou very much indeed for your kind support and
elaborating the issue.I'm Extremely grateful!!! Eliza
Subject: Re: [R] solving simultaneous Equations in R
From: b...@xs4all.nl
Date: Thu, 12 Dec 2013 09:01:02 +0100
CC: r-help@r-project.org
To: eliza_bo...@hotmail.com
On 11-12-2013, at 23:56, eliza botto eliza_bo...@hotmail.com wrote:
Dear Berend,
Thankyou very much indeed for you reply. By taking help from your previous
reply @ http://r.789695.n4.nabble.com/Simultaneous-equations-td2524645.html
i was able to generate the following loop for the calculation of x=x[1] and
y=x[2].
fun - function(x) {
f - numeric(length(x))
# read as:
f[1] -
1-0.514-(gamma(1/x[1])*gamma(2/x[1]-1/x[2]))/(gamma(2/x[1])*gamma(1/x[1]-1/x[2]))
f[2] - 0.57-
(gamma(1/x[1]-1/x[2])/gamma(1/x[1])-3*gamma(2/x[1]-1/x[2])/gamma(2/x[1])+2*gamma(3/x[1]-1/x[2])/gamma(3/x[1]))/(gamma(1/x[1]-1/x[2])/gamma(1
/x[1])-gamma(2/x[1]-1/x[2])/gamma(2/x[1]))
f
}
startx - c(0.1,0.15) # start the answer search here
answers-as.data.frame(nleqslv(startx,fun))
answers
What i cant understand is the concept involved for setting startx. my
x[1] should always be smaller than x[2] and they both should be less than
1. how can i demonstrate it to startx command line?
I dont quite understand what you you mean.
Your starting values obey the restrictions you specify (if that is what you
meant).
The solution vector has all elements 1. But it is a solution.
Do you mean that you want a solution satisfying the constraints you mention?
I cannot tell if that is possible.
Generally speaking a square system of equations is solved or not.
Sometimes you can vary the starting values to get a different solution that
obeys the specified constraints.
If these constraints are necessary you are not solving a system of equations
but trying to find a parameter set that satisfies certain criteria. Together
with a criterion (sum of squares of function values for example) you could
use an optimizing algorithm (optim, nlmin, constrOptim to name a few).
Berend
thanks for your help. I m grateful.
Eliza
Subject: Re: [R] solving simultaneous Equations in R
From: b...@xs4all.nl
Date: Wed, 11 Dec 2013 12:43:02 +0100
CC: r-help@r-project.org
To: eliza_bo...@hotmail.com
On 11-12-2013, at 12:16, eliza botto eliza_bo...@hotmail.com wrote:
Dear users of R,
I'm trying to solve the following 2 equations simultaneously in R for
x and y. I couldn't get through due to my limited knowledge of R.
3=1-[(x-1)!(2x-y-1)!/(2x-1)!(x-y-1)!]
6={[(x-y-1)!/(x-1)!]-[3(2x-y-1)!/(2x-1)!]+[2(3x-y-1)!/(3x-1)!]}/{[(x-y-1)!/(x-1)!]-[(2x-y-1)!/(2x-1)!]}
obviously, ! is factorial.
kindly help me out on it or at least suggest something.
I'll be extremely grateful.
There are several packages that solve a system of equations.
ktsolve, nleqslv, BB, which you can find in CRAN Task views: Numerical
Mathematics and Optimization.
You will have to write your equations in standard R notation.
I cant tell if your system is solvable.
Berend
Eliza
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
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
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] Logistic Regression with 200K features in R?
I thought so (with all the limitations due to collinearity and so on),
but actually there is a limit for the maximum size of an array which is
independent of your memory size and is due to the way arrays are
indexed. You can't create an object with more than 2^31-1 = 2147483647
elements.
https://stat.ethz.ch/pipermail/r-help/2007-June/133238.html
cheers
Am 12.12.2013 12:34, schrieb Romeo Kienzler:
ok, so 200K predictors an 10M observations would work?
On 12/12/2013 12:12 PM, Eik Vettorazzi wrote:
it is simply because you can't do a regression with more predictors than
observations.
Cheers.
Am 12.12.2013 09:00, schrieb Romeo Kienzler:
Dear List,
I'm quite new to R and want to do logistic regression with a 200K
feature data set (around 150 training examples).
I'm aware that I should use Naive Bayes but I have a more general
question about the capability of R handling very high dimensional data.
Please consider the following R code where mygenestrain.tab is a 150
by 20 matrix:
traindata - read.table('mygenestrain.tab');
mylogit - glm(V1 ~ ., data = traindata, family = binomial);
When executing this code I get the following error:
Error in terms.formula(formula, data = data) :
allocMatrix: too many elements specified
Calls: glm ... model.frame - model.frame.default - terms -
terms.formula
Execution halted
Is this because R can't handle 200K features or am I doing something
completely wrong here?
Thanks a lot for your help!
best Regards,
Romeo
__
R-help@r-project.org mailing list
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.
--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Besuchen Sie uns auf: www.uke.de
_
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Christian Gerloff (Vertreter des Vorsitzenden),
Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Rainer Schoppik
_
SAVE PAPER - THINK BEFORE PRINTING
__
R-help@r-project.org mailing list
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] Logistic Regression with 200K features in R?
On 13-12-12 6:51 AM, Eik Vettorazzi wrote:
I thought so (with all the limitations due to collinearity and so on),
but actually there is a limit for the maximum size of an array which is
independent of your memory size and is due to the way arrays are
indexed. You can't create an object with more than 2^31-1 = 2147483647
elements.
https://stat.ethz.ch/pipermail/r-help/2007-June/133238.html
That post is from 2007. The limits were raised considerably when R
3.0.0 was released, and it is now 2^48 for disk-based operations, 2^52
for working in memory.
Duncan Murdoch
cheers
Am 12.12.2013 12:34, schrieb Romeo Kienzler:
ok, so 200K predictors an 10M observations would work?
On 12/12/2013 12:12 PM, Eik Vettorazzi wrote:
it is simply because you can't do a regression with more predictors than
observations.
Cheers.
Am 12.12.2013 09:00, schrieb Romeo Kienzler:
Dear List,
I'm quite new to R and want to do logistic regression with a 200K
feature data set (around 150 training examples).
I'm aware that I should use Naive Bayes but I have a more general
question about the capability of R handling very high dimensional data.
Please consider the following R code where mygenestrain.tab is a 150
by 20 matrix:
traindata - read.table('mygenestrain.tab');
mylogit - glm(V1 ~ ., data = traindata, family = binomial);
When executing this code I get the following error:
Error in terms.formula(formula, data = data) :
allocMatrix: too many elements specified
Calls: glm ... model.frame - model.frame.default - terms -
terms.formula
Execution halted
Is this because R can't handle 200K features or am I doing something
completely wrong here?
Thanks a lot for your help!
best Regards,
Romeo
__
R-help@r-project.org mailing list
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
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] Logistic Regression with 200K features in R?
thanks Duncan for this clarification.
A double precision matrix with 2e11 elements (as the op wanted) would
need about 1.5 TB memory, that's more than a standard (windows 64bit)
computer can handle.
Cheers.
Am 12.12.2013 13:00, schrieb Duncan Murdoch:
On 13-12-12 6:51 AM, Eik Vettorazzi wrote:
I thought so (with all the limitations due to collinearity and so on),
but actually there is a limit for the maximum size of an array which is
independent of your memory size and is due to the way arrays are
indexed. You can't create an object with more than 2^31-1 = 2147483647
elements.
https://stat.ethz.ch/pipermail/r-help/2007-June/133238.html
That post is from 2007. The limits were raised considerably when R
3.0.0 was released, and it is now 2^48 for disk-based operations, 2^52
for working in memory.
Duncan Murdoch
cheers
Am 12.12.2013 12:34, schrieb Romeo Kienzler:
ok, so 200K predictors an 10M observations would work?
On 12/12/2013 12:12 PM, Eik Vettorazzi wrote:
it is simply because you can't do a regression with more predictors
than
observations.
Cheers.
Am 12.12.2013 09:00, schrieb Romeo Kienzler:
Dear List,
I'm quite new to R and want to do logistic regression with a 200K
feature data set (around 150 training examples).
I'm aware that I should use Naive Bayes but I have a more general
question about the capability of R handling very high dimensional
data.
Please consider the following R code where mygenestrain.tab is a 150
by 20 matrix:
traindata - read.table('mygenestrain.tab');
mylogit - glm(V1 ~ ., data = traindata, family = binomial);
When executing this code I get the following error:
Error in terms.formula(formula, data = data) :
allocMatrix: too many elements specified
Calls: glm ... model.frame - model.frame.default - terms -
terms.formula
Execution halted
Is this because R can't handle 200K features or am I doing something
completely wrong here?
Thanks a lot for your help!
best Regards,
Romeo
__
R-help@r-project.org mailing list
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.
--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Besuchen Sie uns auf: www.uke.de
_
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Christian Gerloff (Vertreter des Vorsitzenden),
Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Rainer Schoppik
_
SAVE PAPER - THINK BEFORE PRINTING
__
R-help@r-project.org mailing list
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] Solving a normal distribution pnorm for q
Hi, I found follwowing example of pnorm here: http://www.r-tutor.com/elementary-statistics/probability-distributions/normal-distribution Problem Assume that the test scores of a college entrance exam fits a normal distribution. Furthermore, the mean test score is 72, and the standard deviation is 15.2. What is the percentage of students scoring 84 or more in the exam? Solution pnorm(84, mean=72, sd=15.2, lower.tail=FALSE) [1] 0.21492 That is straight forward, however what if I want to know the score the best 30% students are reaching at least. So I know the solution of pnorm but want to know its q. How can that be achieved in R? Any suggestions? /Johannes [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] Solving a normal distribution pnorm for q
On 12-12-2013, at 14:56, Johannes Radinger johannesradin...@gmail.com wrote: Hi, I found follwowing example of pnorm here: http://www.r-tutor.com/elementary-statistics/probability-distributions/normal-distribution Problem Assume that the test scores of a college entrance exam fits a normal distribution. Furthermore, the mean test score is 72, and the standard deviation is 15.2. What is the percentage of students scoring 84 or more in the exam? Solution pnorm(84, mean=72, sd=15.2, lower.tail=FALSE) [1] 0.21492 That is straight forward, however what if I want to know the score the best 30% students are reaching at least. So I know the solution of pnorm but want to know its q. How can that be achieved in R? Any suggestions? Do ?pnorm in R. You are directed to a help page where you will also see “qnorm”. Then try qnorm. Berend __ R-help@r-project.org mailing list 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] Solving a normal distribution pnorm for q
Looks like homework. Try ?qnorm Feladó: r-help-boun...@r-project.org [r-help-boun...@r-project.org] ; meghatalmaz#243;: Johannes Radinger [johannesradin...@gmail.com] Küldve: 2013. december 12. 14:56 To: R help Tárgy: [R] Solving a normal distribution pnorm for q Hi, I found follwowing example of pnorm here: http://www.r-tutor.com/elementary-statistics/probability-distributions/normal-distribution Problem Assume that the test scores of a college entrance exam fits a normal distribution. Furthermore, the mean test score is 72, and the standard deviation is 15.2. What is the percentage of students scoring 84 or more in the exam? Solution pnorm(84, mean=72, sd=15.2, lower.tail=FALSE) [1] 0.21492 That is straight forward, however what if I want to know the score the best 30% students are reaching at least. So I know the solution of pnorm but want to know its q. How can that be achieved in R? Any suggestions? /Johannes [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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 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] Solving a normal distribution pnorm for q
?pnorm ... carefully... --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Johannes Radinger johannesradin...@gmail.com wrote: Hi, I found follwowing example of pnorm here: http://www.r-tutor.com/elementary-statistics/probability-distributions/normal-distribution Problem Assume that the test scores of a college entrance exam fits a normal distribution. Furthermore, the mean test score is 72, and the standard deviation is 15.2. What is the percentage of students scoring 84 or more in the exam? Solution pnorm(84, mean=72, sd=15.2, lower.tail=FALSE) [1] 0.21492 That is straight forward, however what if I want to know the score the best 30% students are reaching at least. So I know the solution of pnorm but want to know its q. How can that be achieved in R? Any suggestions? /Johannes [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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 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] labels on right y-axis
OS X R 3.0.2 Colleagues, I am displaying two sets of values changing over time; as a result, I have two y-axes. I add a label for the right-side Y axis with mtext(side=3, line=1.2, TEXT). The text is parallel to the axis -- so far, so good. However, the text is rotated counterclockwise from horizontal, whereas I would like it rotated clockwise. Neither srt nor las fixes this. I guess that one convoluted approach would be to use text rather than mtext. However, positioning would be a nuisance and the xpd option would be needed. Is there any simpler approach? Dennis Dennis Fisher MD P (The P Less Than Company) Phone: 1-866-PLessThan (1-866-753-7784) Fax: 1-866-PLessThan (1-866-753-7784) www.PLessThan.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] Solving a normal distribution pnorm for q
Thanks for the fast response. Of course I totally overlooked qnorm as I had a more complex task in my head. I wanted to reverse following equation: r=0.67 q=-150 sd1=100 sd2=1000 X - pnorm(q=q,sd=sd1)*r+pnorm(q=q,sd=sd2)*(1-r) Maybe its mathematically really easy, but somehow I don't get it how to do reverse and provide X and get q especially with the presence of r respectively as weighting factors for the two distributions. /J On Thu, Dec 12, 2013 at 3:09 PM, Dániel Kehl ke...@ktk.pte.hu wrote: Looks like homework. Try ?qnorm Feladó: r-help-boun...@r-project.org [r-help-boun...@r-project.org] ; meghatalmaz#243;: Johannes Radinger [johannesradin...@gmail.com] Küldve: 2013. december 12. 14:56 To: R help Tárgy: [R] Solving a normal distribution pnorm for q Hi, I found follwowing example of pnorm here: http://www.r-tutor.com/elementary-statistics/probability-distributions/normal-distribution Problem Assume that the test scores of a college entrance exam fits a normal distribution. Furthermore, the mean test score is 72, and the standard deviation is 15.2. What is the percentage of students scoring 84 or more in the exam? Solution pnorm(84, mean=72, sd=15.2, lower.tail=FALSE) [1] 0.21492 That is straight forward, however what if I want to know the score the best 30% students are reaching at least. So I know the solution of pnorm but want to know its q. How can that be achieved in R? Any suggestions? /Johannes [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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 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] radial.plot shaded region
From: Jim Lemon [j...@bitwrit.com.au] Sent: Tuesday, December 10, 2013 21:57 To: Hodge, Steven Cc: r-help@r-project.org Subject: Re: [R] radial.plot shaded region On 12/11/2013 11:24 AM, Jim Lemon wrote: ... There may be a solution using the radial.pie function, and if I find it, I'll post it. Hi Steve, Here it is. Just call radial.pie twice to get the annulus, then call radial.grid with the appropriate arguments, and then you can add whatever radial.plot you want, using the add argument. radial.pie(3.5,sector.colors=lightblue) radial.pie(3,sector.colors=white,add=TRUE) radial.grid(radial.lim=c(0,3.5),grid.pos=seq(0,3.5,length.out=8)) Jim Thank you very much! Both of your solutions work for me. Steve __ R-help@r-project.org mailing list 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] Solving a normal distribution pnorm for q
Assuming r, sd1 and sd2 as known(fixed) components, this works: X - pnorm(q=q,sd=sd1)*r+pnorm(q=q,sd=sd2)*(1-r) uniroot(function(x)pnorm(q=x,sd=sd1)*r+pnorm(q=x,sd=sd2)*(1-r)-X,c(-1e6,1e6)) Cheers Am 12.12.2013 15:58, schrieb Johannes Radinger: Thanks for the fast response. Of course I totally overlooked qnorm as I had a more complex task in my head. I wanted to reverse following equation: r=0.67 q=-150 sd1=100 sd2=1000 X - pnorm(q=q,sd=sd1)*r+pnorm(q=q,sd=sd2)*(1-r) Maybe its mathematically really easy, but somehow I don't get it how to do reverse and provide X and get q especially with the presence of r respectively as weighting factors for the two distributions. /J On Thu, Dec 12, 2013 at 3:09 PM, Dániel Kehl ke...@ktk.pte.hu wrote: Looks like homework. Try ?qnorm Feladó: r-help-boun...@r-project.org [r-help-boun...@r-project.org] ; meghatalmaz#243;: Johannes Radinger [johannesradin...@gmail.com] Küldve: 2013. december 12. 14:56 To: R help Tárgy: [R] Solving a normal distribution pnorm for q Hi, I found follwowing example of pnorm here: http://www.r-tutor.com/elementary-statistics/probability-distributions/normal-distribution Problem Assume that the test scores of a college entrance exam fits a normal distribution. Furthermore, the mean test score is 72, and the standard deviation is 15.2. What is the percentage of students scoring 84 or more in the exam? Solution pnorm(84, mean=72, sd=15.2, lower.tail=FALSE) [1] 0.21492 That is straight forward, however what if I want to know the score the best 30% students are reaching at least. So I know the solution of pnorm but want to know its q. How can that be achieved in R? Any suggestions? /Johannes [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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 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. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Besuchen Sie uns auf: www.uke.de _ Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Christian Gerloff (Vertreter des Vorsitzenden), Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Rainer Schoppik _ SAVE PAPER - THINK BEFORE PRINTING __ R-help@r-project.org mailing list 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] censored counts and glmer/glmmADMB
dear R-users, I have to model counts where all counts above some threshold have been censored. In the same dataset I have too many zeroes for a Poisson or even a negative binomial distribution to make sense, so I would need a zero-inflated-censored negative binomial family for use in glmer (or glmmADMB?). That seems not to exist. my question is : how could I add a custom-built family of distributions that I could call in glmer/glmmADMM ? if it's not possible, I am considering imputing fake values to replace the censored ones, but I am unsure whether this is bad or very bad... Eric Elguero MIVEGEC (UM1- UM2 -CNRS 5290-IRD 224) Maladies infectieuses et vecteurs : écologie, génétique, évolution et contrôle Centre IRD de Montpellier 911 Av Agropolis - BP 64501 34394 Montpellier Cedex __ R-help@r-project.org mailing list 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] Solving a normal distribution pnorm for q
Thanks Eik... ..the uniroot() was the function I was looking for. Best regards, /J On Thu, Dec 12, 2013 at 4:27 PM, Eik Vettorazzi e.vettora...@uke.de wrote: Assuming r, sd1 and sd2 as known(fixed) components, this works: X - pnorm(q=q,sd=sd1)*r+pnorm(q=q,sd=sd2)*(1-r) uniroot(function(x)pnorm(q=x,sd=sd1)*r+pnorm(q=x,sd=sd2)*(1-r)-X,c(-1e6,1e6)) Cheers Am 12.12.2013 15:58, schrieb Johannes Radinger: Thanks for the fast response. Of course I totally overlooked qnorm as I had a more complex task in my head. I wanted to reverse following equation: r=0.67 q=-150 sd1=100 sd2=1000 X - pnorm(q=q,sd=sd1)*r+pnorm(q=q,sd=sd2)*(1-r) Maybe its mathematically really easy, but somehow I don't get it how to do reverse and provide X and get q especially with the presence of r respectively as weighting factors for the two distributions. /J On Thu, Dec 12, 2013 at 3:09 PM, Dániel Kehl ke...@ktk.pte.hu wrote: Looks like homework. Try ?qnorm Feladó: r-help-boun...@r-project.org [r-help-boun...@r-project.org] ; meghatalmaz#243;: Johannes Radinger [johannesradin...@gmail.com] Küldve: 2013. december 12. 14:56 To: R help Tárgy: [R] Solving a normal distribution pnorm for q Hi, I found follwowing example of pnorm here: http://www.r-tutor.com/elementary-statistics/probability-distributions/normal-distribution Problem Assume that the test scores of a college entrance exam fits a normal distribution. Furthermore, the mean test score is 72, and the standard deviation is 15.2. What is the percentage of students scoring 84 or more in the exam? Solution pnorm(84, mean=72, sd=15.2, lower.tail=FALSE) [1] 0.21492 That is straight forward, however what if I want to know the score the best 30% students are reaching at least. So I know the solution of pnorm but want to know its q. How can that be achieved in R? Any suggestions? /Johannes [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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 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. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Besuchen Sie uns auf: www.uke.de _ Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Christian Gerloff (Vertreter des Vorsitzenden), Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Rainer Schoppik _ SAVE PAPER - THINK BEFORE PRINTING [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] censored counts and glmer/glmmADMB
This post should go the r-sig-mixed-models list, where you are much more likely to get useful help than here, which is a general R programming help list. Cheers, Bert On Thu, Dec 12, 2013 at 7:41 AM, Eric Elguero eric.elgu...@ird.fr wrote: dear R-users, I have to model counts where all counts above some threshold have been censored. In the same dataset I have too many zeroes for a Poisson or even a negative binomial distribution to make sense, so I would need a zero-inflated-censored negative binomial family for use in glmer (or glmmADMB?). That seems not to exist. my question is : how could I add a custom-built family of distributions that I could call in glmer/glmmADMM ? if it's not possible, I am considering imputing fake values to replace the censored ones, but I am unsure whether this is bad or very bad... Eric Elguero MIVEGEC (UM1- UM2 -CNRS 5290-IRD 224) Maladies infectieuses et vecteurs : écologie, génétique, évolution et contrôle Centre IRD de Montpellier 911 Av Agropolis - BP 64501 34394 Montpellier Cedex __ R-help@r-project.org mailing list 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. -- Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 __ R-help@r-project.org mailing list 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] labels on right y-axis
I don't see an alternative to text(), but the positioning is not that difficult: oldpar - par(mar=c(5.1, 4.1, 4.1, 4.1)) plot(0) axis(4) text(par(usr)[2], mean(par(usr)[3:4]), TEXT, srt=-90, adj=c(.5,-4), xpd=TRUE) par(oldpar) - David L Carlson Department of Anthropology Texas AM University College Station, TX 77840-4352 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Fisher Dennis Sent: Thursday, December 12, 2013 8:32 AM To: r-h...@stat.math.ethz.ch Subject: [R] labels on right y-axis OS X R 3.0.2 Colleagues, I am displaying two sets of values changing over time; as a result, I have two y-axes. I add a label for the right-side Y axis with mtext(side=3, line=1.2, TEXT). The text is parallel to the axis -- so far, so good. However, the text is rotated counterclockwise from horizontal, whereas I would like it rotated clockwise. Neither srt nor las fixes this. I guess that one convoluted approach would be to use text rather than mtext. However, positioning would be a nuisance and the xpd option would be needed. Is there any simpler approach? Dennis Dennis Fisher MD P (The P Less Than Company) Phone: 1-866-PLessThan (1-866-753-7784) Fax: 1-866-PLessThan (1-866-753-7784) www.PLessThan.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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 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] Logistic Regression with 200K features in R?
On 12/12/2013 7:08 AM, Eik Vettorazzi wrote:
thanks Duncan for this clarification.
A double precision matrix with 2e11 elements (as the op wanted) would
need about 1.5 TB memory, that's more than a standard (windows 64bit)
computer can handle.
According to Microsoft's Memory Limits web page (currently at
http://msdn.microsoft.com/en-us/library/windows/desktop/aa366778%28v=vs.85%29.aspx#memory_limits,
but these things tend to move around), the limit is 8 TB for virtual
memory.(The same page lists a variety of smaller physical memory
limits, depending on the Windows version, but R doesn't need physical
memory, virtual is good enough. )
R would be very slow if it was working with objects bigger than physical
memory, but it could conceivably work.
Duncan Murdoch
Cheers.
Am 12.12.2013 13:00, schrieb Duncan Murdoch:
On 13-12-12 6:51 AM, Eik Vettorazzi wrote:
I thought so (with all the limitations due to collinearity and so on),
but actually there is a limit for the maximum size of an array which is
independent of your memory size and is due to the way arrays are
indexed. You can't create an object with more than 2^31-1 = 2147483647
elements.
https://stat.ethz.ch/pipermail/r-help/2007-June/133238.html
That post is from 2007. The limits were raised considerably when R
3.0.0 was released, and it is now 2^48 for disk-based operations, 2^52
for working in memory.
Duncan Murdoch
cheers
Am 12.12.2013 12:34, schrieb Romeo Kienzler:
ok, so 200K predictors an 10M observations would work?
On 12/12/2013 12:12 PM, Eik Vettorazzi wrote:
it is simply because you can't do a regression with more predictors
than
observations.
Cheers.
Am 12.12.2013 09:00, schrieb Romeo Kienzler:
Dear List,
I'm quite new to R and want to do logistic regression with a 200K
feature data set (around 150 training examples).
I'm aware that I should use Naive Bayes but I have a more general
question about the capability of R handling very high dimensional
data.
Please consider the following R code where mygenestrain.tab is a 150
by 20 matrix:
traindata - read.table('mygenestrain.tab');
mylogit - glm(V1 ~ ., data = traindata, family = binomial);
When executing this code I get the following error:
Error in terms.formula(formula, data = data) :
allocMatrix: too many elements specified
Calls: glm ... model.frame - model.frame.default - terms -
terms.formula
Execution halted
Is this because R can't handle 200K features or am I doing something
completely wrong here?
Thanks a lot for your help!
best Regards,
Romeo
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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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.
[R] function inside a function
Dear users of R,
I am trying to inculcate a function inside a function. For that to be done, i
copied following function from internet.
library(nleqslv) fun - function(x) {
f - numeric(length(x))
f[1] - A[,1]+x[2] - 1/x[1]
f[2] - A[,2]+x[2] - sin(x[1])
f
}
x.start - c(1,1)
nleqslv(x.start,fun)
I have a matrix A with dimension 124 rows and 2 columns. In f[1] line, in
place of A[,1] i want to inculcate each value (each row) of
column 1 of matrix A. while doing it, In f[2] line, in place of A[,2] i want to
inculcate each value (each row) of column 2 of matrix A.
For suppose A has following rows
0.6772941 0.5962983
0.4348938 0.4563702
0.4481236 0.4418828
0.4213013 0.3944993
0.4682232 0.4485623
0.4798529 0.4477387
0.7029005 0.5533228
While using 0.66772941 in f[1], use 0.5962983 in f[2].
How can i do this in R.
Thanks in advance,
Eliza
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] labels on right y-axis
Might axis() be useful here? ?axis Cheers, Bert On Thu, Dec 12, 2013 at 8:02 AM, David Carlson dcarl...@tamu.edu wrote: I don't see an alternative to text(), but the positioning is not that difficult: oldpar - par(mar=c(5.1, 4.1, 4.1, 4.1)) plot(0) axis(4) text(par(usr)[2], mean(par(usr)[3:4]), TEXT, srt=-90, adj=c(.5,-4), xpd=TRUE) par(oldpar) - David L Carlson Department of Anthropology Texas AM University College Station, TX 77840-4352 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Fisher Dennis Sent: Thursday, December 12, 2013 8:32 AM To: r-h...@stat.math.ethz.ch Subject: [R] labels on right y-axis OS X R 3.0.2 Colleagues, I am displaying two sets of values changing over time; as a result, I have two y-axes. I add a label for the right-side Y axis with mtext(side=3, line=1.2, TEXT). The text is parallel to the axis -- so far, so good. However, the text is rotated counterclockwise from horizontal, whereas I would like it rotated clockwise. Neither srt nor las fixes this. I guess that one convoluted approach would be to use text rather than mtext. However, positioning would be a nuisance and the xpd option would be needed. Is there any simpler approach? Dennis Dennis Fisher MD P (The P Less Than Company) Phone: 1-866-PLessThan (1-866-753-7784) Fax: 1-866-PLessThan (1-866-753-7784) www.PLessThan.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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 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. -- Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 __ R-help@r-project.org mailing list 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] Heatmap, and heatmap.2 gave different figures for the same dataset
I have a huge dataset(15k X 18) and tried to use the heatmap in R to examine the patterns. However, I found that heatmap and heatmap.2 gave me completely different outputs. Here are the codes: dim(as.matrix(data.dcpm)) [1] 15462 18 heatmap(as.matrix(data.dcpm), col=topo.colors(100)) heatmap.2(as.matrix(data.dcpm), col=topo.colors(100), key=TRUE, density.inf=none,trace=none, scale=none) --- The outputs are attached here. Could anyone help me figure out why? Thanks a lot:) __ R-help@r-project.org mailing list 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] Logistic Regression with 200K features in R?
ok, so 200K predictors an 10M observations would work?
On 12/12/2013 12:12 PM, Eik Vettorazzi wrote:
it is simply because you can't do a regression with more predictors than
observations.
Cheers.
Am 12.12.2013 09:00, schrieb Romeo Kienzler:
Dear List,
I'm quite new to R and want to do logistic regression with a 200K
feature data set (around 150 training examples).
I'm aware that I should use Naive Bayes but I have a more general
question about the capability of R handling very high dimensional data.
Please consider the following R code where mygenestrain.tab is a 150
by 20 matrix:
traindata - read.table('mygenestrain.tab');
mylogit - glm(V1 ~ ., data = traindata, family = binomial);
When executing this code I get the following error:
Error in terms.formula(formula, data = data) :
allocMatrix: too many elements specified
Calls: glm ... model.frame - model.frame.default - terms - terms.formula
Execution halted
Is this because R can't handle 200K features or am I doing something
completely wrong here?
Thanks a lot for your help!
best Regards,
Romeo
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[R] boxcox transformations
Hi, I am new to R. I need help with regards to box cox transformation. I have phenotypic data for e.g. plant height. data is non-normal. Skewness is 0.34. Could you please help me? Regards, Yogi -- View this message in context: http://r.789695.n4.nabble.com/boxcox-transformations-tp4682077.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] Logistic Regression with 200K features in R?
Dear Eik,
thank you so much for your help!
best Regards,
Romeo
On 12/12/2013 12:51 PM, Eik Vettorazzi wrote:
I thought so (with all the limitations due to collinearity and so on),
but actually there is a limit for the maximum size of an array which is
independent of your memory size and is due to the way arrays are
indexed. You can't create an object with more than 2^31-1 = 2147483647
elements.
https://stat.ethz.ch/pipermail/r-help/2007-June/133238.html
cheers
Am 12.12.2013 12:34, schrieb Romeo Kienzler:
ok, so 200K predictors an 10M observations would work?
On 12/12/2013 12:12 PM, Eik Vettorazzi wrote:
it is simply because you can't do a regression with more predictors than
observations.
Cheers.
Am 12.12.2013 09:00, schrieb Romeo Kienzler:
Dear List,
I'm quite new to R and want to do logistic regression with a 200K
feature data set (around 150 training examples).
I'm aware that I should use Naive Bayes but I have a more general
question about the capability of R handling very high dimensional data.
Please consider the following R code where mygenestrain.tab is a 150
by 20 matrix:
traindata - read.table('mygenestrain.tab');
mylogit - glm(V1 ~ ., data = traindata, family = binomial);
When executing this code I get the following error:
Error in terms.formula(formula, data = data) :
allocMatrix: too many elements specified
Calls: glm ... model.frame - model.frame.default - terms -
terms.formula
Execution halted
Is this because R can't handle 200K features or am I doing something
completely wrong here?
Thanks a lot for your help!
best Regards,
Romeo
__
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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.
__
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Re: [R] Wilcoxon test output as a table
Hello there Was looking up a similar problem yesterday and came across the gMWT package (Generalized Mann-Whitney Type Tests). It addresses your problem, although I am still trying to figure out how it works. Kind regards, David -- View this message in context: http://r.789695.n4.nabble.com/Wilcoxon-test-output-as-a-table-tp2244565p4682058.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] function inside a function
On 12-12-2013, at 17:10, eliza botto eliza_bo...@hotmail.com wrote:
Dear users of R,
I am trying to inculcate a function inside a function. For that to be done, i
copied following function from internet.
library(nleqslv) fun - function(x) {
f - numeric(length(x))
f[1] - A[,1]+x[2] - 1/x[1]
f[2] - A[,2]+x[2] - sin(x[1])
f
}
x.start - c(1,1)
nleqslv(x.start,fun)
I have a matrix A with dimension 124 rows and 2 columns. In f[1] line, in
place of A[,1] i want to inculcate each value (each row) of
column 1 of matrix A. while doing it, In f[2] line, in place of A[,2] i want
to inculcate each value (each row) of column 2 of matrix A.
For suppose A has following rows
0.6772941 0.5962983
0.4348938 0.4563702
0.4481236 0.4418828
0.4213013 0.3944993
0.4682232 0.4485623
0.4798529 0.4477387
0.7029005 0.5533228
While using 0.66772941 in f[1], use 0.5962983 in f[2].
How can i do this in R.
Thanks in advance,
It is not clear what you actually want.
Maybe something like this
library(nleqslv)
A - matrix(c(0.6772941, 0.5962983,
0.4348938, 0.4563702,
0.4481236, 0.4418828,
0.4213013, 0.3944993,
0.4682232, 0.4485623,
0.4798529, 0.4477387,
0.7029005, 0.5533228), byrow=TRUE,ncol=2)
fun - function(x, krow) {
f - numeric(length(x))
f[1] - A[krow,1] + x[2]- 1/x[1]
f[2] - A[krow,2] +x[2] - sin(x[1])
f
}
x.start - c(1,1)
nleqslv(x.start, fun, krow=1)
nleqslv(x.start, fun, krow=2)
Eliza
[[alternative HTML version deleted]]
Please do not post in HTML. You should know by now.
Berend
__
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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
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] function inside a function
Dear Berend,Thankyou very much for your reply. I actually wanted to inserts
each row value of A in f[1] anf f[2], with column 1 value in f[1] and column 2
values in f[2]. Once that been done, i should have in the end, 124 vaues of
x[1] for column 1 and 124 for column 2.I hope i am clear this time.Thanks for
your help.Eliza
Subject: Re: [R] function inside a function
From: b...@xs4all.nl
Date: Thu, 12 Dec 2013 19:35:58 +0100
CC: r-help@r-project.org
To: eliza_bo...@hotmail.com
On 12-12-2013, at 17:10, eliza botto eliza_bo...@hotmail.com wrote:
Dear users of R,
I am trying to inculcate a function inside a function. For that to be done,
i copied following function from internet.
library(nleqslv) fun - function(x) {
f - numeric(length(x))
f[1] - A[,1]+x[2] - 1/x[1]
f[2] - A[,2]+x[2] - sin(x[1])
f
}
x.start - c(1,1)
nleqslv(x.start,fun)
I have a matrix A with dimension 124 rows and 2 columns. In f[1] line, in
place of A[,1] i want to inculcate each value (each row) of
column 1 of matrix A. while doing it, In f[2] line, in place of A[,2] i
want to inculcate each value (each row) of column 2 of matrix A.
For suppose A has following rows
0.6772941 0.5962983
0.4348938 0.4563702
0.4481236 0.4418828
0.4213013 0.3944993
0.4682232 0.4485623
0.4798529 0.4477387
0.7029005 0.5533228
While using 0.66772941 in f[1], use 0.5962983 in f[2].
How can i do this in R.
Thanks in advance,
It is not clear what you actually want.
Maybe something like this
library(nleqslv)
A - matrix(c(0.6772941, 0.5962983,
0.4348938, 0.4563702,
0.4481236, 0.4418828,
0.4213013, 0.3944993,
0.4682232, 0.4485623,
0.4798529, 0.4477387,
0.7029005, 0.5533228), byrow=TRUE,ncol=2)
fun - function(x, krow) {
f - numeric(length(x))
f[1] - A[krow,1] + x[2]- 1/x[1]
f[2] - A[krow,2] +x[2] - sin(x[1])
f
}
x.start - c(1,1)
nleqslv(x.start, fun, krow=1)
nleqslv(x.start, fun, krow=2)
Eliza
[[alternative HTML version deleted]]
Please do not post in HTML. You should know by now.
Berend
__
R-help@r-project.org mailing list
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
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] boxcox transformations
Dear Yogi, try the powerTransform function in the car package and boxcox function in the MASS package. there are nice examples for both functions. HTH, daniel Feladó: r-help-boun...@r-project.org [r-help-boun...@r-project.org] ; meghatalmaz#243;: Yogendra [ykhedi...@gmail.com] Küldve: 2013. december 12. 19:00 To: r-help@r-project.org Tárgy: [R] boxcox transformations Hi, I am new to R. I need help with regards to box cox transformation. I have phenotypic data for e.g. plant height. data is non-normal. Skewness is 0.34. Could you please help me? Regards, Yogi -- View this message in context: http://r.789695.n4.nabble.com/boxcox-transformations-tp4682077.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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 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] Heatmap, and heatmap.2 gave different figures for the same dataset
Read the help page for both and pay particular attention to the scale argument. Kevin Wright On Thu, Dec 12, 2013 at 9:45 AM, capricy gao capri...@yahoo.com wrote: I have a huge dataset(15k X 18) and tried to use the heatmap in R to examine the patterns. However, I found that heatmap and heatmap.2 gave me completely different outputs. Here are the codes: dim(as.matrix(data.dcpm)) [1] 1546218 heatmap(as.matrix(data.dcpm), col=topo.colors(100)) heatmap.2(as.matrix(data.dcpm), col=topo.colors(100), key=TRUE, density.inf=none,trace=none, scale=none) --- The outputs are attached here. Could anyone help me figure out why? Thanks a lot:) __ R-help@r-project.org mailing list 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. -- Kevin Wright [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] labels on right y-axis
On 12/13/2013 01:32 AM, Fisher Dennis wrote: OS X R 3.0.2 Colleagues, I am displaying two sets of values changing over time; as a result, I have two y-axes. I add a label for the right-side Y axis with mtext(side=3, line=1.2, TEXT). The text is parallel to the axis -- so far, so good. However, the text is rotated counterclockwise from horizontal, whereas I would like it rotated clockwise. Neither srt nor las fixes this. I guess that one convoluted approach would be to use text rather than mtext. However, positioning would be a nuisance and the xpd option would be needed. Is there any simpler approach? Dennis Hi Dennis, This sounds like a job for twoord.plot (plotrix). I think I can leave the warnings, disclaimers and miscellaneous dissuasion to others. Jim __ R-help@r-project.org mailing list 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] function inside a function
Dear Arun,
Thankyou very much for your reply. You always come up to save the day.
So kind of you...!!!
Eliza
Date: Thu, 12 Dec 2013 12:36:18 -0800
From: smartpink...@yahoo.com
Subject: Re: [R] function inside a function
To: eliza_bo...@hotmail.com
Hi Eliza,
Actually, I used the modified function by Berend.
fun - function(x, krow) {
f - numeric(length(x))
f[1] - A[krow,1] + x[2]- 1/x[1]
f[2] - A[krow,2] +x[2] - sin(x[1])
f
}
x.start - c(1,1)
t(sapply(seq_len(nrow(A)),function(i) nleqslv(x.start,fun,krow=i)$x))
[,1] [,2]
[1,] 1.052828 0.2725286
[2,] 1.131672 0.4487542
[3,] 1.109174 0.4534483
[4,] 1.093078 0.4935468
[5,] 1.098613 0.4420152
[6,] 1.088997 0.4384237
[7,] 1.005843 0.2912907
A.K.
On Thursday, December 12, 2013 3:16 PM, eliza botto eliza_bo...@hotmail.com
wrote:
Dear Arun,
Thankyou very much for your reply.
I m getting the following error.
May be i m not running the full code.
i used it in the following way.
library(nleqslv)
fun - function(x) {
f - numeric(length(x))
f[1] - A[,1]+x[2] - 1/x[1]
f[2] - A[,2]+x[2] -
sin(x[1])
f
}
x.start - c(1,1)
t(sapply(seq_len(nrow(A)),function(i) nleqslv(x.start,fun,krow=i)$x))
Kindly see, if you could help.
Eliza
Date: Thu, 12 Dec 2013 11:01:47 -0800
From: smartpink...@yahoo.com
Subject: Re: [R] function inside a function
To: r-help@r-project.org
CC: b...@xs4all.nl; eliza_bo...@hotmail.com
Hi,
May be this helps:
t(sapply(seq_len(nrow(A)),function(i) nleqslv(x.start,fun,krow=i)$x))
A.K.
On Thursday, December 12, 2013 1:54 PM, eliza botto
eliza_bo...@hotmail.com wrote:
Dear Berend,Thankyou very much for your reply. I actually wanted to inserts
each row value of A in f[1] anf f[2], with column 1 value in f[1] and
column 2 values in f[2]. Once that been done, i should have in the end, 124
vaues of x[1] for column 1 and 124 for column 2.I hope i am clear this
time.Thanks for your help.Eliza
Subject: Re: [R] function inside a function
From: b...@xs4all.nl
Date: Thu, 12 Dec 2013 19:35:58 +0100
CC: r-help@r-project.org
To: eliza_bo...@hotmail.com
On 12-12-2013, at 17:10, eliza botto eliza_bo...@hotmail.com wrote:
Dear users of R,
I am trying to inculcate a function inside a function. For that to be
done, i copied following function from internet.
library(nleqslv) fun - function(x) {
á á f - numeric(length(x))
á á f[1] -á A[,1]+x[2] - 1/x[1]
á á f[2] -á A[,2]+x[2] - sin(x[1])
á á f
}
x.start - c(1,1)
nleqslv(x.start,fun)
I have a matrix A with dimension 124 rows and 2 columns. In f[1]
line, in place of A[,1] i want to inculcate each value (each row) of
column 1 of matrix A. while doing it, In f[2] line, in place of A[,2] i
want to inculcate each value (each row) of column 2 of matrix A.á
For suppose A has following rows
0.6772941á á á á á á á á á á á á á á á 0.5962983
0.4348938á á á á á á á á á á á á á á á 0.4563702
0.4481236á á á á á á á á á á á á á á á 0.4418828
0.4213013á á á á á á á á á á á á á á á 0.3944993
0.4682232á á á á á á á á á á á á á á á 0.4485623
0.4798529á á á á á á á á á á á á á á á 0.4477387
0.7029005á á á á á á á á á á á á á á á 0.5533228
While using 0.66772941 in f[1], use 0.5962983 in f[2].
How can i do this in R.
Thanks in advance,
It is not clear what you actually want.
Maybe something like this
library(nleqslv)
A - matrix(c(0.6772941,á 0.5962983,
á á á á á á á 0.4348938,á 0.4563702,
á á á á á á á 0.4481236,á 0.4418828,
á á á á á á á 0.4213013,á 0.3944993,
á á á á á á á 0.4682232,á 0.4485623,
á á á á á á á 0.4798529,á 0.4477387,
á á á á á á á 0.7029005,á 0.5533228), byrow=TRUE,ncol=2)
fun - function(x, krow) {
á á f - numeric(length(x))
á á f[1] - A[krow,1] + x[2]- 1/x[1]
á á f[2] - A[krow,2] +x[2] - sin(x[1])
á á f
}
x.start - c(1,1)
nleqslv(x.start, fun, krow=1)
nleqslv(x.start, fun, krow=2)
Eliza ááá ááá ááá á ááá ááá á
ááá [[alternative HTML version deleted]]
Please do not post in HTML. You should know by now.
Berend
__
R-help@r-project.org mailing list
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.
ááá ááá ááá á ááá ááá á
ááá [[alternative HTML version deleted]]
__
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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] method default for hclust function
I could not figure out what was the default when I ran hclust() without specifying the method. For example: I just have a code like: hclust(dist(data)) Any input would be appreciated:) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] method default for hclust function
Absolute distance is the default distance in hclust. v-c(1,2,3,4,5,6) dist(v) 2 1 3 2 1 4 3 2 1 5 4 3 2 1 6 5 4 3 2 1 Eliza Date: Thu, 12 Dec 2013 15:09:19 -0800 From: capri...@yahoo.com To: r-help@r-project.org Subject: [R] method default for hclust function I could not figure out what was the default when I ran hclust() without specifying the method. For example: I just have a code like: hclust(dist(data)) Any input would be appreciated:) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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 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] method default for hclust function
On Thu, Dec 12, 2013 at 3:09 PM, capricy gao capri...@yahoo.com wrote: I could not figure out what was the default when I ran hclust() without specifying the method. According to help(hclust), the default method is complete linkage. HTH, Peter __ R-help@r-project.org mailing list 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] Heatmap, and heatmap.2 gave different figures for the same dataset
Thank you very much for the hints. I fixed the problem. On Thursday, December 12, 2013 2:08 PM, Kevin Wright kw.s...@gmail.com wrote: Read the help page for both and pay particular attention to the scale argument. Kevin Wright On Thu, Dec 12, 2013 at 9:45 AM, capricy gao capri...@yahoo.com wrote: I have a huge dataset(15k X 18) and tried to use the heatmap in R to examine the patterns. However, I found that heatmap and heatmap.2 gave me completely different outputs. Here are the codes: dim(as.matrix(data.dcpm)) [1] 15462 18 heatmap(as.matrix(data.dcpm), col=topo.colors(100)) heatmap.2(as.matrix(data.dcpm), col=topo.colors(100), key=TRUE, density.inf=none,trace=none, scale=none) --- The outputs are attached here. Could anyone help me figure out why? Thanks a lot:) __ R-help@r-project.org mailing list 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. -- Kevin Wright [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] Coxme time-dependent covariate
Hi all, I am using the coxme function to fit random effects survival models. The base survival models show a typical J shaped curve where survival drops sharply then plateaus over time interval. I want to include a time covariate which could account for this non-linear decrease in survival over time. However, I have not found a clear way to do this in the coxme models. Any suggestions would be greatly appreciated. Below is some sample code and data. -Jared selection=coxme(Surv(start,stop,status)~prob+(1|year); fawn status start stop prob year 3 0 0 86 0.83 2009 4 1 0 9 0.6 2009 5 0 0 355 0.63 2009 6 1 0 6 0.52 2010 7 0 0 68 0.79 2010 8 0 0 354 0.69 2010 9 1 0 3 0.72 2010 10 1 0 4 0.77 2010 11 1 0 12 0.62 2010 13 0 0 82 0.7 2011 14 1 0 7 0.67 2011 15 1 0 74 0.49 2011 16 0 0 351 0.64 2011 17 0 0 351 0.74 2011 18 0 0 82 0.72 2011 19 1 0 342 0.69 2011 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] function inside a function
Hi,
May be this helps:
t(sapply(seq_len(nrow(A)),function(i) nleqslv(x.start,fun,krow=i)$x))
A.K.
On Thursday, December 12, 2013 1:54 PM, eliza botto eliza_bo...@hotmail.com
wrote:
Dear Berend,Thankyou very much for your reply. I actually wanted to inserts
each row value of A in f[1] anf f[2], with column 1 value in f[1] and column 2
values in f[2]. Once that been done, i should have in the end, 124 vaues of
x[1] for column 1 and 124 for column 2.I hope i am clear this time.Thanks for
your help.Eliza
Subject: Re: [R] function inside a function
From: b...@xs4all.nl
Date: Thu, 12 Dec 2013 19:35:58 +0100
CC: r-help@r-project.org
To: eliza_bo...@hotmail.com
On 12-12-2013, at 17:10, eliza botto eliza_bo...@hotmail.com wrote:
Dear users of R,
I am trying to inculcate a function inside a function. For that to be done,
i copied following function from internet.
library(nleqslv) fun - function(x) {
f - numeric(length(x))
f[1] - A[,1]+x[2] - 1/x[1]
f[2] - A[,2]+x[2] - sin(x[1])
f
}
x.start - c(1,1)
nleqslv(x.start,fun)
I have a matrix A with dimension 124 rows and 2 columns. In f[1] line, in
place of A[,1] i want to inculcate each value (each row) of
column 1 of matrix A. while doing it, In f[2] line, in place of A[,2] i
want to inculcate each value (each row) of column 2 of matrix A.
For suppose A has following rows
0.6772941 0.5962983
0.4348938 0.4563702
0.4481236 0.4418828
0.4213013 0.3944993
0.4682232 0.4485623
0.4798529 0.4477387
0.7029005 0.5533228
While using 0.66772941 in f[1], use 0.5962983 in f[2].
How can i do this in R.
Thanks in advance,
It is not clear what you actually want.
Maybe something like this
library(nleqslv)
A - matrix(c(0.6772941, 0.5962983,
0.4348938, 0.4563702,
0.4481236, 0.4418828,
0.4213013, 0.3944993,
0.4682232, 0.4485623,
0.4798529, 0.4477387,
0.7029005, 0.5533228), byrow=TRUE,ncol=2)
fun - function(x, krow) {
f - numeric(length(x))
f[1] - A[krow,1] + x[2]- 1/x[1]
f[2] - A[krow,2] +x[2] - sin(x[1])
f
}
x.start - c(1,1)
nleqslv(x.start, fun, krow=1)
nleqslv(x.start, fun, krow=2)
Eliza
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Berend
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[R] charToRaw(Œ) is not 8C in R console
in http://www.ascii-code.com/, you can see the the hex value of Å is 8C, why in my R console ? charToRaw(Å) [1] c5 92 is not 8C ? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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.
