[R] apply, lapply and data.frame in R 2.5
Hello everyone, A recent (in 2.5 I suspect) change in R is giving me trouble. I want to apply a function (tolower) to all the columns of a data.frame and get a data.frame in return. Currently, on a data.frame, both apply (for arrays) and lapply (for lists) work, but each returns its native class (resp. matrix and list): apply(mydat,2,tolower) # gives a matrix lapply(mydat,tolower) # gives a list and sapply(mydat,tolower) # gives a matrix If I remember well, apply did not used to work on data.frames and lapply returned a data.frame when it was provided with one, with the same properties (columns classes etc). At least this is what my code written with R 2.4.* suggests. The solution would be: as.data.frame(apply(mydat,2,tolower)) or as.data.frame(lapply(mydat,tolower)) But this does not keep columns attributes (all columns are reinterpreted, for example strings are converted to factors etc). For my particular use stringsAsFactors=FALSE does what I need, but I am wondering wether there is a more general solution to apply a function on all elements of a data.frame and get a similar data.frame in return. Indeed data.frames are probably the most common object in R and applying a function to each of its columns/variables appears to me as something one would want to do quite often. Thank you in advance. JiHO --- http://jo.irisson.free.fr/ __ R-help@stat.math.ethz.ch 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] apply, lapply and data.frame in R 2.5
On Mon, 30 Jul 2007, jiho wrote: Hello everyone, A recent (in 2.5 I suspect) change in R is giving me trouble. I want to apply a function (tolower) to all the columns of a data.frame and get a data.frame in return. Currently, on a data.frame, both apply (for arrays) and lapply (for lists) work, but each returns its native class (resp. matrix and list): apply(mydat,2,tolower)# gives a matrix lapply(mydat,tolower) # gives a list and sapply(mydat,tolower) # gives a matrix which is exactly what R 2.0.0 did, so no recent(ish) change at all. If I remember well, apply did not used to work on data.frames and lapply returned a data.frame when it was provided with one, with the same properties (columns classes etc). At least this is what my code written with R 2.4.* suggests. apply has coerced data frames for many years and lapply always returned a list. The solution has always been mydat[] - lapply(mydat,tolower) The solution would be: as.data.frame(apply(mydat,2,tolower)) or as.data.frame(lapply(mydat,tolower)) But this does not keep columns attributes (all columns are reinterpreted, for example strings are converted to factors etc). For my particular use stringsAsFactors=FALSE does what I need, but I am wondering wether there is a more general solution to apply a function on all elements of a data.frame and get a similar data.frame in return. Indeed data.frames are probably the most common object in R and applying a function to each of its columns/variables appears to me as something one would want to do quite often. Thank you in advance. JiHO --- http://jo.irisson.free.fr/ __ R-help@stat.math.ethz.ch 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. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch 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] apply, lapply and data.frame in R 2.5
On 2007-July-30 , at 12:20 , Prof Brian Ripley wrote: On Mon, 30 Jul 2007, jiho wrote: A recent (in 2.5 I suspect) change in R is giving me trouble. I want to apply a function (tolower) to all the columns of a data.frame and get a data.frame in return. Currently, on a data.frame, both apply (for arrays) and lapply (for lists) work, but each returns its native class (resp. matrix and list): apply(mydat,2,tolower) # gives a matrix lapply(mydat,tolower)# gives a list and sapply(mydat,tolower)# gives a matrix which is exactly what R 2.0.0 did, so no recent(ish) change at all. If I remember well, apply did not used to work on data.frames and lapply returned a data.frame when it was provided with one, with the same properties (columns classes etc). At least this is what my code written with R 2.4.* suggests. apply has coerced data frames for many years and lapply always returned a list. The solution has always been mydat[] - lapply(mydat,tolower) sorry about that, my previous code was misleading and indeed your code above does exactly what I need. I should have tested this a bit further before posting. I was just afraid to install two different R versions I guess. thank you again. JiHO --- http://jo.irisson.free.fr/ __ R-help@stat.math.ethz.ch 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] apply incompatible dimensions error
Hi,
I've created the following two matrices (mat1 and mat2) and a function
(f) to calculate the correlations between the two on a row by row basis.
mat1 - matrix(sample(1:500,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
mat2 - matrix(sample(501:1000,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
f - function(x,y) cor(x,y)
When the matrices are squares (# rows = # columns) I have no problems.
However, when they are not (as in the example above with 5 columns and
10 rows), I get the following error:
apply(mat1, 1, f, y=mat2)
Error in cor(x, y, na.method, method == kendall) :
incompatible dimensions
Any help would be appreciated. Thanks!
- Bruce
**
Please be aware that, notwithstanding the fact that the pers...{{dropped}}
__
R-help@stat.math.ethz.ch 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] apply incompatible dimensions error
Your apply is trying to take the correlations of the rows of mat1 with the
columns of mat2 which, of course, does not work if they have different
numbers of columns. I think you mean to take the correlations of the columns
of mat1 with the columns of mat2. For example, to take the correlations
of the 5 columns of mat1 with the first 4 columns of mat2 try:
cor(mat1, mat2[,1:4])
col1 col2 col3 col4
col1 -0.34624254 -0.2669519 -0.2705077 0.2183249
col2 -0.26553255 -0.2687643 -0.0865895 0.1819025
col3 0.19474613 -0.2334986 0.1746522 0.2326915
col4 0.09328338 0.5117784 0.2413143 -0.3374916
col5 0.27519716 0.1605331 -0.4057137 0.3282105
On 7/24/07, Bernzweig, Bruce (Consultant) [EMAIL PROTECTED] wrote:
Hi,
I've created the following two matrices (mat1 and mat2) and a function
(f) to calculate the correlations between the two on a row by row basis.
mat1 - matrix(sample(1:500,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
mat2 - matrix(sample(501:1000,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
f - function(x,y) cor(x,y)
When the matrices are squares (# rows = # columns) I have no problems.
However, when they are not (as in the example above with 5 columns and
10 rows), I get the following error:
apply(mat1, 1, f, y=mat2)
Error in cor(x, y, na.method, method == kendall) :
incompatible dimensions
Any help would be appreciated. Thanks!
- Bruce
**
Please be aware that, notwithstanding the fact that the pers...{{dropped}}
__
R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] apply incompatible dimensions error
are you positive that your function is doing what you expect it to do?
it looks like you want something like:
sapply(1:10, function(i) cor(mat1[i,], mat2[i,]))
b
On Jul 24, 2007, at 11:05 AM, Bernzweig, Bruce ((Consultant)) wrote:
Hi,
I've created the following two matrices (mat1 and mat2) and a function
(f) to calculate the correlations between the two on a row by row
basis.
mat1 - matrix(sample(1:500,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
mat2 - matrix(sample(501:1000,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
f - function(x,y) cor(x,y)
When the matrices are squares (# rows = # columns) I have no problems.
However, when they are not (as in the example above with 5 columns and
10 rows), I get the following error:
apply(mat1, 1, f, y=mat2)
Error in cor(x, y, na.method, method == kendall) :
incompatible dimensions
Any help would be appreciated. Thanks!
- Bruce
**
Please be aware that, notwithstanding the fact that the pers...
{{dropped}}
__
R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] apply incompatible dimensions error
Thanks Gabor!
You state that my apply is taking rows of mat1 with columns of mat2.
Is this because I have the y=mat2 parameter?
apply(mat1, 1, f, y=mat2)
Actually, what I would like is to run the correlations on a row against
row basis: mat1 row1 x mat2 row1, etc.
Thanks again,
- Bruce
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent: Tuesday, July 24, 2007 11:31 AM
To: Bernzweig, Bruce (Consultant)
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] apply incompatible dimensions error
Your apply is trying to take the correlations of the rows of mat1 with
the
columns of mat2 which, of course, does not work if they have different
numbers of columns. I think you mean to take the correlations of the
columns
of mat1 with the columns of mat2. For example, to take the correlations
of the 5 columns of mat1 with the first 4 columns of mat2 try:
cor(mat1, mat2[,1:4])
col1 col2 col3 col4
col1 -0.34624254 -0.2669519 -0.2705077 0.2183249
col2 -0.26553255 -0.2687643 -0.0865895 0.1819025
col3 0.19474613 -0.2334986 0.1746522 0.2326915
col4 0.09328338 0.5117784 0.2413143 -0.3374916
col5 0.27519716 0.1605331 -0.4057137 0.3282105
On 7/24/07, Bernzweig, Bruce (Consultant) [EMAIL PROTECTED] wrote:
Hi,
I've created the following two matrices (mat1 and mat2) and a function
(f) to calculate the correlations between the two on a row by row
basis.
mat1 - matrix(sample(1:500,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
mat2 - matrix(sample(501:1000,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
f - function(x,y) cor(x,y)
When the matrices are squares (# rows = # columns) I have no problems.
However, when they are not (as in the example above with 5 columns and
10 rows), I get the following error:
apply(mat1, 1, f, y=mat2)
Error in cor(x, y, na.method, method == kendall) :
incompatible dimensions
Any help would be appreciated. Thanks!
- Bruce
**
Please be aware that, notwithstanding the fact that the
pers...{{dropped}}
__
R-help@stat.math.ethz.ch 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.
**
Please be aware that, notwithstanding the fact that the pers...{{dropped}}
__
R-help@stat.math.ethz.ch 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] apply incompatible dimensions error
Then try this:
cor(t(mat1), t(mat2))
Also note
1. the above implies that mat1 and mat2 must have the same
number of columns since if x and y are vectors cor(x,y) only makes
sense if they have the same length.
2. the usual convention is that variables are stored as columns
andt that rows correspond to cases so typically you would have
(in terms of your mat1 and mat2):
Mat1 - t(mat1)
Mat2 - t(mat2)
and then use Mat1 and Mat2, e.g. cor(Mat1, Mat2)
On 7/24/07, Bernzweig, Bruce (Consultant) [EMAIL PROTECTED] wrote:
Thanks Gabor!
You state that my apply is taking rows of mat1 with columns of mat2.
Is this because I have the y=mat2 parameter?
apply(mat1, 1, f, y=mat2)
Actually, what I would like is to run the correlations on a row against
row basis: mat1 row1 x mat2 row1, etc.
Thanks again,
- Bruce
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent: Tuesday, July 24, 2007 11:31 AM
To: Bernzweig, Bruce (Consultant)
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] apply incompatible dimensions error
Your apply is trying to take the correlations of the rows of mat1 with
the
columns of mat2 which, of course, does not work if they have different
numbers of columns. I think you mean to take the correlations of the
columns
of mat1 with the columns of mat2. For example, to take the correlations
of the 5 columns of mat1 with the first 4 columns of mat2 try:
cor(mat1, mat2[,1:4])
col1 col2 col3 col4
col1 -0.34624254 -0.2669519 -0.2705077 0.2183249
col2 -0.26553255 -0.2687643 -0.0865895 0.1819025
col3 0.19474613 -0.2334986 0.1746522 0.2326915
col4 0.09328338 0.5117784 0.2413143 -0.3374916
col5 0.27519716 0.1605331 -0.4057137 0.3282105
On 7/24/07, Bernzweig, Bruce (Consultant) [EMAIL PROTECTED] wrote:
Hi,
I've created the following two matrices (mat1 and mat2) and a function
(f) to calculate the correlations between the two on a row by row
basis.
mat1 - matrix(sample(1:500,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
mat2 - matrix(sample(501:1000,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
f - function(x,y) cor(x,y)
When the matrices are squares (# rows = # columns) I have no problems.
However, when they are not (as in the example above with 5 columns and
10 rows), I get the following error:
apply(mat1, 1, f, y=mat2)
Error in cor(x, y, na.method, method == kendall) :
incompatible dimensions
Any help would be appreciated. Thanks!
- Bruce
**
Please be aware that, notwithstanding the fact that the
pers...{{dropped}}
__
R-help@stat.math.ethz.ch 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.
**
Please be aware that, notwithstanding the fact that the person sending
this communication has an address in Bear Stearns' e-mail system, this
person is not an employee, agent or representative of Bear Stearns.
Accordingly, this person has no power or authority to represent, make
any recommendation, solicitation, offer or statements or disclose
information on behalf of or in any way bind Bear Stearns or any of its
affiliates.
**
__
R-help@stat.math.ethz.ch 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] apply incompatible dimensions error
that's garbor's suggestion then.
sorry for the misunderstanding. :-)
b
On Jul 24, 2007, at 11:35 AM, Bernzweig, Bruce ((Consultant)) wrote:
Thanks Benilton,
I know what I want to do, just not sure how to do it using R. The
help
documentation is not very clear.
What I am trying to do is calculate correlations on a row against row
basis: mat1 row1 x mat2 row1, mat1 row1 x mat2 row2, ... mat1 row1 x
mat2 row-n, mat1 row-n, mat2 row-n
- Bruce
-Original Message-
From: Benilton Carvalho [mailto:[EMAIL PROTECTED]
Sent: Tuesday, July 24, 2007 11:31 AM
To: Bernzweig, Bruce (Consultant)
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] apply incompatible dimensions error
are you positive that your function is doing what you expect it to do?
it looks like you want something like:
sapply(1:10, function(i) cor(mat1[i,], mat2[i,]))
b
On Jul 24, 2007, at 11:05 AM, Bernzweig, Bruce ((Consultant)) wrote:
Hi,
I've created the following two matrices (mat1 and mat2) and a
function
(f) to calculate the correlations between the two on a row by row
basis.
mat1 - matrix(sample(1:500,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
mat2 - matrix(sample(501:1000,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
f - function(x,y) cor(x,y)
When the matrices are squares (# rows = # columns) I have no
problems.
However, when they are not (as in the example above with 5 columns
and
10 rows), I get the following error:
apply(mat1, 1, f, y=mat2)
Error in cor(x, y, na.method, method == kendall) :
incompatible dimensions
Any help would be appreciated. Thanks!
- Bruce
*
*
Please be aware that, notwithstanding the fact that the pers...
{{dropped}}
__
R-help@stat.math.ethz.ch 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.
**
Please be aware that, notwithstanding the fact that the person sending
this communication has an address in Bear Stearns' e-mail system, this
person is not an employee, agent or representative of Bear Stearns.
Accordingly, this person has no power or authority to represent, make
any recommendation, solicitation, offer or statements or disclose
information on behalf of or in any way bind Bear Stearns or any of its
affiliates.
**
__
R-help@stat.math.ethz.ch 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] apply incompatible dimensions error
Thanks for the explanation.
As for the rows/columns thing, the data I receive is given to me in that
way. I currently read it in using read.csv. Is there a function I
should look at that can take that and transpose it or should I just
process the data first outside of R?
Thanks,
- Bruce
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent: Tuesday, July 24, 2007 11:43 AM
To: Bernzweig, Bruce (Consultant)
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] apply incompatible dimensions error
Then try this:
cor(t(mat1), t(mat2))
Also note
1. the above implies that mat1 and mat2 must have the same
number of columns since if x and y are vectors cor(x,y) only makes
sense if they have the same length.
2. the usual convention is that variables are stored as columns
andt that rows correspond to cases so typically you would have
(in terms of your mat1 and mat2):
Mat1 - t(mat1)
Mat2 - t(mat2)
and then use Mat1 and Mat2, e.g. cor(Mat1, Mat2)
On 7/24/07, Bernzweig, Bruce (Consultant) [EMAIL PROTECTED] wrote:
Thanks Gabor!
You state that my apply is taking rows of mat1 with columns of mat2.
Is this because I have the y=mat2 parameter?
apply(mat1, 1, f, y=mat2)
Actually, what I would like is to run the correlations on a row
against
row basis: mat1 row1 x mat2 row1, etc.
Thanks again,
- Bruce
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent: Tuesday, July 24, 2007 11:31 AM
To: Bernzweig, Bruce (Consultant)
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] apply incompatible dimensions error
Your apply is trying to take the correlations of the rows of mat1 with
the
columns of mat2 which, of course, does not work if they have different
numbers of columns. I think you mean to take the correlations of the
columns
of mat1 with the columns of mat2. For example, to take the
correlations
of the 5 columns of mat1 with the first 4 columns of mat2 try:
cor(mat1, mat2[,1:4])
col1 col2 col3 col4
col1 -0.34624254 -0.2669519 -0.2705077 0.2183249
col2 -0.26553255 -0.2687643 -0.0865895 0.1819025
col3 0.19474613 -0.2334986 0.1746522 0.2326915
col4 0.09328338 0.5117784 0.2413143 -0.3374916
col5 0.27519716 0.1605331 -0.4057137 0.3282105
On 7/24/07, Bernzweig, Bruce (Consultant) [EMAIL PROTECTED] wrote:
Hi,
I've created the following two matrices (mat1 and mat2) and a
function
(f) to calculate the correlations between the two on a row by row
basis.
mat1 - matrix(sample(1:500,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
mat2 - matrix(sample(501:1000,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
f - function(x,y) cor(x,y)
When the matrices are squares (# rows = # columns) I have no
problems.
However, when they are not (as in the example above with 5 columns
and
10 rows), I get the following error:
apply(mat1, 1, f, y=mat2)
Error in cor(x, y, na.method, method == kendall) :
incompatible dimensions
Any help would be appreciated. Thanks!
- Bruce
**
Please be aware that, notwithstanding the fact that the
pers...{{dropped}}
__
R-help@stat.math.ethz.ch 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.
**
Please be aware that, notwithstanding the fact that the person sending
this communication has an address in Bear Stearns' e-mail system, this
person is not an employee, agent or representative of Bear Stearns.
Accordingly, this person has no power or authority to represent, make
any recommendation, solicitation, offer or statements or disclose
information on behalf of or in any way bind Bear Stearns or any of its
affiliates.
**
**
Please be aware that, notwithstanding the fact that the pers...{{dropped}}
__
R-help@stat.math.ethz.ch 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] apply incompatible dimensions error
Thanks Benilton,
I know what I want to do, just not sure how to do it using R. The help
documentation is not very clear.
What I am trying to do is calculate correlations on a row against row
basis: mat1 row1 x mat2 row1, mat1 row1 x mat2 row2, ... mat1 row1 x
mat2 row-n, mat1 row-n, mat2 row-n
- Bruce
-Original Message-
From: Benilton Carvalho [mailto:[EMAIL PROTECTED]
Sent: Tuesday, July 24, 2007 11:31 AM
To: Bernzweig, Bruce (Consultant)
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] apply incompatible dimensions error
are you positive that your function is doing what you expect it to do?
it looks like you want something like:
sapply(1:10, function(i) cor(mat1[i,], mat2[i,]))
b
On Jul 24, 2007, at 11:05 AM, Bernzweig, Bruce ((Consultant)) wrote:
Hi,
I've created the following two matrices (mat1 and mat2) and a function
(f) to calculate the correlations between the two on a row by row
basis.
mat1 - matrix(sample(1:500,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
mat2 - matrix(sample(501:1000,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
f - function(x,y) cor(x,y)
When the matrices are squares (# rows = # columns) I have no problems.
However, when they are not (as in the example above with 5 columns and
10 rows), I get the following error:
apply(mat1, 1, f, y=mat2)
Error in cor(x, y, na.method, method == kendall) :
incompatible dimensions
Any help would be appreciated. Thanks!
- Bruce
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[R] apply a function to loop through
Dear R users, I have a dataset generated as follows, myDat - data.frame(matrix(c(rep(LETTERS[1:3], each=5), rnorm(5,mean=3,sd=0.03), rnorm(5,12,1), rnorm(5,1,0.5)), ncol=2, dimnames=list(c(letters[1:15]), c(tissue,Scores myDat$Grp -c(gene1) There is one level gene1 in $Grp in my data step. I'd like to do pairwise.wilcox.test on $tissue while going throug $Grp if there are more levels with gene2, gene3. I tried to loop through $Grp using apply with an error message Error in sort(unique.default(x), na.last = TRUE) : 'x' must be atomic. mytry - apply(as.matrix(myDat), 1, function(Grp)pairwise.wilcox.test(Grp$Scores, Grp$tissue, p.adjust.method = none)$p.value) I could not find any similar stuffs in forum. Could anyone here give a hand? Thanks a bunch. kevin __ R-help@stat.math.ethz.ch 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] apply a function to loop through
Hi,
first of all, your Scores are factors instead of numeric, you need to change it;
I made a new myDat for test purpose by
myDat = as.data.frame(rbind(myDat, myDat))
myDat[16:30, 3] - gene2
myDat
tissueScores Grp
a A 3.01494535196933 gene1
b A 2.99647624484379 gene1
c A 3.00527533284709 gene1
d A 3.02321059636917 gene1
e A 3.04694197334289 gene1
f B 11.1464974692841 gene1
g B 12.2372838904939 gene1
h B 11.7277801841221 gene1
i B 11.6405683605147 gene1
j B 11.1631961720026 gene1
k C 0.67528704974662 gene1
l C 1.21000950157251 gene1
m C 0.843722400594721 gene1
n C 0.881706314004343 gene1
o C 1.43670211710054 gene1
a1 A 3.01494535196933 gene2
b1 A 2.99647624484379 gene2
c1 A 3.00527533284709 gene2
d1 A 3.02321059636917 gene2
e1 A 3.04694197334289 gene2
f1 B 11.1464974692841 gene2
g1 B 12.2372838904939 gene2
h1 B 11.7277801841221 gene2
i1 B 11.6405683605147 gene2
j1 B 11.1631961720026 gene2
k1 C 0.67528704974662 gene2
l1 C 1.21000950157251 gene2
m1 C 0.843722400594721 gene2
n1 C 0.881706314004343 gene2
o1 C 1.43670211710054 gene2
mytry - by(myDat, INDICES=as.factor(myDat[,3]), FUN=function(x) {
+ pairwise.wilcox.test(as.numeric(as.character(x$Scores)),
+
+ x$tissue,
+
+ p.adjust.method = none)$p.value
+ })
mytry
as.factor(myDat[, 3]): gene1
A B
B 0.007936508 NA
C 0.007936508 0.007936508
as.factor(myDat[, 3]): gene2
A B
B 0.007936508 NA
C 0.007936508 0.007936508
HTH,
Weiwei
On 7/17/07, Hai Lin [EMAIL PROTECTED] wrote:
Dear R users,
I have a dataset generated as follows,
myDat - data.frame(matrix(c(rep(LETTERS[1:3],
each=5),
rnorm(5,mean=3,sd=0.03),
rnorm(5,12,1),
rnorm(5,1,0.5)),
ncol=2,
dimnames=list(c(letters[1:15]),
c(tissue,Scores
myDat$Grp -c(gene1)
There is one level gene1 in $Grp in my data step.
I'd like to do pairwise.wilcox.test on $tissue while
going throug $Grp if there are more levels with gene2,
gene3.
I tried to loop through $Grp using apply with an error
message Error in sort(unique.default(x), na.last =
TRUE) :
'x' must be atomic.
mytry - apply(as.matrix(myDat),
1,
function(Grp)pairwise.wilcox.test(Grp$Scores,
Grp$tissue,
p.adjust.method = none)$p.value)
I could not find any similar stuffs in forum. Could
anyone here give a hand?
Thanks a bunch.
kevin
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Re: [R] apply( )
Thank you all for your answers... To summarize,
1. with(X, ifelse(V1 V2 | V1 V3, 1, 0))
2. ifelse((foo$x foo$y) | (foo$x foo$z), 1, 0)
3. with(foo, (x y) * (x z))
were the responses to my question (see below).
Kindly,
Greg
--- Original post ---
I have a question that must have a simple answer (but eludes me).
I need a row-by-row logical comparison across three numeric variables
in a data frame: foo$x, foo$y, foo$z. The logic is
if( x y || x z ) 1 else 0
for a particular row.
It is simple and very inefficient to use for(i in 1:length(foo$x)){ }
loops. How can I accomplish this using sappy( ) / lapply( ) /
apply( ) or some other more efficient method?
Thank you in advance,
Greg
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Re: [R] apply( )
Hi
[EMAIL PROTECTED] napsal dne 10.05.2007 17:59:22:
or
with(foo, (x y) * (x z))
Should not it be
with(foo, ((x y) | (x z))*1)
Regards
Petr
On 5/10/07, jim holtman [EMAIL PROTECTED] wrote:
You don't need apply. Just do
foo$result - ifelse((foo$x foo$y) | (foo$x foo$z), 1, 0)
On 5/10/07, Greg Tarpinian [EMAIL PROTECTED] wrote:
I have a question that must have a simple answer (but eludes me).
I need a row-by-row logical comparison across three numeric
variables
in
a data frame: foo$x, foo$y, foo$z. The logic is
if( x y || x z ) 1 else 0
for a particular row.
It is simple and very inefficient to use for(i in 1:length(foo$x)){
}
loops. How can I accomplish this using sappy( ) / lapply( ) /
apply( )
or some other more efficient method?
Thank you in advance,
Greg
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[R] apply( )
I have a question that must have a simple answer (but eludes me).
I need a row-by-row logical comparison across three numeric variables
in
a data frame: foo$x, foo$y, foo$z. The logic is
if( x y || x z ) 1 else 0
for a particular row.
It is simple and very inefficient to use for(i in 1:length(foo$x)){ }
loops. How can I accomplish this using sappy( ) / lapply( ) / apply( )
or some other more efficient method?
Thank you in advance,
Greg
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Re: [R] apply( )
You don't need apply. Just do
foo$result - ifelse((foo$x foo$y) | (foo$x foo$z), 1, 0)
On 5/10/07, Greg Tarpinian [EMAIL PROTECTED] wrote:
I have a question that must have a simple answer (but eludes me).
I need a row-by-row logical comparison across three numeric variables
in
a data frame: foo$x, foo$y, foo$z. The logic is
if( x y || x z ) 1 else 0
for a particular row.
It is simple and very inefficient to use for(i in 1:length(foo$x)){ }
loops. How can I accomplish this using sappy( ) / lapply( ) / apply( )
or some other more efficient method?
Thank you in advance,
Greg
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+1 513 646 9390
What is the problem you are trying to solve?
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Re: [R] apply( )
Greg Tarpinian wrote:
I have a question that must have a simple answer (but eludes me).
I need a row-by-row logical comparison across three numeric variables
in
a data frame: foo$x, foo$y, foo$z. The logic is
if( x y || x z ) 1 else 0
for a particular row.
It is simple and very inefficient to use for(i in 1:length(foo$x)){ }
loops. How can I accomplish this using sappy( ) / lapply( ) / apply( )
or some other more efficient method?
X - as.data.frame(matrix(rnorm(30), ncol=3))
X
V1 V2 V3
1 -0.48026236 0.8629789 -1.2600858
2 -1.32408219 -0.5590268 1.1310638
3 0.02717575 -0.5661402 0.7824019
4 0.80783373 0.2300440 -0.4477275
5 1.24518907 -0.3778392 1.7546530
6 -0.39254125 -1.0388962 -0.4436296
7 -1.44473455 1.8606963 0.4253889
8 -0.63543047 -1.6408418 -1.0409473
9 0.81075970 0.3914066 -1.0361739
10 1.66021280 -1.6694101 -0.4810839
with(X, ifelse(V1 V2 | V1 V3, 1, 0))
[1] 1 1 0 1 0 1 1 1 1 1
Thank you in advance,
Greg
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Re: [R] apply( )
or
with(foo, (x y) * (x z))
On 5/10/07, jim holtman [EMAIL PROTECTED] wrote:
You don't need apply. Just do
foo$result - ifelse((foo$x foo$y) | (foo$x foo$z), 1, 0)
On 5/10/07, Greg Tarpinian [EMAIL PROTECTED] wrote:
I have a question that must have a simple answer (but eludes me).
I need a row-by-row logical comparison across three numeric variables
in
a data frame: foo$x, foo$y, foo$z. The logic is
if( x y || x z ) 1 else 0
for a particular row.
It is simple and very inefficient to use for(i in 1:length(foo$x)){ }
loops. How can I accomplish this using sappy( ) / lapply( ) / apply( )
or some other more efficient method?
Thank you in advance,
Greg
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+1 513 646 9390
What is the problem you are trying to solve?
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Re: [R] apply( )
ifelse(((x y) | (x z)), 1, 0)
Note in particular the use of | instead of || for elementwise comparisons.
Petr
Greg Tarpinian napsal(a):
I have a question that must have a simple answer (but eludes me).
I need a row-by-row logical comparison across three numeric variables
in
a data frame: foo$x, foo$y, foo$z. The logic is
if( x y || x z ) 1 else 0
for a particular row.
It is simple and very inefficient to use for(i in 1:length(foo$x)){ }
loops. How can I accomplish this using sappy( ) / lapply( ) / apply( )
or some other more efficient method?
Thank you in advance,
Greg
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Charles University in Prague
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Re: [R] apply( )
Sorry, you wanted or, not and.
with(foo, pmax(x y, x z))
with(foo, as.numeric(x y | x z))
with(foo, 1*(x y | x z))
On 5/10/07, Gabor Grothendieck [EMAIL PROTECTED] wrote:
or
with(foo, (x y) * (x z))
On 5/10/07, jim holtman [EMAIL PROTECTED] wrote:
You don't need apply. Just do
foo$result - ifelse((foo$x foo$y) | (foo$x foo$z), 1, 0)
On 5/10/07, Greg Tarpinian [EMAIL PROTECTED] wrote:
I have a question that must have a simple answer (but eludes me).
I need a row-by-row logical comparison across three numeric variables
in
a data frame: foo$x, foo$y, foo$z. The logic is
if( x y || x z ) 1 else 0
for a particular row.
It is simple and very inefficient to use for(i in 1:length(foo$x)){ }
loops. How can I accomplish this using sappy( ) / lapply( ) / apply( )
or some other more efficient method?
Thank you in advance,
Greg
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+1 513 646 9390
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[R] apply problem
Dear R-Help I am running apply on a data.frame containing factors and numeric columns. It appears to convert are columns into as.character? Does it convert data.frame into matrix? Is this expected? I wish it to recognise numerical columns and round numbers. Can I use another function instead of apply, or should I use a for loop in the case? summary(xmat) A B C D Min. : 1.0 414: 1 Stage 2: 5 Min. :-0.075369 1st Qu.:113.8 422: 1 Stage 3: 6 1st Qu.:-0.018102 Median :226.5 426: 1 Stage 4:441 Median :-0.003033 Mean :226.5 436: 1 Mean : 0.008007 3rd Qu.:339.2 460: 1 3rd Qu.: 0.015499 Max. :452.0 462: 1 Max. : 0.400578 (Other):446 EFG Min. :0.2345 Min. :0.9808 Min. :0.01558 1st Qu.:0.2840 1st Qu.:0.9899 1st Qu.:0.02352 Median :0.3265 Median :0.9965 Median :0.02966 Mean :0.3690 Mean :1.0079 Mean :0.03580 3rd Qu.:0.3859 3rd Qu.:1.0129 3rd Qu.:0.03980 Max. :2.0422 Max. :1.3742 Max. :0.20062 for(i in 1:7) print(class(xmat[,i])) [1] integer [1] factor [1] factor [1] numeric [1] numeric [1] numeric [1] numeric apply(xmat, 2, class) A B C D E F character character character character character character G character Thanks for your help Aedin __ [EMAIL PROTECTED] 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] apply problem
On Fri, 13 Apr 2007, aedin culhane wrote: Dear R-Help I am running apply on a data.frame containing factors and numeric columns. It appears to convert are columns into as.character? Does it convert data.frame into matrix? Is this expected? I wish it to recognise Yes, and quite explicit on the help page Arguments: X: the array to be used. ^ If 'X' is not an array but has a dimension attribute, 'apply' attempts to coerce it to an array via 'as.matrix' if it is two-dimensional (e.g., data frames) or via 'as.array'. I am baffled as to how you managed to miss this, as it is part of the homework the posting guide asked you to do *before* posting. numerical columns and round numbers. Can I use another function instead of apply, or should I use a for loop in the case? You haven;t told us _what_ you want to do, but lapply works by column. summary(xmat) A B C D Min. : 1.0 414: 1 Stage 2: 5 Min. :-0.075369 1st Qu.:113.8 422: 1 Stage 3: 6 1st Qu.:-0.018102 Median :226.5 426: 1 Stage 4:441 Median :-0.003033 Mean :226.5 436: 1 Mean : 0.008007 3rd Qu.:339.2 460: 1 3rd Qu.: 0.015499 Max. :452.0 462: 1 Max. : 0.400578 (Other):446 EFG Min. :0.2345 Min. :0.9808 Min. :0.01558 1st Qu.:0.2840 1st Qu.:0.9899 1st Qu.:0.02352 Median :0.3265 Median :0.9965 Median :0.02966 Mean :0.3690 Mean :1.0079 Mean :0.03580 3rd Qu.:0.3859 3rd Qu.:1.0129 3rd Qu.:0.03980 Max. :2.0422 Max. :1.3742 Max. :0.20062 for(i in 1:7) print(class(xmat[,i])) [1] integer [1] factor [1] factor [1] numeric [1] numeric [1] numeric [1] numeric Better, sapply(xmat, class). apply(xmat, 2, class) A B C D E F character character character character character character G character Well, all columns of a matrix are of the same class. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://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] apply problem
?apply says
If X is not an array but has a dimension attribute, apply attempts to coerce
it to an array via as.matrix if it is two-dimensional (e.g., data frames). .
.
It would probably be easiest with a FOR-LOOP, but you could also try
something like the code below (and insert your operations in #...).
myfunc - function(x,classOfX) {
x - as.data.frame(t(x))
factvars - which(classOfX==factor)
x[,factvars] - lapply(x[,factvars],factor)
for( i in seq(along=x) ) x[,i] - as(x[,i],Class=classOfX[i])
# ...
return(x)
}
x - data.frame(a=as.integer(1:10),b=factor(letters[1:10]),c=runif(10))
Fold - function(f,x,L) for(e in L) x - f(x,e)
y - Fold(rbind,vector(),apply(x,1,myfunc,rapply(x,class)))
rapply(x,class)
a b c
integer factor numeric
rapply(y,class)
a b c
integer factor numeric
--- aedin culhane [EMAIL PROTECTED] wrote:
Dear R-Help
I am running apply on a data.frame containing factors and numeric
columns. It appears to convert are columns into as.character? Does it
convert data.frame into matrix? Is this expected? I wish it to recognise
numerical columns and round numbers. Can I use another function instead
of apply, or should I use a for loop in the case?
summary(xmat)
A B C D
Min. : 1.0 414: 1 Stage 2: 5 Min. :-0.075369
1st Qu.:113.8 422: 1 Stage 3: 6 1st Qu.:-0.018102
Median :226.5 426: 1 Stage 4:441 Median :-0.003033
Mean :226.5 436: 1 Mean : 0.008007
3rd Qu.:339.2 460: 1 3rd Qu.: 0.015499
Max. :452.0 462: 1 Max. : 0.400578
(Other):446
EFG
Min. :0.2345 Min. :0.9808 Min. :0.01558
1st Qu.:0.2840 1st Qu.:0.9899 1st Qu.:0.02352
Median :0.3265 Median :0.9965 Median :0.02966
Mean :0.3690 Mean :1.0079 Mean :0.03580
3rd Qu.:0.3859 3rd Qu.:1.0129 3rd Qu.:0.03980
Max. :2.0422 Max. :1.3742 Max. :0.20062
for(i in 1:7) print(class(xmat[,i]))
[1] integer
[1] factor
[1] factor
[1] numeric
[1] numeric
[1] numeric
[1] numeric
apply(xmat, 2, class)
A B C D E F
character character character character character character
G
character
Thanks for your help
Aedin
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] apply problem
aedin culhane [EMAIL PROTECTED] writes: Dear R-Help I am running apply on a data.frame containing factors and numeric columns. It appears to convert are columns into as.character? Does it convert data.frame into matrix? Is this expected? I wish it to recognise numerical columns and round numbers. Can I use another function instead of apply, or should I use a for loop in the case? If you want to modify the data.frame object, a for loop will likely be the best bet. As noted in other replies, lapply will operate on the columns of a data.frame since a data.frame is a list. But the return value will be a list, not a data.frame. I think for loops get a bad wrap. There are times when they are appropriate and even optimal in R programming. + seth -- Seth Falcon | Computational Biology | Fred Hutchinson Cancer Research Center http://bioconductor.org __ [EMAIL PROTECTED] 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] apply ? function doesnt create object
hello,
i have written a function to extract certain lines from a matrix. the
result is a matrix with 6 cols, named dynamically according to the
functions arguments.
the problem is now, that i'm not able to return the resultmatrix for
further use. the object is not being created.
example from my console:
getans(27,27)
[...]
[189,] 3969 161 27101
[190,] 2142 87 27101
[191,] 1318 52 27101
[192,] 2881 120 27101
[193,] 3669 152 27101
[194,] 5653 229 27101
[195,] 2308 95 27102
[196,] 4942 200 27101
the matrix ansblock27 contains answers to block 27
ansblock27
error: object ansblock27 not found
the code of the function:
getans = function(x=qids,bnr=1,type=block)
{
#generate name of matrix
matnam=paste(ans,type,as.character(bnr),sep=)
#display result matrix
show(assign(matnam,matrix(as.numeric(as.matrix(allans[(allans[, 3] %in
% x), , drop = FALSE])),ncol=dim(allans)[2])))
#create result matrix
assign(matnam,matrix(as.numeric(as.matrix(allans[(allans[, 3] %in%
x), , drop = FALSE])),ncol=dim(allans)[2]))
#print info
cat(the matrix,matnam,contains answers to,type,as.character(bnr))
}
i have tried to use return, but had no success.
what else can i do?
thx in advance, and have a nice weekend
regards,
[[alternative HTML version deleted]]
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Re: [R] apply ? function doesnt create object
On Mar 3, 2007, at 4:28 PM, bunny , lautloscrew.com wrote:
Please use - for assignments instead of = :
getans = function(x=qids,bnr=1,type=block)
{
#generate name of matrix
matnam=paste(ans,type,as.character(bnr),sep=)
#display result matrix
show(assign(matnam,matrix(as.numeric(as.matrix(allans[(allans[, 3] %in
% x), , drop = FALSE])),ncol=dim(allans)[2])))
You are assigning things twice here.
#create result matrix
assign(matnam,matrix(as.numeric(as.matrix(allans[(allans[, 3] %in%
x), , drop = FALSE])),ncol=dim(allans)[2]))
The documentation for assign makes it pretty clear that the
assignment happens by default in the current environment, so it will
be local to the function unless you alter the call. The description
there, along with the examples, and a study of environments, should
provide you with the answer.
#print info
cat(the matrix,matnam,contains answers to,type,as.character
(bnr))
}
Haris Skiadas
Department of Mathematics and Computer Science
Hanover College
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[R] apply for list or list - array
Dear all, I have the following list, aa, composed of two 3*3 tables. I would like to use apply function to summarize it, but apply cannot handle list. I want to do it without using any interation. 1. Is there any function like apply for list? 2. Is there any way to transform that list to a 2*3*3 array? thanks cahn - aa [[1]] [,1] [,2] [,3] [1,]131 [2,]201 [3,]110 [[2]] [,1] [,2] [,3] [1,]131 [2,]201 [3,]110 I would like to avoid any iteration - Ahn, Chaehyung (cahn) Ph.D. Alpha-Gamma Technologies Inc. My Home: http://blog.naver.com/cahn88 __ R-help@stat.math.ethz.ch 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] apply for list or list - array
Ahn ChaeHyung wrote: Dear all, I have the following list, aa, composed of two 3*3 tables. I would like to use apply function to summarize it, but apply cannot handle list. I want to do it without using any interation. 1. Is there any function like apply for list? 2. Is there any way to transform that list to a 2*3*3 array? Use lapply() or sapply(). Example to transpose both matrices: lapply(a, t) Uwe Ligges thanks cahn - aa [[1]] [,1] [,2] [,3] [1,]131 [2,]201 [3,]110 [[2]] [,1] [,2] [,3] [1,]131 [2,]201 [3,]110 I would like to avoid any iteration - Ahn, Chaehyung (cahn) Ph.D. Alpha-Gamma Technologies Inc. My Home: http://blog.naver.com/cahn88 __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] apply for list or list - array
On Mon, 18 Sep 2006, Ahn ChaeHyung wrote: Dear all, I have the following list, aa, composed of two 3*3 tables. I would like to use apply function to summarize it, but apply cannot handle list. I want to do it without using any interation. 1. Is there any function like apply for list? ?lapply, ?sapply 2. Is there any way to transform that list to a 2*3*3 array? Among other possibilities: library(abind) aa - list(m1=matrix(rnorm(9), 3, 3), m2=matrix(runif(9), 3, 3)) a1 - abind(aa, along=3) str(a1) dim(a1) thanks cahn - aa [[1]] [,1] [,2] [,3] [1,]131 [2,]201 [3,]110 [[2]] [,1] [,2] [,3] [1,]131 [2,]201 [3,]110 I would like to avoid any iteration - Ahn, Chaehyung (cahn) Ph.D. Alpha-Gamma Technologies Inc. My Home: http://blog.naver.com/cahn88 __ R-help@stat.math.ethz.ch 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. -- Roger Bivand Economic Geography Section, Department of Economics, Norwegian School of Economics and Business Administration, Helleveien 30, N-5045 Bergen, Norway. voice: +47 55 95 93 55; fax +47 55 95 95 43 e-mail: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch 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] apply for list or list - array
for lists look at ?lapply() and ?sapply(); for your 2nd question try something like: array(unlist(aa), dim = c(3, 3, 2)) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm - Original Message - From: Ahn ChaeHyung [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Tuesday, September 19, 2006 3:32 AM Subject: [R] apply for list or list - array Dear all, I have the following list, aa, composed of two 3*3 tables. I would like to use apply function to summarize it, but apply cannot handle list. I want to do it without using any interation. 1. Is there any function like apply for list? 2. Is there any way to transform that list to a 2*3*3 array? thanks cahn - aa [[1]] [,1] [,2] [,3] [1,]131 [2,]201 [3,]110 [[2]] [,1] [,2] [,3] [1,]131 [2,]201 [3,]110 I would like to avoid any iteration - Ahn, Chaehyung (cahn) Ph.D. Alpha-Gamma Technologies Inc. My Home: http://blog.naver.com/cahn88 __ R-help@stat.math.ethz.ch 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. Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@stat.math.ethz.ch 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] apply least angle regression to generalized linear models
Hello list, I've been searching around trying to find whether somebody has written such a package of least angle regression on generalized linear models, like what Lasso2 package does. The extension to generalized linear models is briefly discussed in the comment by D. Madigan and G. Ridgeway. Is such a package available? Thanks, Mike [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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] apply least angle regression to generalized linear models
On Fri, 2006-08-18 at 11:17 -0400, Mike Wolfgang wrote: Hello list, I've been searching around trying to find whether somebody has written such a package of least angle regression on generalized linear models, like what Lasso2 package does. The extension to generalized linear models is briefly discussed in the comment by D. Madigan and G. Ridgeway. Is such a package available? Thanks, Mike See the aptly named 'lars' package on CRAN and the attendant paper here: http://www-stat.stanford.edu/~hastie/Papers/LARS/LeastAngle_2002.pdf You might also want to review Professor Hastie's presentation at useR! 2006 this past spring: http://www.r-project.org/useR-2006/Slides/Hastie.pdf HTH, Marc Schwartz __ R-help@stat.math.ethz.ch 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] apply least angle regression to generalized linear models
I believe `lars' does not currently fit glms. For that you'll probably need to look at `glar', at: http://www.insightful.com/Hesterberg/glars/default.asp HTH, Andy From: Marc Schwartz On Fri, 2006-08-18 at 11:17 -0400, Mike Wolfgang wrote: Hello list, I've been searching around trying to find whether somebody has written such a package of least angle regression on generalized linear models, like what Lasso2 package does. The extension to generalized linear models is briefly discussed in the comment by D. Madigan and G. Ridgeway. Is such a package available? Thanks, Mike See the aptly named 'lars' package on CRAN and the attendant paper here: http://www-stat.stanford.edu/~hastie/Papers/LARS/LeastAngle_2002.pdf You might also want to review Professor Hastie's presentation at useR! 2006 this past spring: http://www.r-project.org/useR-2006/Slides/Hastie.pdf HTH, Marc Schwartz __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] apply least angle regression to generalized linear models
Andy,
Upon further review of the documentation for lars, you are correct.
Thanks for the pointer to the work by Tim et al.
Regards,
Marc
On Fri, 2006-08-18 at 12:48 -0400, Liaw, Andy wrote:
I believe `lars' does not currently fit glms. For that you'll probably need
to look at `glar', at:
http://www.insightful.com/Hesterberg/glars/default.asp
HTH,
Andy
From: Marc Schwartz
On Fri, 2006-08-18 at 11:17 -0400, Mike Wolfgang wrote:
Hello list,
I've been searching around trying to find whether somebody
has written
such a package of least angle regression on generalized
linear models,
like what
Lasso2 package does. The extension to generalized linear models is
briefly discussed in the comment by D. Madigan and G. Ridgeway. Is
such a package available? Thanks,
Mike
See the aptly named 'lars' package on CRAN and the attendant
paper here:
http://www-stat.stanford.edu/~hastie/Papers/LARS/LeastAngle_2002.pdf
You might also want to review Professor Hastie's presentation at useR!
2006 this past spring:
http://www.r-project.org/useR-2006/Slides/Hastie.pdf
HTH,
Marc Schwartz
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[R] apply a function to several lists' components
Dear R-user I have 100 lists. Each list has several components. For example, data1 $a [1] 1 2 $b [1] 3 4 $c [1] 5 There are data1, data2,, data100. All lists have the same number and the same name of components. Is there any function I can use for applying to only a specific component across 100 lists? (e.g., taking mean of $c acorss 100 lists) or do I need to write my own function for that? Thank you. Taka, __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply a function to several lists' components
Maybe this helps
( data1 = list(a=c(1,2), b=c(3,4), c=c(5,6,7)) )
( data2 = list(a=c(10,11), b=c(30,40), c=c(70,80)) )
cc - NULL
for(data in ls(pattern=^data[0-9]+$)) {
cc - c(cc, with(get(data), c))
}
mean(cc)
JeeBee.
On Fri, 30 Jun 2006 09:50:51 -0500, Taka Matzmoto wrote:
Dear R-user
I have 100 lists.
Each list has several components.
For example,
data1
$a
[1] 1 2
$b
[1] 3 4
$c
[1] 5
There are data1, data2,, data100. All lists have the same number and the
same name of components.
Is there any function I can use for applying to only a specific component
across 100 lists?
(e.g., taking mean of $c acorss 100 lists) or do I need to write my own
function for that?
Thank you.
Taka,
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[R] apply, apply speed vs. traditional looping mechanisms
Can someone tell me why apply (and apply) are faster in performing repeated operations than a for (or do) loop? I am looking for a technical explanation. Thanks, John John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics Baltimore VA Medical Center GRECC and University of Maryland School of Medicine Claude Pepper OAIC University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 410-605-7119 [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply, apply speed vs. traditional looping mechanisms
On Sun, 14 May 2006, John Sorkin wrote: Can someone tell me why apply (and apply) are faster in performing repeated operations than a for (or do) loop? I am looking for a technical explanation. apply() is just a wrapper for a for loop. So it is not faster that at least one implementation using a for loop: it may be neater and easier to understand than an explicit for loop. I don't understand why you used 'apply' twice here. lapply() can be faster than a carefully crafted for() loop (since C-level code is more efficient in memory allocation), but its main rationale is clarity (and especially to avoid traps like ans - vector(list, n) for(i in 1:n) ans[[i]] - fun(i) if n turns out to be zero or fun(i) to be NULL). -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply, apply speed vs. traditional looping mechanisms
Prof Ripley, Many thanks for your reply. I did not mean to use apply twice, I meant to ask about apply and lapply. I understand your comment re: apply. I assume your comment about lapply is meant to mean that lapply is implemented in C code and therefore should be faster than a loop written in R. I would also like to take this opportunity to thank you for your may contributions to R and this list. Thank you, John Sorkin John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics Baltimore VA Medical Center GRECC and University of Maryland School of Medicine Claude Pepper OAIC University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 410-605-7119 [EMAIL PROTECTED] Prof Brian Ripley [EMAIL PROTECTED] 05/14/06 4:26 PM On Sun, 14 May 2006, John Sorkin wrote: Can someone tell me why apply (and apply) are faster in performing repeated operations than a for (or do) loop? I am looking for a technical explanation. apply() is just a wrapper for a for loop. So it is not faster that at least one implementation using a for loop: it may be neater and easier to understand than an explicit for loop. I don't understand why you used 'apply' twice here. lapply() can be faster than a carefully crafted for() loop (since C-level code is more efficient in memory allocation), but its main rationale is clarity (and especially to avoid traps like ans - vector(list, n) for(i in 1:n) ans[[i]] - fun(i) if n turns out to be zero or fun(i) to be NULL). -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] apply
Dear Sir, Iam a new user of R Software. I thank you very much for this Software. By running the R 2.3.0 software I have encountred aproblem: I have downloaded the tseriesChaos package from CRAN. When I try to run any function of the the tseriesChaos package for example, the c2 function the R software diplays on the screen error, impossible to find the c2 function). Please help me! What shall I do to add on this package and run it correctly? Why the Software displays error, impossible to find the c2 function? Is it caused by a damaged version of the R 2.3.0? I thank you in advance and I am waiting forward your e-mail This is my e-mail: [EMAIL PROTECTED] Accédez au courrier électronique de La Poste : www.laposte.net ; 3615 LAPOSTENET (0,34 /mn) ; tél : 08 92 68 13 50 (0,34/mn) [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply
On Sat, 2006-05-13 at 23:02 +0200, karim.regh wrote: Dear Sir, Iam a new user of R Software. I thank you very much for this Software. By running the R 2.3.0 software I have encountred aproblem: I have downloaded the tseriesChaos package from CRAN. When I try to run any function of the the tseriesChaos package for example, the c2 function the R software diplays on the screen error, impossible to find the c2 function). Please help me! What shall I do to add on this package and run it correctly? Why the Software displays error, impossible to find the c2 function? Is it caused by a damaged version of the R 2.3.0? I thank you in advance and I am waiting forward your e-mail Did you use: library(tseriesChaos) before trying to use the functions that are contained in it? See R FAQ: 7.30 I installed a package but the functions are not there Reading the Posting Guide as noted at the bottom of each e-mail coming through the list would enable you to solve many basic problems such as this. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] apply(table) miss factor structure
Hi, all. I didn't find something similar to this problem in past list. I have a data frame (named restr) where some columns are factors, like you can see: table(restr[,p1]) 0 1 2 3 4 5 0 26 1 0 1 0 table(restr[,p2]) 0 1 2 3 4 5 6 0 13 11 1 2 1 0 When I use apply, the factor structure is missed: apply(restr[,c(p1,p2)], 2, table) $p1 1 2 4 26 1 1 $p2 1 2 3 4 5 13 11 1 2 1 Can I use a matricial (like apply) manner to do this holding the factor structure (the zero-counts must be displayed)? Thanks, Cezar Freitas Unicamp - Brasil __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply(table) miss factor structure
you should use: lapply(restr[c(p1,p2)], table) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://www.med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm - Original Message - From: Cézar Freitas [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Wednesday, April 19, 2006 3:50 PM Subject: [R] apply(table) miss factor structure Hi, all. I didn't find something similar to this problem in past list. I have a data frame (named restr) where some columns are factors, like you can see: table(restr[,p1]) 0 1 2 3 4 5 0 26 1 0 1 0 table(restr[,p2]) 0 1 2 3 4 5 6 0 13 11 1 2 1 0 When I use apply, the factor structure is missed: apply(restr[,c(p1,p2)], 2, table) $p1 1 2 4 26 1 1 $p2 1 2 3 4 5 13 11 1 2 1 Can I use a matricial (like apply) manner to do this holding the factor structure (the zero-counts must be displayed)? Thanks, Cezar Freitas Unicamp - Brasil __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] apply(ing) to sum subset of a vector
Dear R users I am trying to sum selective elements of a vector but my solution is not cutting it. Example: g - 1:5; from - 1:3; to - 3:5; from to 1 3 2 4 3 5 so I expect 3 sums from g 1+2+3 that is 1 to 3 of g 2+3+4 that is 2 to 4 of g 3+4+5 that is 3 to 5 of g my solution will not work. sum.em - function(g, c1, c2) sum(g[c1:c2]) apply(g, 1, sum.em, ...) I don't think so because apply is not aware of the from and to. and if I f - list(g, from, to) that will not fit with the second arg of apply. thank you - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply(ing) to sum subset of a vector
create a matrix and then use apply: g - 1:5; from - 1:3; to - 3:5; index - cbind(from,to) apply(index, 1, function(x) sum(g[x[1]:x[2]])) [1] 6 9 12 On 3/27/06, Fred J. [EMAIL PROTECTED] wrote: Dear R users I am trying to sum selective elements of a vector but my solution is not cutting it. Example: g - 1:5; from - 1:3; to - 3:5; from to 1 3 2 4 3 5 so I expect 3 sums from g 1+2+3 that is 1 to 3 of g 2+3+4 that is 2 to 4 of g 3+4+5 that is 3 to 5 of g my solution will not work. sum.em - function(g, c1, c2) sum(g[c1:c2]) apply(g, 1, sum.em, ...) I don't think so because apply is not aware of the from and to. and if I f - list(g, from, to) that will not fit with the second arg of apply. thank you - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Jim Holtman Cincinnati, OH +1 513 646 9390 (Cell) +1 513 247 0281 (Home) What the problem you are trying to solve? [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply(ing) to sum subset of a vector
apply(cbind(from,to), 1, function(x) sum(g[x[1]:x[2]])) Fred J. a écrit : Dear R users I am trying to sum selective elements of a vector but my solution is not cutting it. Example: g - 1:5; from - 1:3; to - 3:5; from to 1 3 2 4 3 5 so I expect 3 sums from g 1+2+3 that is 1 to 3 of g 2+3+4 that is 2 to 4 of g 3+4+5 that is 3 to 5 of g my solution will not work. sum.em - function(g, c1, c2) sum(g[c1:c2]) apply(g, 1, sum.em, ...) I don't think so because apply is not aware of the from and to. and if I f - list(g, from, to) that will not fit with the second arg of apply. thank you - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Apply and more argumnts function
Dear useRs,
shame on me, but I have no idea how to apply two arguments function on
my data. I have 2 vectors, 'n' and 'm' and the function below:
n - c(10,30,50,1000)
m - c(10,50,100,200)
MonteCarlo - function(n,m){
temp - NULL
for(i in 1:m){
temp - c(temp,walk(n)) # walk is external function
}
return(c(n,m,mean(temp),sd(temp)))
}
Now I have to calculate mean(temp) and sd(temp) for each pair of
corresponding elements in vectors.
Thanks in advance, Andrej
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[R] apply
Hello,
I have a dataset d which is
d
pop catch
1 66462.01 10807.757
2 87486.73 46257.885
3 57211.64 9345.058
4 71321.62 4892.868
5 100024.89 27334.248
6 104504.91 48535.092
7 95295.51 39348.195
8 93737.35 34343.489
9 89375.05 28750.743
10 95312.65 30755.064
11 100888.17 55404.370
12 84790.23 37751.074
13 81129.82 29277.443
14 69797.09 21500.398
15 61690.34 18199.664
16 60671.08 19349.051
17 62852.57 16300.982
18 90646.32 34793.148
And a function opt1:
opt1 - function (x) {
Z - x + M
c.hat - (x/Z)*pop*(1-exp(-Z))
y - (catch - c.hat)^2
return(y)
}
And define M = 0.25
For each row I want to return a value F that is a minimization of the
opt1 function
I have tried many variations of:
d$F-apply(d,2,optim(0.3,opt1,method=BFGS)
and even:
For (I in 1:length(pop))apply(d,2,optim(0.3,opt1,method=BFGS))
Every time I get the same error message
Error in optim(0.3, opt1, method = BFGS) :
objective function in optim evaluates to length 18 not 1
So the apply function is returning 18 values of F which I want but the
function only wants to return 1 value.
Any Suggestions.
Thanks,
Cam
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Re: [R] apply
You are not calling apply() properly. Please read relevant reference
material carefully. You might also wish to pick up one of the several good
books on R (check CRAN website) -- I like VR's S PROGRAMMING.
I did not go through your example in detail, but your apply() call should be
of the form
apply(d,2,function(x)optim(x,...))
You may or may not get into scoping problems with the opt1 argument and have
to pass it in explicitly -- I can't remember how things work with optim.
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
The business of the statistician is to catalyze the scientific learning
process. - George E. P. Box
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
Guenther, Cameron
Sent: Friday, January 13, 2006 11:03 AM
To: [EMAIL PROTECTED]
Subject: [R] apply
Hello,
I have a dataset d which is
d
pop catch
1 66462.01 10807.757
2 87486.73 46257.885
3 57211.64 9345.058
4 71321.62 4892.868
5 100024.89 27334.248
6 104504.91 48535.092
7 95295.51 39348.195
8 93737.35 34343.489
9 89375.05 28750.743
10 95312.65 30755.064
11 100888.17 55404.370
12 84790.23 37751.074
13 81129.82 29277.443
14 69797.09 21500.398
15 61690.34 18199.664
16 60671.08 19349.051
17 62852.57 16300.982
18 90646.32 34793.148
And a function opt1:
opt1 - function (x) {
Z - x + M
c.hat - (x/Z)*pop*(1-exp(-Z))
y - (catch - c.hat)^2
return(y)
}
And define M = 0.25
For each row I want to return a value F that is a minimization of the
opt1 function
I have tried many variations of:
d$F-apply(d,2,optim(0.3,opt1,method=BFGS)
and even:
For (I in 1:length(pop))apply(d,2,optim(0.3,opt1,method=BFGS))
Every time I get the same error message
Error in optim(0.3, opt1, method = BFGS) :
objective function in optim evaluates to length 18 not 1
So the apply function is returning 18 values of F which I want but the
function only wants to return 1 value.
Any Suggestions.
Thanks,
Cam
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Re: [R] apply
In addition the error message explicitly says:
Error in optim(0.3, opt1, method = BFGS) :
objective function in optim evaluates to length 18 not 1
And from the objective function we see
opt1 - function (x) {
Z - x + M
c.hat - (x/Z)*pop*(1-exp(-Z))
y - (catch - c.hat)^2
return(y)
}
The last line should be replaced by:
sum(y)
rather than return(y). See ?optim, ?apply, and the reference Bert
gives below.
HTH,
--sundar
Berton Gunter wrote:
You are not calling apply() properly. Please read relevant reference
material carefully. You might also wish to pick up one of the several good
books on R (check CRAN website) -- I like VR's S PROGRAMMING.
I did not go through your example in detail, but your apply() call should be
of the form
apply(d,2,function(x)optim(x,...))
You may or may not get into scoping problems with the opt1 argument and have
to pass it in explicitly -- I can't remember how things work with optim.
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
The business of the statistician is to catalyze the scientific learning
process. - George E. P. Box
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
Guenther, Cameron
Sent: Friday, January 13, 2006 11:03 AM
To: [EMAIL PROTECTED]
Subject: [R] apply
Hello,
I have a dataset d which is
d
pop catch
1 66462.01 10807.757
2 87486.73 46257.885
3 57211.64 9345.058
4 71321.62 4892.868
5 100024.89 27334.248
6 104504.91 48535.092
7 95295.51 39348.195
8 93737.35 34343.489
9 89375.05 28750.743
10 95312.65 30755.064
11 100888.17 55404.370
12 84790.23 37751.074
13 81129.82 29277.443
14 69797.09 21500.398
15 61690.34 18199.664
16 60671.08 19349.051
17 62852.57 16300.982
18 90646.32 34793.148
And a function opt1:
opt1 - function (x) {
Z - x + M
c.hat - (x/Z)*pop*(1-exp(-Z))
y - (catch - c.hat)^2
return(y)
}
And define M = 0.25
For each row I want to return a value F that is a minimization of the
opt1 function
I have tried many variations of:
d$F-apply(d,2,optim(0.3,opt1,method=BFGS)
and even:
For (I in 1:length(pop))apply(d,2,optim(0.3,opt1,method=BFGS))
Every time I get the same error message
Error in optim(0.3, opt1, method = BFGS) :
objective function in optim evaluates to length 18 not 1
So the apply function is returning 18 values of F which I want but the
function only wants to return 1 value.
Any Suggestions.
Thanks,
Cam
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[R] apply() and dropped dimensions
Hi
I am having difficulty with apply(). I want apply() to return a
matrix, but sometimes a vector is returned.
Toy example follows.
Function jj() takes a couple of matrices m1 and m2 as arguments
and returns a matrix with r rows and c columns where r=nrow(m2)
and c=nrow(m1).
jj - function(m1,m2,f,...){
apply(m1, 1, function(y) {
apply(m2, 1, function(x) {
f(x, y, ...)
})
})
}
R jj(matrix(1:20,4,5),matrix(1:10,2,5),f=sum)
[,1] [,2] [,3] [,4]
[1,] 70 75 80 85
[2,] 75 80 85 90
intended behaviour: matrix returned. [eg, 70 = sum(m1[1,],m2[1,]) ]
R jj(matrix(1:20,4,5),matrix(1:5,1,5),f=sum)
[1] 60 65 70 75
R
not intended behaviour: vector returned.
How do I rewrite jj() so that it consistently returns a matrix?
--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
tel 023-8059-7743
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Re: [R] apply() and dropped dimensions
G'day Robin, RH == Robin Hankin [EMAIL PROTECTED] writes: RH How do I rewrite jj() so that it consistently returns a RH matrix? How about explicitly returning a matrix with the desired dimensions? jj function(m1,m2,f,...) matrix(apply(m1, 1, function(y) apply(m2, 1, function(x) f(x, y, ...) )), nrow=nrow(m2), ncol=nrow(m1)) jj(matrix(1:20,4,5),matrix(1:10,2,5),f=sum) [,1] [,2] [,3] [,4] [1,] 70 75 80 85 [2,] 75 80 85 90 jj(matrix(1:20,4,5),matrix(1:5,1,5),f=sum) [,1] [,2] [,3] [,4] [1,] 60 65 70 75 Or, using the outer command: jj function(m1,m2,f,...) outer(1:nrow(m2), 1:nrow(m1), function(i,j) apply(cbind(i,j), 1, function(ii) f(m2[ii[1],], m1[ii[2],], ...))) jj(matrix(1:20,4,5),matrix(1:10,2,5),f=sum) [,1] [,2] [,3] [,4] [1,] 70 75 80 85 [2,] 75 80 85 90 jj(matrix(1:20,4,5),matrix(1:5,1,5),f=sum) [,1] [,2] [,3] [,4] [1,] 60 65 70 75 Cheers, Berwin __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] apply and plot
R-help, I use the code below to plot some data by applying apply function. But I don't know how I can get the argument type or col on the plot function to distinguish the different lines in the graph: apply ( my.data, 2, function ( x ) lines ( dimnames ( my.data ) [[1]] , x ) ) Thank you in advance __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply and plot
On Thu, 2005-10-13 at 14:50 +0100, Luis Ridao Cruz wrote: R-help, I use the code below to plot some data by applying apply function. But I don't know how I can get the argument type or col on the plot function to distinguish the different lines in the graph: apply ( my.data, 2, function ( x ) lines ( dimnames ( my.data ) [[1]] , x ) ) Thank you in advance Rather than trying the use the construct above, take a look at ?matlines and/or ?matplot (both on the same help page with ?matpoints.) I think that you will find these purposefully designed functions better suited to what I believe you are trying to do here. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Apply a function for each Row
Dear All, I wonder how to apply a given function to each row of a data frame. I've seen this function before but don't remember its name Thank you, Bernard - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Apply a function for each Row
Marc Bernard wrote: Dear All, I wonder how to apply a given function to each row of a data frame. I've seen this function before but don't remember its name You've just said it twice! 'apply'! Baz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Apply a function for each Row
From: Barry Rowlingson Marc Bernard wrote: Dear All, I wonder how to apply a given function to each row of a data frame. I've seen this function before but don't remember its name You've just said it twice! 'apply'! A small catch: Marc wants to apply the function to rows of a data frame, but apply() expects a matrix or array, and will coerce to such if given a data frame, which may (or may not) be problematic... Andy Baz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply and arrays
On Mon, 25 Jul 2005, Uwe Ligges wrote: Laura Holt wrote: Hi R! I have a 3 dimensional array, which is 21 x 3 x 3 I want to use apply to sum on each 21x3 matrix, which is fine. Is there a way that I can do this in 1 step instead of a loop (3), please? Don't know which direction you mean, I guess one of the following: apply(x, c(1,2), sum) apply(x, 3, sum) Or use rowSums or colSums for clarity (and speed, irrelevant here). -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] apply and arrays
Hi R! I have a 3 dimensional array, which is 21 x 3 x 3 I want to use apply to sum on each 21x3 matrix, which is fine. Is there a way that I can do this in 1 step instead of a loop (3), please? thanks, Laura Holt mailto: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] apply the function factor to multiple columns
I have a case where I would like to change multiple columns containing numbers to factors. I can change each column one at a time as in: TEMP.FACT$EXPOS01-factor(TEMP.FACT$EXPOS01,levels=c(1,2,3),labels=c(No ne,Low Impact,MedHigh Imp)) TEMP.FACT$EXPOS02-factor(TEMP.FACT$EXPOS02,levels=c(1,2,3),labels=c(No ne,Low Impact,MedHigh Imp)) TEMP.FACT$EXPOS03-factor(TEMP.FACT$EXPOS03,levels=c(1,2,3),labels=c(No ne,Low Impact,MedHigh Imp)) summary(TEMP.FACT[,1:3]) EXPOS01 EXPOS02 EXPOS03 None :219 None :432 None :377 Low Impact :428 Low Impact :248 Low Impact :297 MedHigh Imp:108 MedHigh Imp: 77 MedHigh Imp: 83 NA's : 25 NA's : 23 NA's : 23 It would be much easier, however to use apply as in: TEMP.FACT [,1:3]-apply(TEMP.FACT[,1:3],2,factor,labels=c(None,Low Impact,MedHigh Imp)) This appears to work (no error messages); however, this does not actually change the variables to factors. That is they are still treated as numbers: summary(TEMP.FACT[,1:3]) EXPOS01 EXPOS02 EXPOS03 Min. : 1.000 Min. : 1.000 Min. : 1.000 1st Qu.: 1.000 1st Qu.: 1.000 1st Qu.: 1.000 Median : 2.000 Median : 1.000 Median : 2.000 Mean : 1.853 Mean : 1.531 Mean : 1.612 3rd Qu.: 2.000 3rd Qu.: 2.000 3rd Qu.: 2.000 Max. : 3.000 Max. : 3.000 Max. : 3.000 NA's :25.000 NA's :23.000 NA's :23.000 Any ideas on how I could efficiently change a lot of columns to factors? Thanks, PB [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply the function factor to multiple columns
you could try this way: dat - data.frame(V1 = factor(1:3), V2 = factor(1:3), V3 = factor(1:3)) dat[1:3] - lapply(dat[1:3], factor, labels = c(None, Low Impact, MedHigh Imp)) dat I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/16/336899 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat/ http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: Bliese, Paul D LTC USAMH [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Tuesday, May 31, 2005 10:42 AM Subject: [R] apply the function factor to multiple columns I have a case where I would like to change multiple columns containing numbers to factors. I can change each column one at a time as in: TEMP.FACT$EXPOS01-factor(TEMP.FACT$EXPOS01,levels=c(1,2,3),labels=c(No ne,Low Impact,MedHigh Imp)) TEMP.FACT$EXPOS02-factor(TEMP.FACT$EXPOS02,levels=c(1,2,3),labels=c(No ne,Low Impact,MedHigh Imp)) TEMP.FACT$EXPOS03-factor(TEMP.FACT$EXPOS03,levels=c(1,2,3),labels=c(No ne,Low Impact,MedHigh Imp)) summary(TEMP.FACT[,1:3]) EXPOS01 EXPOS02 EXPOS03 None :219 None :432 None :377 Low Impact :428 Low Impact :248 Low Impact :297 MedHigh Imp:108 MedHigh Imp: 77 MedHigh Imp: 83 NA's : 25 NA's : 23 NA's : 23 It would be much easier, however to use apply as in: TEMP.FACT [,1:3]-apply(TEMP.FACT[,1:3],2,factor,labels=c(None,Low Impact,MedHigh Imp)) This appears to work (no error messages); however, this does not actually change the variables to factors. That is they are still treated as numbers: summary(TEMP.FACT[,1:3]) EXPOS01 EXPOS02 EXPOS03 Min. : 1.000 Min. : 1.000 Min. : 1.000 1st Qu.: 1.000 1st Qu.: 1.000 1st Qu.: 1.000 Median : 2.000 Median : 1.000 Median : 2.000 Mean : 1.853 Mean : 1.531 Mean : 1.612 3rd Qu.: 2.000 3rd Qu.: 2.000 3rd Qu.: 2.000 Max. : 3.000 Max. : 3.000 Max. : 3.000 NA's :25.000 NA's :23.000 NA's :23.000 Any ideas on how I could efficiently change a lot of columns to factors? Thanks, PB [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] apply question
Dear R users, I´ve got a simple question but somehow I can´t find the solution: I have a data frame with columns 1-5 containing one set of integer values, and columns 6-10 containing another set of integer values. Columns 6-10 contain NA´s at some places. I now want to calculate (1) the number of values in each row of columns 6-10 that were NA´s (2) the sum of all values on columns 1-5 for which there were no missing values in the corresponding cells of columns 6-10. Example: (let´s call the data frame data) Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8 Col9 Col10 1 2 5 2 3 NA 5 NA1 4 3 1 4 5 2 6 NA 4 NA 1 The result would then be (for the first row) (1) There were 2 NA´s in columns 6-10. (2) The mean of Columns 1-5 was 2+2+3=7 (because there were NA´s in the 1st and 3rd position in rows 6-10) So far, I know how to calculate the rowSums for the data.frame, but I don´t know how to condition these on the values of columns 6-10 rowSums(data[,1:5]) #that´s straightforward apply(data[,6:19],1,function(x)sum(is.na(x))) #this also works fine But I don´t know how to select just the desired values of columns 1-5 (as described above) Can anyone help me? Thanks a lot in advance! Best regards Christoph __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply question
- Original Message - From: Christoph Scherber [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Monday, May 02, 2005 10:52 AM Subject: [R] apply question Dear R users, I´ve got a simple question but somehow I can´t find the solution: I have a data frame with columns 1-5 containing one set of integer values, and columns 6-10 containing another set of integer values. Columns 6-10 contain NA´s at some places. I now want to calculate (1) the number of values in each row of columns 6-10 that were NA´s (2) the sum of all values on columns 1-5 for which there were no missing values in the corresponding cells of columns 6-10. Example: (let´s call the data frame data) Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8 Col9 Col10 1 2 5 2 3 NA 5 NA1 4 3 1 4 5 2 6 NA 4 NA 1 The result would then be (for the first row) (1) There were 2 NA´s in columns 6-10. (2) The mean of Columns 1-5 was 2+2+3=7 (because there were NA´s in the 1st and 3rd position in rows 6-10) So far, I know how to calculate the rowSums for the data.frame, but I don´t know how to condition these on the values of columns 6-10 rowSums(data[,1:5]) #that´s straightforward apply(data[,6:19],1,function(x)sum(is.na(x))) #this also works fine But I don´t know how to select just the desired values of columns 1-5 (as described above) tmp - rowSums(data[apply(data[,6:19],1,function(x) sum(is.na(x)))==0,1:5]) Now, tmp contains only the rowsums for the rows with no NAs in the other columns. Sean __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] apply question
Try: ## Number of NAs in columns 6-10. colSums(is.na(data[6:10])) Col6 Col7 Col8 Col9 Col10 1 1 1 1 0 ## Number of NAs in each row of columns 6-10. rowSums(is.na(data[6:10])) 1 2 2 2 ## Sums of rows 1-5 omitting corresponding NAs in cols 6-10. rowSums(data[,1:5] * !is.na(data[,6:10])) 1 2 7 9 If all entries are numeric, it'd be easier to use matrices instead of data frames. HTH, Andy From: Christoph Scherber Dear R users, I´ve got a simple question but somehow I can´t find the solution: I have a data frame with columns 1-5 containing one set of integer values, and columns 6-10 containing another set of integer values. Columns 6-10 contain NA´s at some places. I now want to calculate (1) the number of values in each row of columns 6-10 that were NA´s (2) the sum of all values on columns 1-5 for which there were no missing values in the corresponding cells of columns 6-10. Example: (let´s call the data frame data) Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8 Col9 Col10 1 2 5 2 3 NA 5 NA1 4 3 1 4 5 2 6 NA 4 NA 1 The result would then be (for the first row) (1) There were 2 NA´s in columns 6-10. (2) The mean of Columns 1-5 was 2+2+3=7 (because there were NA´s in the 1st and 3rd position in rows 6-10) So far, I know how to calculate the rowSums for the data.frame, but I don´t know how to condition these on the values of columns 6-10 rowSums(data[,1:5]) #that´s straightforward apply(data[,6:19],1,function(x)sum(is.na(x))) #this also works fine But I don´t know how to select just the desired values of columns 1-5 (as described above) Can anyone help me? Thanks a lot in advance! Best regards Christoph __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply question
you could try something like this: dat - rbind(c(1, 2, 5, 2, 3, NA, 5, NA, 1, 4), c(3, 1, 4, 5, 2, 6, NA, 4, NA, 1)) ## # (1) rowSums(is.na(dat[, 6:10])) ## (2) dat. - dat[, 1:5] dat.[is.na(dat[, 6:10])] - NA rowSums(dat., na.rm=TRUE) rowMeans(dat., na.rm=TRUE) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/16/336899 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat/ http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: Christoph Scherber [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Monday, May 02, 2005 4:52 PM Subject: [R] apply question Dear R users, I´ve got a simple question but somehow I can´t find the solution: I have a data frame with columns 1-5 containing one set of integer values, and columns 6-10 containing another set of integer values. Columns 6-10 contain NA´s at some places. I now want to calculate (1) the number of values in each row of columns 6-10 that were NA´s (2) the sum of all values on columns 1-5 for which there were no missing values in the corresponding cells of columns 6-10. Example: (let´s call the data frame data) Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8 Col9 Col10 1 2 5 2 3 NA 5 NA1 4 3 1 4 5 2 6 NA 4 NA 1 The result would then be (for the first row) (1) There were 2 NA´s in columns 6-10. (2) The mean of Columns 1-5 was 2+2+3=7 (because there were NA´s in the 1st and 3rd position in rows 6-10) So far, I know how to calculate the rowSums for the data.frame, but I don´t know how to condition these on the values of columns 6-10 rowSums(data[,1:5]) #that´s straightforward apply(data[,6:19],1,function(x)sum(is.na(x))) #this also works fine But I don´t know how to select just the desired values of columns 1-5 (as described above) Can anyone help me? Thanks a lot in advance! Best regards Christoph __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply question
On 5/2/05, Christoph Scherber [EMAIL PROTECTED] wrote: Dear R users, I´ve got a simple question but somehow I can´t find the solution: I have a data frame with columns 1-5 containing one set of integer values, and columns 6-10 containing another set of integer values. Columns 6-10 contain NA´s at some places. I now want to calculate (1) the number of values in each row of columns 6-10 that were NA´s Supposing our data is called DF, rowSums(!is.na(DF[,6:10])) (2) the sum of all values on columns 1-5 for which there were no missing values in the corresponding cells of columns 6-10. In the expression below 1 + 0 *DF[,6:10] is like DF[,6:10] except all non-NAs are replaced by 1. Multiplying DF[,1:5] by that effectively replaces each element in DF[,1:5] with an NA if the corresponding DF[,6:10] contained an NA. rowSums( DF[,1:5] * (1 + 0 * DF[,6:10]), na.rm = TRUE ) Example: (let´s call the data frame data) Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8 Col9 Col10 1 2 5 2 3 NA 5 NA1 4 3 1 4 5 2 6 NA 4 NA 1 The result would then be (for the first row) (1) There were 2 NA´s in columns 6-10. (2) The mean of Columns 1-5 was 2+2+3=7 (because there were NA´s in the 1st and 3rd position in rows 6-10) I guess you meant sum when you referred to mean in (2). If you really do want the mean replace rowSums with rowMeans in the expression given above in the answer to (2). __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] apply vs sapply vs loop - lm() call appl(y)ied on array
Dear useRs
(Code of the now mentioned small example is below)
I have 7 * 8 * 9 = 504 series of data (each length 5). For each of
theses series I want to compute a lm(), where the designmatrx X is the
same for all these computations.
The 504 series are in an array of dimension d.dim - c(5, 7, 8, 9)
means, the first dimension holds the data-series.
The lm computation needs performance optimization, since in fact the
dimensions are much larger. I compared the following approaches:
using a for-loop. using apply, and using sapply. All of these require
roughly the same time of computation. I was astonished since I expected
at least sapply to outperfomr the for-loop.
Do you have me another solution, which is faster? many thanks
here is the code
## --
t.length - 5
d.dim - c(t.length,7,8,9) # dimesions: time, x, y, z
Y - array( rep(1:t.length, prod(d.dim)) + rnorm(prod(d.dim), 0, 0.1),
d.dim)
X - c(1,3,2,4,5)
## performance tests
## using for loop
date()
z - rep(0, prod(d.dim[2:4]))
l - 0
for (i in 1:dim(Y)[4])
for (j in 1:dim(Y)[3])
for (k in 1:dim(Y)[2]) {
l - l + 1
z[l] - unlist(summary(lm(Y[,k, j, i] ~ X)))$r.squared
}
date()
## using apply
date()
z - apply(Y, 2:4, function(x) unlist(summary(lm(x ~ X)))$r.squared)
date()
## using sapply
date()
fac - rep(1:prod(d.dim[2:4]), rep(t.length, prod(d.dim[2:4])))
z - sapply(split(as.vector(Y), fac), FUN = function(x)
unlist(summary(lm(x ~ X)))$r.squared)
dim(z) - d.dim[2:4]
date()
## --
--
Christoph LehmannPhone: ++41 31 930 93 83
Department of Psychiatric NeurophysiologyMobile: ++41 76 570 28 00
University Hospital of Clinical Psychiatry Fax:++41 31 930 99 61
Waldau[EMAIL PROTECTED]
CH-3000 Bern 60 http://www.puk.unibe.ch/cl/pn_ni_cv_cl_04.html
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RE: [R] apply vs sapply vs loop - lm() call appl(y)ied on array
Christoph --
There was just a thread on this earlier this week. You can search in the
archives for the title: refitting lm() with same x, different y.
(Actually, it doesn't turn up in the R site search yet, at least for me.
But if you just go to the archive of recent messages, available through
CRAN, you can search on refitting and find it. The original post was from
William Valdar, on April 19.)
Hope this helps,
Matt Wiener
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Christoph Lehmann
Sent: Thursday, April 21, 2005 9:24 AM
To: R-help@stat.math.ethz.ch
Subject: [R] apply vs sapply vs loop - lm() call appl(y)ied on array
Dear useRs
(Code of the now mentioned small example is below)
I have 7 * 8 * 9 = 504 series of data (each length 5). For each of
theses series I want to compute a lm(), where the designmatrx X is the
same for all these computations.
The 504 series are in an array of dimension d.dim - c(5, 7, 8, 9)
means, the first dimension holds the data-series.
The lm computation needs performance optimization, since in fact the
dimensions are much larger. I compared the following approaches:
using a for-loop. using apply, and using sapply. All of these require
roughly the same time of computation. I was astonished since I expected
at least sapply to outperfomr the for-loop.
Do you have me another solution, which is faster? many thanks
here is the code
## --
t.length - 5
d.dim - c(t.length,7,8,9) # dimesions: time, x, y, z
Y - array( rep(1:t.length, prod(d.dim)) + rnorm(prod(d.dim), 0, 0.1),
d.dim)
X - c(1,3,2,4,5)
## performance tests
## using for loop
date()
z - rep(0, prod(d.dim[2:4]))
l - 0
for (i in 1:dim(Y)[4])
for (j in 1:dim(Y)[3])
for (k in 1:dim(Y)[2]) {
l - l + 1
z[l] - unlist(summary(lm(Y[,k, j, i] ~ X)))$r.squared
}
date()
## using apply
date()
z - apply(Y, 2:4, function(x) unlist(summary(lm(x ~ X)))$r.squared)
date()
## using sapply
date()
fac - rep(1:prod(d.dim[2:4]), rep(t.length, prod(d.dim[2:4])))
z - sapply(split(as.vector(Y), fac), FUN = function(x)
unlist(summary(lm(x ~ X)))$r.squared)
dim(z) - d.dim[2:4]
date()
## --
--
Christoph LehmannPhone: ++41 31 930 93 83
Department of Psychiatric NeurophysiologyMobile: ++41 76 570 28 00
University Hospital of Clinical Psychiatry Fax:++41 31 930 99 61
Waldau[EMAIL PROTECTED]
CH-3000 Bern 60 http://www.puk.unibe.ch/cl/pn_ni_cv_cl_04.html
__
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Re: [R] apply vs sapply vs loop - lm() call appl(y)ied on array
Ok thanks to a hint of Matthew to a former post with a similar request I
have now three faster solutions (see below), the last one being the
fastest, but the former two also faster than the for-loop,
apply(lm(formula)) and sapply(lm(formula)) versions in my last mail:
one problem only: using lsfit I can't get directly measures such as
r.squared ...
---
## using lm with a matrix response (recommended by BDR)
date()
rsq -unlist(summary(lm(array(c(Y), dim = c(t.length, prod(d.dim[2:4])))
~ X)))[seq(22, prod(d.dim[2:4]) * 30, by = 30)] #get r.squared list-element
names(rsq) - prod(d.dim[2:4])
rsq - array(rsq, dim = d.dim[2:4])
date()
## using sapply and lsfit instead of lm (recommended by Kevin Wright)
date()
fac - rep(1:prod(d.dim[2:4]), rep(t.length, prod(d.dim[2:4])))
z - sapply(split(as.vector(Y), fac), FUN = function(x) lsfit(X, x)$coef[2])
dim(z) - d.dim[2:4]
date()
## using lsfit with a matrix response:
date()
rsq -lsfit(X, array(c(Y), dim = c(t.length, prod(d.dim[2:4]$coef[2,]
names(rsq) - prod(d.dim[2:4])
rsq - array(rsq, dim = d.dim[2:4])
date()
--
thanks
Christoph
Wiener, Matthew wrote:
Christoph --
There was just a thread on this earlier this week. You can search in the
archives for the title: refitting lm() with same x, different y.
(Actually, it doesn't turn up in the R site search yet, at least for me.
But if you just go to the archive of recent messages, available through
CRAN, you can search on refitting and find it. The original post was from
William Valdar, on April 19.)
Hope this helps,
Matt Wiener
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Christoph Lehmann
Sent: Thursday, April 21, 2005 9:24 AM
To: R-help@stat.math.ethz.ch
Subject: [R] apply vs sapply vs loop - lm() call appl(y)ied on array
Dear useRs
(Code of the now mentioned small example is below)
I have 7 * 8 * 9 = 504 series of data (each length 5). For each of
theses series I want to compute a lm(), where the designmatrx X is the
same for all these computations.
The 504 series are in an array of dimension d.dim - c(5, 7, 8, 9)
means, the first dimension holds the data-series.
The lm computation needs performance optimization, since in fact the
dimensions are much larger. I compared the following approaches:
using a for-loop. using apply, and using sapply. All of these require
roughly the same time of computation. I was astonished since I expected
at least sapply to outperfomr the for-loop.
Do you have me another solution, which is faster? many thanks
here is the code
## --
t.length - 5
d.dim - c(t.length,7,8,9) # dimesions: time, x, y, z
Y - array( rep(1:t.length, prod(d.dim)) + rnorm(prod(d.dim), 0, 0.1),
d.dim)
X - c(1,3,2,4,5)
## performance tests
## using for loop
date()
z - rep(0, prod(d.dim[2:4]))
l - 0
for (i in 1:dim(Y)[4])
for (j in 1:dim(Y)[3])
for (k in 1:dim(Y)[2]) {
l - l + 1
z[l] - unlist(summary(lm(Y[,k, j, i] ~ X)))$r.squared
}
date()
## using apply
date()
z - apply(Y, 2:4, function(x) unlist(summary(lm(x ~ X)))$r.squared)
date()
## using sapply
date()
fac - rep(1:prod(d.dim[2:4]), rep(t.length, prod(d.dim[2:4])))
z - sapply(split(as.vector(Y), fac), FUN = function(x)
unlist(summary(lm(x ~ X)))$r.squared)
dim(z) - d.dim[2:4]
date()
## --
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[R] apply
Hi, simple question I guess: the following line works well: aveBehav=c(apply(sdata, 2, mean)) However, I would like to pass an argument to the function mean, namely na.rm=TRUE Does anyone knows how to do this? Thanks in advance, Jan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply
On Apr 7, 2005, at 8:27 AM, malte wrote:
Hi,
simple question I guess:
the following line works well:
aveBehav=c(apply(sdata, 2, mean))
However, I would like to pass an argument to the function mean, namely
na.rm=TRUE
apply(sdata,2,function(x) {mean(x,na.rm=TRUE)})
Sean
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RE: [R] apply
From: malte Hi, simple question I guess: the following line works well: aveBehav=c(apply(sdata, 2, mean)) However, I would like to pass an argument to the function mean, namely na.rm=TRUE Does anyone knows how to do this? aveBehav - apply(sdata, 2, mean, na.rm=TRUE) or more efficiently: aveBehav - colMeans(sdata, na.rm=TRUE) Read ?apply and look at the ... argument. If you don't understand how it works, try the example on that page. Andy Thanks in advance, Jan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply
try, apply(sdata, 2, mean, na.rm=TRUE) or # assuming `sdata' is a matrix colMeans(sdata, na.rm=TRUE) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/16/336899 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat/ http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: malte [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Thursday, April 07, 2005 2:27 PM Subject: [R] apply Hi, simple question I guess: the following line works well: aveBehav=c(apply(sdata, 2, mean)) However, I would like to pass an argument to the function mean, namely na.rm=TRUE Does anyone knows how to do this? Thanks in advance, Jan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply
malte [EMAIL PROTECTED] writes: aveBehav=c(apply(sdata, 2, mean)) aveBehav= apply(sdata, 2, mean, na.rm=TRUE) and ?apply will tell you about this. + seth __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply
On 7 Apr 2005 at 14:27, malte wrote: Hi, simple question I guess: the following line works well: aveBehav=c(apply(sdata, 2, mean)) Hallo try aveBehav=c(apply(sdata, 2, mean, na.rm=T)) Cheers Petr However, I would like to pass an argument to the function mean, namely na.rm=TRUE Does anyone knows how to do this? Thanks in advance, Jan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Petr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] apply a function to a rolling subset of a vector
Does anyone know an easy way to calculate the rolling 20 period average or sum of a vector? For instance: x - rnorm(1000) y - apply.subset(x,20,fun=sum) The first element of y would contain the sum of elements 1 to 20, the second element of y would contain the sum of elements 2:21, and so on. I thought I had seen this on the list a year or so ago, but I couldn't find anything in the archives. Thanks in advance, Whit [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] apply a function to a rolling subset of a vector
Try this: ?convolve x-rnorm(1000) y-rep(1,20) z-convolve(x,y,type=filter) plot(x,type=l) str(z) num [1:981] 6.31 7.28 8.16 7.39 4.65 ... lines(c(rep(0,10),z,rep(0,10)),col=yellow,lwd=3) lines(c(rep(0,10),z,rep(0,10))/length(y),col=red,lwd=3) #running mean You wrote: Does anyone know an easy way to calculate the rolling 20 period average or sum of a vector? For instance: x - rnorm(1000) y - apply.subset(x,20,fun=sum) The first element of y would contain the sum of elements 1 to 20, the second element of y would contain the sum of elements 2:21, and so on. I thought I had seen this on the list a year or so ago, but I couldn't find anything in the archives. Thanks in advance, Whit [[alternative HTML version deleted]] Ken Knoblauch Inserm U 371 Cerveau et Vision 18 avenue du Doyen Lepine 69675 Bron cedex France tel: +33 (0)4 72 91 34 77 fax: +33 (0)4 72 91 34 61 portable: 06 84 10 64 10 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply a function to a rolling subset of a vector
On Wed, 2 Mar 2005 17:22:43 -0500, Whit Armstrong [EMAIL PROTECTED] wrote : Does anyone know an easy way to calculate the rolling 20 period average or sum of a vector? For instance: x - rnorm(1000) y - apply.subset(x,20,fun=sum) The first element of y would contain the sum of elements 1 to 20, the second element of y would contain the sum of elements 2:21, and so on. I thought I had seen this on the list a year or so ago, but I couldn't find anything in the archives. I don't know of a general purpose function, but filter() (in the stats package) can do the example you give, or any other linear filter. e.g. x - rnorm(1000) y - filter(x, rep(1,20)) puts 20 element sums into y. The vector ends up the same length as x, with NAs at the beginning and end (by default). Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply a function to a rolling subset of a vector
On Wed, 2005-03-02 at 17:22 -0500, Whit Armstrong wrote: Does anyone know an easy way to calculate the rolling 20 period average or sum of a vector? For instance: x - rnorm(1000) y - apply.subset(x,20,fun=sum) The first element of y would contain the sum of elements 1 to 20, the second element of y would contain the sum of elements 2:21, and so on. I thought I had seen this on the list a year or so ago, but I couldn't find anything in the archives. You can use the running() function in the gtools package, which is in the gregmisc bundle: x - rnorm(1000) running(x, fun = sum, width = 20) 1:20 2:21 3:22 4:23 5:24 -2.009684610 -2.205737077 -1.410810606 -2.226661837 -1.684604289 6:25 7:26 8:27 9:28 10:29 -4.492008605 -3.816273719 -5.348364598 -6.444591766 -5.263013812 11:30 12:31 13:32 14:33 15:34 -4.609829115 -5.935537291 -6.909232329 -4.881021777 -5.803659103 ... See ?running for more information, after installing gregmisc from CRAN. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply a function to a rolling subset of a vector
Whit Armstrong whit at twinfieldscapital.com writes: : : Does anyone know an easy way to calculate the rolling 20 period average : or sum of a vector? : : For instance: : x - rnorm(1000) : : y - apply.subset(x,20,fun=sum) : : The first element of y would contain the sum of elements 1 to 20, the : second element of y : would contain the sum of elements 2:21, and so on. : : I thought I had seen this on the list a year or so ago, but I couldn't : find anything in the archives. : Look at ?filter . Also ?embed and gtools::running . filter is the fastest. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] apply a function to a rolling subset of a vector
Thanks, everyone, for all the suggestions. The rollFun turs out to be just what I needed. Cheers, Whit -Original Message- From: Kjetil Brinchmann Halvorsen [mailto:[EMAIL PROTECTED] Sent: Wednesday, March 02, 2005 5:45 PM To: Whit Armstrong Cc: r-help@stat.math.ethz.ch Subject: Re: [R] apply a function to a rolling subset of a vector Whit Armstrong wrote: Does anyone know an easy way to calculate the rolling 20 period average or sum of a vector? For instance: x - rnorm(1000) y - apply.subset(x,20,fun=sum) help.search(rolling) gives me (among others) RollingAnalysis(fSeries) Rolling Analysis so trying library(fSeries) x - rnorm(1000) y - rollFun(x, 20, mean) Kjetil The first element of y would contain the sum of elements 1 to 20, the second element of y would contain the sum of elements 2:21, and so on. I thought I had seen this on the list a year or so ago, but I couldn't find anything in the archives. Thanks in advance, Whit [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Kjetil Halvorsen. Peace is the most effective weapon of mass construction. -- Mahdi Elmandjra -- No virus found in this outgoing message. Checked by AVG Anti-Virus. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] apply for nested lists
apply() statements **are** disguised loops and therefore are **not**
necessarily more efficient than explicit looping. Their principal advantage
is usually code readability.
As another readability issue, note that x[[i]][[j]][[k]] can be abbreviated
to x[[c(i,j,k]].
I leave to others the pleasure of deciphering and improving your code,
however.
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
The business of the statistician is to catalyze the scientific learning
process. - George E. P. Box
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
Alexandre Sanchez Pla
Sent: Wednesday, January 26, 2005 8:50 AM
To: r-help@stat.math.ethz.ch
Subject: [R] apply for nested lists
Hi,
I am working with lists whose terms are lists whose terms are
lists. Although
the real ones contain locuslink identifiers and GO
annotations (I work with the
Bioconductor GO) package, I have prepared an simplified
example of what I have
and what I would like to do with it:
Imagine I have a list such as:
tst.list-list(1=list(1A=list(ID=1A,VAL=172),1B=list
(ID=1B,VAL=134),1C=list(ID=1C,VAL=0)),2=list(2A=NA),
3=list(3A=list
(ID=3A,VAL=33),3B=list(ID=3B,VAL=2)))
I would like, for instance, to be able to extract some values
such as the
content of the VAL field, which may sometimes not be available.
I may do it using a nested for such as:
x-character(0)
for (i in 1:length(tst.list)){
if (!is.na(tst.list[[i]][[1]][[1]])){
for (j in 1:length(tst.list[[i]]))
{x-c(x,tst.list[[i]][[j]]$VAL)}}
else
{x-c(x, NA)}}
which gives me what I need
x
[1] 172 134 0 NA 33 2
According to most R documents this may be done more
efficiently using apply
instructions, but I have failed in my temptatives to obtain the same
Thanks for any help.
Alex
--
Dr.Alex Sánchez
Departament d'Estadística.
Universitat de Barcelona
[EMAIL PROTECTED]
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Re: [R] apply for nested lists
Alexandre Sanchez Pla asanchez at ub.edu writes:
:
: Hi,
:
: I am working with lists whose terms are lists whose terms are lists.
Although
: the real ones contain locuslink identifiers and GO annotations (I work with
the
: Bioconductor GO) package, I have prepared an simplified example of what I
have
: and what I would like to do with it:
:
: Imagine I have a list such as:
:
: tst.list-list(1=list(1A=list(ID=1A,VAL=172),1B=list
: (ID=1B,VAL=134),1C=list(ID=1C,VAL=0)),2=list(2A=NA),3=list
(3A=list
: (ID=3A,VAL=33),3B=list(ID=3B,VAL=2)))
:
: I would like, for instance, to be able to extract some values such as the
: content of the VAL field, which may sometimes not be available.
: I may do it using a nested for such as:
:
: x-character(0)
: for (i in 1:length(tst.list)){
: if (!is.na(tst.list[[i]][[1]][[1]])){
: for (j in 1:length(tst.list[[i]]))
: {x-c(x,tst.list[[i]][[j]]$VAL)}}
: else
: {x-c(x, NA)}}
:
: which gives me what I need
:
: x
: [1] 172 134 0 NA 33 2
:
: According to most R documents this may be done more efficiently using apply
: instructions, but I have failed in my temptatives to obtain the same
:
: Thanks for any help.
:
Try this:
f - function(x) if (is.null(x$VAL)) NA else x$VAL
unlist(lapply(tst.list, lapply, f))
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Re: [R] apply for nested lists
Actually, what you want is sapply.
sapply(tst.list, [[, VAL)
Kevin
Alexandre Sanchez Pla wrote:
Hi,
I am working with lists whose terms are lists whose terms are lists. Although
the real ones contain locuslink identifiers and GO annotations (I work with the
Bioconductor GO) package, I have prepared an simplified example of what I have
and what I would like to do with it:
Imagine I have a list such as:
tst.list-list(1=list(1A=list(ID=1A,VAL=172),1B=list
(ID=1B,VAL=134),1C=list(ID=1C,VAL=0)),2=list(2A=NA),3=list(3A=list
(ID=3A,VAL=33),3B=list(ID=3B,VAL=2)))
I would like, for instance, to be able to extract some values such as the
content of the VAL field, which may sometimes not be available.
I may do it using a nested for such as:
x-character(0)
for (i in 1:length(tst.list)){
if (!is.na(tst.list[[i]][[1]][[1]])){
for (j in 1:length(tst.list[[i]]))
{x-c(x,tst.list[[i]][[j]]$VAL)}}
else
{x-c(x, NA)}}
which gives me what I need
x
[1] 172 134 0 NA 33 2
According to most R documents this may be done more efficiently using apply
instructions, but I have failed in my temptatives to obtain the same
Thanks for any help.
Alex
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[R] apply function
Hi all, I have a question about apply function. Is that possible to pass some non-default arguments in the function we want to apply ? For example: if mat is a matrix and I want to use the tabulate function on its row. The command apply(mat,1,tabulate) works but I have problem with this one apply(mat, 1, tabulate(nbins=4)). Any clue ? Thanks, Eric -- Eric Pellegrini, PhD Computer-Chemie-Centrum University of Erlangen-Nürnberg Nägelbachstraße, 25 D-91052 Erlangen Germany __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] apply function
Try apply(mat, 1, tabulate, nbins=4). HTH, Andy From: Eric Pellegrini Hi all, I have a question about apply function. Is that possible to pass some non-default arguments in the function we want to apply ? For example: if mat is a matrix and I want to use the tabulate function on its row. The command apply(mat,1,tabulate) works but I have problem with this one apply(mat, 1, tabulate(nbins=4)). Any clue ? Thanks, Eric -- Eric Pellegrini, PhD Computer-Chemie-Centrum University of Erlangen-Nürnberg Nägelbachstraße, 25 D-91052 Erlangen Germany __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply function
Eric Pellegrini [EMAIL PROTECTED] writes: Hi all, I have a question about apply function. Is that possible to pass some non-default arguments in the function we want to apply ? For example: if mat is a matrix and I want to use the tabulate function on its row. The command apply(mat,1,tabulate) works but I have problem with this one apply(mat, 1, tabulate(nbins=4)). Any clue ? You might have gotten one by R'ing TFHP apply(mat, 1, tabulate, nbins=4) There's even an example! -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] apply ( , , table)
a - matrix (c( 7, 1, 1, 2, 6, 3, 4, 0, 1, 4, 5, 1, 8, 4, 4, 6, 1, 1, 2, 5), nrow=4, byrow=TRUE) b - apply (a, 1, table) apply documentation says clearly that if the rows of the result of FUN are the same length, then an array will be returned. And column-major would be the appropriate order in R. But b above is pretty opaque compared to what one would expect, and what one would get from apply ( , , table) if the rows were not of equal length. One needs to do something like n - matrix (apply (a, 1, function (x) unique (sort (x))), nrow=nrow(a)) to get the corresponding names of b to figure out the counts. Denis White __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] apply ( , , table)
apply() tries to be a bit smart about what it does (sometimes maybe too smart), but it actually is pretty useful a lot of the time. It's extremely widely used, so changing the behavior is not an option -- changing the behavior would break a lot of existing code. (Personally, I'd prefer it if apply() put its dimensions back together in a slightly more intelligent way, i.e., if apply(x, 1, c) and apply(x, 2, c) returned the same thing, but apply is how it is.) In situations where you don't want apply() to try to construct a matrix from your results, you can wrap the results in a list, to force apply() to return just a list of results, e.g. (the outer lapply() strips off an unnecessary level of list depth): b2 - lapply(apply (a, 1, function(x) list(table(x))), [[, 1) length(b2) [1] 4 b2[[1]] x 1 2 6 7 2 1 1 1 attributes(b2[[1]]) $dim [1] 4 $dimnames $dimnames$x [1] 1 2 6 7 $class [1] table Your particular case might benefit from more information given to table, which allows it to provide results in a more uniform format, e.g.: b1 - apply (a, 1, function(x) table(factor(x, levels=0:9))) b1 [,1] [,2] [,3] [,4] 00100 12112 21001 30100 40220 50011 61001 71000 80010 90000 hope this helps, Tony Plate At Tuesday 10:42 AM 8/24/2004, [EMAIL PROTECTED] wrote: a - matrix (c( 7, 1, 1, 2, 6, 3, 4, 0, 1, 4, 5, 1, 8, 4, 4, 6, 1, 1, 2, 5), nrow=4, byrow=TRUE) b - apply (a, 1, table) apply documentation says clearly that if the rows of the result of FUN are the same length, then an array will be returned. And column-major would be the appropriate order in R. But b above is pretty opaque compared to what one would expect, and what one would get from apply ( , , table) if the rows were not of equal length. One needs to do something like n - matrix (apply (a, 1, function (x) unique (sort (x))), nrow=nrow(a)) to get the corresponding names of b to figure out the counts. Denis White __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] apply on a data frame
Hi R People: There are 2 data sets of the iris data: one is iris3, which is a 3-d array, and the other is iris, which is a data frame with 150 rows and 6 variables. Getting means is straightforward from the 3-day iris3 set: apply(iris3,c(2,3),mean) Setosa Versicolor Virginica Sepal L. 5.006 5.936 6.588 Sepal W. 3.428 2.770 2.974 Petal L. 1.462 4.260 5.552 Petal W. 0.246 1.326 2.026 Is there is similar way to obtain those values from the iris data frame, please? thanks in advance R Version 1.9.1 Windows. Sincerely, Laura Holt mailto: [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] apply and data frames
Whoops! Just found it: by(iris[,1:4],Species,mean) Sorry for the inconvenience. Laura. Security. http://clinic.mcafee.com/clinic/ibuy/campaign.asp?cid=3963 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] apply/looping query
I have several large matrices each having perhaps one or two
data points missing. For instance one point in 2881 is missing. As I want to perform
various analyses on these matrices I feel it is not
unreasonable to linearly interpolate over the missing points. I want to basically
fill in
the gaps of the original matrix - the following piece of code
doesn't work and i'm not sure where i'm going wrong:
x=1:2881
my.new.matrix-matrix(nrow=2881,ncol=20)
for(i in 1:20){
my.new.matrix[[i]]-approx(x,my.matrix[,i],n=2881)
}
the error message says:
Error: more elements supplied than there are to replace
where am I going wrong??
Thanks in advance..
Laura
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Re: [R] apply/looping query
Laura Quinn wrote:
x=1:2881
my.new.matrix-matrix(nrow=2881,ncol=20)
for(i in 1:20){
my.new.matrix[[i]]-approx(x,my.matrix[,i],n=2881)
}
the error message says:
Error: more elements supplied than there are to replace
where am I going wrong??
approx() returns a list with $x and $y components
I'd use [,x] instead of [[i]] to replace columns.
Hence:
for(i in 1:20){
my.new.matrix[,i]-approx(x,my.matrix[,i],n=2881)$y
}
Baz
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[R] Apply a function to each cell of a ragged matrix
R-Helpers: There are a matrix x and a factor f. nrow(x) == length(f), e.g.: x - matrix(1:6, nrow = 3) f - factor(c(daytime, daytime, night)) I want the sum of all elements of rows of x for each corresponding level in factor f, In this case, I want output like: daytime [1] x[1,1]+x[2,1]+x[1,2]+x[2,2] night [2] x[3,1]+x[3,2] But, tapply(x,f,sum) or by(x,f,sum) do not work. What other functions can I use? Thank you very much Xiao __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Apply a function to each cell of a ragged matrix
rowsum(x,f) --- Date: Tue, 17 Feb 2004 17:38:46 -0500 From: XIAO LIU [EMAIL PROTECTED] To: R Help [EMAIL PROTECTED] Subject: [R] Apply a function to each cell of a ragged matrix R-Helpers: There are a matrix x and a factor f. nrow(x) == length(f), e.g.: x - matrix(1:6, nrow = 3) f - factor(c(daytime, daytime, night)) I want the sum of all elements of rows of x for each corresponding level in factor f, In this case, I want output like: daytime [1] x[1,1]+x[2,1]+x[1,2]+x[2,2] night [2] x[3,1]+x[3,2] But, tapply(x,f,sum) or by(x,f,sum) do not work. What other functions can I use? Thank you very much Xiao __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Apply a function to each cell of a ragged matrix
Hi, you could simply use functions such as: time - dim (3) for ( i in 1:3) time [i] - x[i,1]+x[i,2] the result of time will be the sum of rows. best regards.. - Original Message - From: XIAO LIU [EMAIL PROTECTED] To: R Help [EMAIL PROTECTED] Sent: Wednesday, February 18, 2004 12:38 AM Subject: [R] Apply a function to each cell of a ragged matrix R-Helpers: There are a matrix x and a factor f. nrow(x) == length(f), e.g.: x - matrix(1:6, nrow = 3) f - factor(c(daytime, daytime, night)) I want the sum of all elements of rows of x for each corresponding level in factor f, In this case, I want output like: daytime [1] x[1,1]+x[2,1]+x[1,2]+x[2,2] night [2] x[3,1]+x[3,2] But, tapply(x,f,sum) or by(x,f,sum) do not work. What other functions can I use? Thank you very much Xiao __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Apply a function to each cell of a ragged matrix
sapply(split(x,f), sum)
daytime night
12 9
HTH,
Andy
From: XIAO LIU
R-Helpers:
There are a matrix x and a factor f. nrow(x) == length(f), e.g.:
x - matrix(1:6, nrow = 3)
f - factor(c(daytime, daytime, night))
I want the sum of all elements of rows of x for each
corresponding level in factor f,
In this case, I want output like:
daytime [1] x[1,1]+x[2,1]+x[1,2]+x[2,2]
night [2] x[3,1]+x[3,2]
But, tapply(x,f,sum) or by(x,f,sum) do not work. What other
functions can I use?
Thank you very much
Xiao
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RE: [R] Apply a function to each cell of a ragged matrix
I misread your post. Try any of these: rowSums(rowsum(x,f)) rowsum(rowSums(x),f) tapply(rowSums(x),f,sum) by(rowSums(x),f,sum) --- Date: Tue, 17 Feb 2004 19:10:11 -0500 (EST) From: Gabor Grothendieck [EMAIL PROTECTED] To: [EMAIL PROTECTED], [EMAIL PROTECTED] Subject: RE: [R] Apply a function to each cell of a ragged matrix rowsum(x,f) --- Date: Tue, 17 Feb 2004 17:38:46 -0500 From: XIAO LIU [EMAIL PROTECTED] To: R Help [EMAIL PROTECTED] Subject: [R] Apply a function to each cell of a ragged matrix R-Helpers: There are a matrix x and a factor f. nrow(x) == length(f), e.g.: x - matrix(1:6, nrow = 3) f - factor(c(daytime, daytime, night)) I want the sum of all elements of rows of x for each corresponding level in factor f, In this case, I want output like: daytime [1] x[1,1]+x[2,1]+x[1,2]+x[2,2] night [2] x[3,1]+x[3,2] But, tapply(x,f,sum) or by(x,f,sum) do not work. What other functions can I use? __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] apply to multiple arrays simultaneously
Off the top of my head, seems like you can abind() the two together and then
run apply. See the abind package on CRAN.
HTH,
Andy
From: Carlos Soares
Dear R users,
Suppose two arrays which partly have the same dimensions. For
instance,
a1 and a2 with dim(a1) is c(3,4,5,6) and dim(a2) is
c(3,4,7,8). How can
I perform an apply on the equivalent part of the dimensions on both
arrays simultaneously? Something like:
apply(list(a1, a2), c(1,2), function(x,y) {my.function(x,y)})
A useful bonus would be, assuming that my.function always returns an
array withthe same dimensions (e.g., c(2,3,4), that the final result
would be an array wit
h dimensions c(3,4,2,3,4).
With best regards,
Carlos
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