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
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)
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
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=)))
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
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)
-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
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
, 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
-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
- 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
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,
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
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
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
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$z. The logic is
if( x y || x z ) 1 else 0
for a particular row.
It is simple and very inefficient to use for(i in
You don't need apply. Just do
foo$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
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$z. The logic is
if( x y || x z ) 1 else 0
for a particular row.
It is simple and very
or
with(foo, (x y) * (x z))
On 5/10/07, jim holtman [EMAIL PROTECTED] wrote:
You don't need apply. Just do
foo$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).
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:
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.
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
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
?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
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
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
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
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
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
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?
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
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
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
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
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:
Dear R-user
I have 100 lists.
Each list has several components.
For example,
data1
$a
[1] 1 2
$b
[1] 3 4
$c
[1] 5
There are data1, data2,, data100. All lists have the same number and the
same name of components.
Is there any function I can use for applying to only a specific component
Maybe this helps
( data1 = list(a=c(1,2), b=c(3,4), c=c(5,6,7)) )
( data2 = list(a=c(10,11), b=c(30,40), c=c(70,80)) )
cc - NULL
for(data in ls(pattern=^data[0-9]+$)) {
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
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
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
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.
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
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
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,
://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
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
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
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
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
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
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
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
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 -
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,
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
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
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]]
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
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
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
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]
__
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))
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
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
- 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
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.
/~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
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
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,
@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
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
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
__
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
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
: +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
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
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 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.
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
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,
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
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,
: [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
:[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
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
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
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
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
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
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
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
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.
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
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[EMAIL PROTECTED] mailing list
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PLEASE do read the
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
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
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]
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
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
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
] 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
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)
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