Re: [R] Comparing matrices in R - matrixB %in% matrixA
Since both of you seem to have misinterpreted my response, consider the
following for clarification:
A <- matrix(1:1000, 1000, 10)
B <- A[1:100, ]
# my recommended solution
t1 <- system.time({match(as.data.frame(t(B)), as.data.frame(t(A)))})
# similar to John's recommended solution
t2 <- system.time({
+ AA <- as.list(as.data.frame(t(A)))
+ BB <- as.list(as.data.frame(t(B)))
+ which( AA %in% BB )
+ })
t3 <- system.time({
+ lresult <- rep( NA, nrow(A) )
+ for ( ia in seq.int( nrow( A ) ) ) {
+ lres <- FALSE
+ ib <- 0
+ while ( ib < nrow( B ) & !lres ) {
+ ib <- ib + 1
+ lres <- all( A[ ia, ] == B[ ib, ] )
+ }
+ lresult[ ia ] <- lres
+ }
+ which( lresult )
+ })
t4 <- system.time({
+ res<-c()
+ rowsB = length(B[,1])
+ rowsA = length(A[,1])
+ colsB = length(B[1,])
+ colsA = length(A[1,])
+ for (i in 1:rowsB){
+ for (j in 1:colsB){
+ for (k in 1:rowsA){
+ for (l in 1:colsA){
+ if(A[k,l]==B[i,j]){res<-c(res,k)}
+ }
+ }
+ }
+ }
+ unique(sort(res))
+ })
t1
user system elapsed
0.022 0.000 0.020
t2
user system elapsed
0.020.000.02
t3
user system elapsed
0.748 0.000 0.746
t4
user system elapsed
16.612 0.016 16.636
# data.frames are lists, but applying as.list seems to speed up the
# match for some reason
t2[1]/t1[1]
user.self
0.9090909
# intended comparison for learning purposes
t4[1]/t3[1]
user.self
22.20856
I recognize that the reference implementation does not need to be
optimized, but the changes I suggested to it illustrate an incremental
improvement toward "thinking in R" rather than the optimal solution.
On Fri, 31 Oct 2014, John Fox wrote:
Dear Jeff,
For curiosity, I compared your solution with the one I posted earlier this
morning (when I was working on a slower computer, accounting for the somewhat
different timings for my solution):
snip --
A <- matrix(1:1, 1, 10)
B <- A[1:1000, ]
system.time({
+AA <- as.list(as.data.frame(t(A)))
+BB <- as.list(as.data.frame(t(B)))
+print(sum(AA %in% BB))
+ })
[1] 1000
user system elapsed
0.140.010.16
system.time({
+ lresult <- rep( NA, nrow(A) )
+ for ( ia in seq.int( nrow( A ) ) ) {
+ lres <- FALSE
+ ib <- 0
+ while ( ib < nrow( B ) & !lres ) {
+ ib <- ib + 1
+ lres <- all( A[ ia, ] == B[ ib, ] )
+ }
+ lresult[ ia ] <- lres
+ }
+ print(sum( lresult ))
+ })
[1] 1000
user system elapsed
45.760.01 45.77
46/0.16
[1] 287.5
snip --
So the solution using nested loops is more than 2 orders of magnitude slower
for this problem. Of course, for a one-off problem, depending on its size, the
difference may not matter.
Best,
John
---
John Fox, Professor
McMaster University
Hamilton, Ontario, Canada
http://socserv.socsci.mcmaster.ca/jfox/
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Jeff Newmiller
Sent: Friday, October 31, 2014 10:15 AM
To: Charles Novaes de Santana; r-help@r-project.org
Subject: Re: [R] Comparing matrices in R - matrixB %in% matrixA
Thank you for the reproducible example, but posting in HTML can corrupt
your example code so please learn to set your email client mail format
appropriately when posting to this list.
I think this [1] post, found with a quick Google search for "R match
matrix", fits your situation perfectly.
match(data.frame(t(B)), data.frame(t(A)))
Note that concatenating vectors in loops is bad news... a basic
optimization for your code would be to preallocate a logical result
vector and fill in each element with a TRUE/FALSE in the outer loop,
and use the which() function on that completed vector to identify the
index numbers (if you really need that). For example:
lresult <- rep( NA, nrow(A) )
for ( ia in seq.int( nrow( A ) ) ) {
lres <- FALSE
ib <- 0
while ( ib < nrow( B ) & !lres ) {
ib <- ib + 1
lres <- all( A[ ia, ] == B[ ib, ] )
}
lresult[ ia ] <- lres
}
result <- which( lresult )
[1] http://stackoverflow.com/questions/12697122/in-r-match-function-
for-rows-or-columns-of-matrix
---
Jeff NewmillerThe . . Go
Live...
DCN:Basics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#..
Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#.
rocks...1k
---
Sent from my phone. Please excuse my br
Re: [R] Comparing matrices in R - matrixB %in% matrixA
Dear Jeff,
For curiosity, I compared your solution with the one I posted earlier this
morning (when I was working on a slower computer, accounting for the somewhat
different timings for my solution):
snip --
> A <- matrix(1:1, 1, 10)
> B <- A[1:1000, ]
>
> system.time({
+AA <- as.list(as.data.frame(t(A)))
+BB <- as.list(as.data.frame(t(B)))
+print(sum(AA %in% BB))
+ })
[1] 1000
user system elapsed
0.140.010.16
>
>
> system.time({
+ lresult <- rep( NA, nrow(A) )
+ for ( ia in seq.int( nrow( A ) ) ) {
+ lres <- FALSE
+ ib <- 0
+ while ( ib < nrow( B ) & !lres ) {
+ ib <- ib + 1
+ lres <- all( A[ ia, ] == B[ ib, ] )
+ }
+ lresult[ ia ] <- lres
+ }
+ print(sum( lresult ))
+ })
[1] 1000
user system elapsed
45.760.01 45.77
> 46/0.16
[1] 287.5
snip --
So the solution using nested loops is more than 2 orders of magnitude slower
for this problem. Of course, for a one-off problem, depending on its size, the
difference may not matter.
Best,
John
---
John Fox, Professor
McMaster University
Hamilton, Ontario, Canada
http://socserv.socsci.mcmaster.ca/jfox/
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Jeff Newmiller
> Sent: Friday, October 31, 2014 10:15 AM
> To: Charles Novaes de Santana; r-help@r-project.org
> Subject: Re: [R] Comparing matrices in R - matrixB %in% matrixA
>
> Thank you for the reproducible example, but posting in HTML can corrupt
> your example code so please learn to set your email client mail format
> appropriately when posting to this list.
>
> I think this [1] post, found with a quick Google search for "R match
> matrix", fits your situation perfectly.
>
> match(data.frame(t(B)), data.frame(t(A)))
>
> Note that concatenating vectors in loops is bad news... a basic
> optimization for your code would be to preallocate a logical result
> vector and fill in each element with a TRUE/FALSE in the outer loop,
> and use the which() function on that completed vector to identify the
> index numbers (if you really need that). For example:
>
> lresult <- rep( NA, nrow(A) )
> for ( ia in seq.int( nrow( A ) ) ) {
> lres <- FALSE
> ib <- 0
> while ( ib < nrow( B ) & !lres ) {
> ib <- ib + 1
> lres <- all( A[ ia, ] == B[ ib, ] )
> }
> lresult[ ia ] <- lres
> }
> result <- which( lresult )
>
> [1] http://stackoverflow.com/questions/12697122/in-r-match-function-
> for-rows-or-columns-of-matrix
> ---
>
> Jeff NewmillerThe . . Go
> Live...
> DCN:Basics: ##.#. ##.#. Live
> Go...
> Live: OO#.. Dead: OO#..
> Playing
> Research Engineer (Solar/BatteriesO.O#. #.O#. with
> /Software/Embedded Controllers) .OO#. .OO#.
> rocks...1k
> ---
>
> Sent from my phone. Please excuse my brevity.
>
> On October 31, 2014 6:20:38 AM PDT, Charles Novaes de Santana
> wrote:
> >My apologies, because I sent the message before finishing it. i am
> very
> >sorry about this. Please find below my message (I use to write the
> >messages
> >from the end to the beginning... sorry :)).
> >
> >Dear all,
> >
> >I am trying to compare two matrices, in order to find in which rows of
> >a
> >matrix A I can find the same values as in matrix B. I am trying to do
> >it
> >for matrices with around 2500 elements, but please find below a toy
> >example:
> >
> >A = matrix(1:10,nrow=5)
> >B = A[-c(1,2,3),];
> >
> >So
> >> A
> > [,1] [,2]
> >[1,]16
> >[2,]27
> >[3,]38
> >[4,]49
> >[5,]5 10
> >
> >and
> >> B
> > [,1] [,2]
> >[1,]49
> >[2,]5 10
> >
> >I would like to compare A and B in order to find in which rows of A I
> >can
> >find the rows of B. Something similar to %in% with one dimensional
> >arrays.
> >In the example above, the answer should be 4 and 5.
> >
> >I did a function to do it (see it below), it gives me the correct
> >answer
> >for this toy example, but the excess of for-loops makes it extremely
> >slow
> >for larger matrices. I was wondering if there is a better w
Re: [R] Comparing matrices in R - matrixB %in% matrixA
Thank you, Jeff, for your message.
I did a search, but maybe my problem was that I didn't know the correct way
to search my problem (in other words: my vocabulary in R/English is not
good). Because of this I choose to send a message to the list the most
detailed as possible, and with a first solution, that was not the optimum
one.
Thank you very much for taking your time to think about my problem and to
search for a good solution! I much appreciate it! But I think John'
suggestion does what I need and it is much simpler :)
Best,
Charles
P.S.: the HTML issue is probably because I am using a webmail-client and I
did a copy and paste from the first email I sent.
On Fri, Oct 31, 2014 at 3:15 PM, Jeff Newmiller
wrote:
> Thank you for the reproducible example, but posting in HTML can corrupt
> your example code so please learn to set your email client mail format
> appropriately when posting to this list.
>
> I think this [1] post, found with a quick Google search for "R match
> matrix", fits your situation perfectly.
>
> match(data.frame(t(B)), data.frame(t(A)))
>
> Note that concatenating vectors in loops is bad news... a basic
> optimization for your code would be to preallocate a logical result vector
> and fill in each element with a TRUE/FALSE in the outer loop, and use the
> which() function on that completed vector to identify the index numbers (if
> you really need that). For example:
>
> lresult <- rep( NA, nrow(A) )
> for ( ia in seq.int( nrow( A ) ) ) {
> lres <- FALSE
> ib <- 0
> while ( ib < nrow( B ) & !lres ) {
> ib <- ib + 1
> lres <- all( A[ ia, ] == B[ ib, ] )
> }
> lresult[ ia ] <- lres
> }
> result <- which( lresult )
>
> [1]
> http://stackoverflow.com/questions/12697122/in-r-match-function-for-rows-or-columns-of-matrix
> ---
> Jeff NewmillerThe . . Go Live...
> DCN:Basics: ##.#. ##.#. Live
> Go...
> Live: OO#.. Dead: OO#.. Playing
> Research Engineer (Solar/BatteriesO.O#. #.O#. with
> /Software/Embedded Controllers) .OO#. .OO#. rocks...1k
> ---
> Sent from my phone. Please excuse my brevity.
>
> On October 31, 2014 6:20:38 AM PDT, Charles Novaes de Santana <
> charles.sant...@gmail.com> wrote:
> >My apologies, because I sent the message before finishing it. i am very
> >sorry about this. Please find below my message (I use to write the
> >messages
> >from the end to the beginning... sorry :)).
> >
> >Dear all,
> >
> >I am trying to compare two matrices, in order to find in which rows of
> >a
> >matrix A I can find the same values as in matrix B. I am trying to do
> >it
> >for matrices with around 2500 elements, but please find below a toy
> >example:
> >
> >A = matrix(1:10,nrow=5)
> >B = A[-c(1,2,3),];
> >
> >So
> >> A
> > [,1] [,2]
> >[1,]16
> >[2,]27
> >[3,]38
> >[4,]49
> >[5,]5 10
> >
> >and
> >> B
> > [,1] [,2]
> >[1,]49
> >[2,]5 10
> >
> >I would like to compare A and B in order to find in which rows of A I
> >can
> >find the rows of B. Something similar to %in% with one dimensional
> >arrays.
> >In the example above, the answer should be 4 and 5.
> >
> >I did a function to do it (see it below), it gives me the correct
> >answer
> >for this toy example, but the excess of for-loops makes it extremely
> >slow
> >for larger matrices. I was wondering if there is a better way to do
> >this
> >kind of comparison. Any idea? Sorry if it is a stupid question.
> >
> >matbinmata<-function(B,A){
> >res<-c();
> >rowsB = length(B[,1]);
> >rowsA = length(A[,1]);
> >colsB = length(B[1,]);
> >colsA = length(A[1,]);
> >for (i in 1:rowsB){
> >for (j in 1:colsB){
> >for (k in 1:rowsA){
> >for (l in 1:colsA){
> >if(A[k,l]==B[i,j]){res<-c(res,k);}
> >}
> >}
> >}
> >}
> >return(unique(sort(res)));
> >}
> >
> >
> >Best,
> >
> >Charles
> >
> >On Fri, Oct 31, 2014 at 2:12 PM, Charles Novaes de Santana <
> >charles.sant...@gmail.com> wrote:
> >
> >> A = matrix(1:10,nrow=5)
> >> B = A[-c(1,2,3),];
> >>
> >> So
> >> > A
> >> [,1] [,2]
> >> [1,]16
> >> [2,]27
> >> [3,]38
> >> [4,]49
> >> [5,]5 10
> >>
> >> and
> >> > B
> >> [,1] [,2]
> >> [1,]49
> >> [2,]5 10
> >>
> >> I would like to compare A and B in order to find in which rows of A I
> >can
> >> find the rows of B. Something similar to %in% with one dimensional
> >arrays.
> >> In the example above, the answer should be 4 and 5.
> >>
> >> I did a function to do it (see it below), it gives me the correct
> >answer
> >> for this toy example, but the excess of for-loops makes it extremely
> >slow
> >> for larger matric
Re: [R] Comparing matrices in R - matrixB %in% matrixA
Thank you for the reproducible example, but posting in HTML can corrupt your
example code so please learn to set your email client mail format appropriately
when posting to this list.
I think this [1] post, found with a quick Google search for "R match matrix",
fits your situation perfectly.
match(data.frame(t(B)), data.frame(t(A)))
Note that concatenating vectors in loops is bad news... a basic optimization
for your code would be to preallocate a logical result vector and fill in each
element with a TRUE/FALSE in the outer loop, and use the which() function on
that completed vector to identify the index numbers (if you really need that).
For example:
lresult <- rep( NA, nrow(A) )
for ( ia in seq.int( nrow( A ) ) ) {
lres <- FALSE
ib <- 0
while ( ib < nrow( B ) & !lres ) {
ib <- ib + 1
lres <- all( A[ ia, ] == B[ ib, ] )
}
lresult[ ia ] <- lres
}
result <- which( lresult )
[1]
http://stackoverflow.com/questions/12697122/in-r-match-function-for-rows-or-columns-of-matrix
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
On October 31, 2014 6:20:38 AM PDT, Charles Novaes de Santana
wrote:
>My apologies, because I sent the message before finishing it. i am very
>sorry about this. Please find below my message (I use to write the
>messages
>from the end to the beginning... sorry :)).
>
>Dear all,
>
>I am trying to compare two matrices, in order to find in which rows of
>a
>matrix A I can find the same values as in matrix B. I am trying to do
>it
>for matrices with around 2500 elements, but please find below a toy
>example:
>
>A = matrix(1:10,nrow=5)
>B = A[-c(1,2,3),];
>
>So
>> A
> [,1] [,2]
>[1,]16
>[2,]27
>[3,]38
>[4,]49
>[5,]5 10
>
>and
>> B
> [,1] [,2]
>[1,]49
>[2,]5 10
>
>I would like to compare A and B in order to find in which rows of A I
>can
>find the rows of B. Something similar to %in% with one dimensional
>arrays.
>In the example above, the answer should be 4 and 5.
>
>I did a function to do it (see it below), it gives me the correct
>answer
>for this toy example, but the excess of for-loops makes it extremely
>slow
>for larger matrices. I was wondering if there is a better way to do
>this
>kind of comparison. Any idea? Sorry if it is a stupid question.
>
>matbinmata<-function(B,A){
>res<-c();
>rowsB = length(B[,1]);
>rowsA = length(A[,1]);
>colsB = length(B[1,]);
>colsA = length(A[1,]);
>for (i in 1:rowsB){
>for (j in 1:colsB){
>for (k in 1:rowsA){
>for (l in 1:colsA){
>if(A[k,l]==B[i,j]){res<-c(res,k);}
>}
>}
>}
>}
>return(unique(sort(res)));
>}
>
>
>Best,
>
>Charles
>
>On Fri, Oct 31, 2014 at 2:12 PM, Charles Novaes de Santana <
>charles.sant...@gmail.com> wrote:
>
>> A = matrix(1:10,nrow=5)
>> B = A[-c(1,2,3),];
>>
>> So
>> > A
>> [,1] [,2]
>> [1,]16
>> [2,]27
>> [3,]38
>> [4,]49
>> [5,]5 10
>>
>> and
>> > B
>> [,1] [,2]
>> [1,]49
>> [2,]5 10
>>
>> I would like to compare A and B in order to find in which rows of A I
>can
>> find the rows of B. Something similar to %in% with one dimensional
>arrays.
>> In the example above, the answer should be 4 and 5.
>>
>> I did a function to do it (see it below), it gives me the correct
>answer
>> for this toy example, but the excess of for-loops makes it extremely
>slow
>> for larger matrices. I was wondering if there is a better way to do
>this
>> kind of comparison. Any idea? Sorry if it is a stupid question.
>>
>> matbinmata<-function(B,A){
>> res<-c();
>> rowsB = length(B[,1]);
>> rowsA = length(A[,1]);
>> colsB = length(B[1,]);
>> colsA = length(A[1,]);
>> for (i in 1:rowsB){
>> for (j in 1:colsB){
>> for (k in 1:rowsA){
>> for (l in 1:colsA){
>> if(A[k,l]==B[i,j]){res<-c(res,k);}
>> }
>> }
>> }
>> }
>> return(unique(sort(res)));
>> }
>>
>>
>> Best,
>>
>> Charles
>>
>>
>> --
>> Um axé! :)
>>
>> --
>> Charles Novaes de Santana, PhD
>> http://www.imedea.uib-csic.es/~charles
>>
>
>
>
>--
>Um axé! :)
>
>--
>Charles Novaes de Santana, PhD
>http://www.imedea.uib-csic.es/~charles
>
> [[alternative HTML version deleted]]
>
>__
>R-help@r-project.org mailing list
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the po
Re: [R] Comparing matrices in R - matrixB %in% matrixA
Great!! It is perfect!
Thank you, John, for this elegant and fast suggestion!
Best,
Charles
On Fri, Oct 31, 2014 at 2:35 PM, John Fox wrote:
> Dear Charles,
>
> How about the following?
>
> --- snip -
>
> > AA <- as.list(as.data.frame(t(A)))
> > BB <- as.list(as.data.frame(t(B)))
> > which(AA %in% BB)
> [1] 4 5
>
> --- snip -
>
> This seems reasonably fast. For example:
>
> --- snip -
>
> > A <- matrix(1:1, 1, 10)
> > B <- A[1:1000, ]
> >
> > system.time({
> + AA <- as.list(as.data.frame(t(A)))
> + BB <- as.list(as.data.frame(t(B)))
> + print(sum(AA %in% BB))
> + })
> [1] 1000
>user system elapsed
>0.260.000.26
>
> --- snip -
>
> I hope this helps,
> John
>
>
> John Fox, Professor
> McMaster University
> Hamilton, Ontario, Canada
> http://socserv.mcmaster.ca/jfox/
>
>
>
>
> On Fri, 31 Oct 2014 14:20:38 +0100
> Charles Novaes de Santana wrote:
> > My apologies, because I sent the message before finishing it. i am very
> > sorry about this. Please find below my message (I use to write the
> messages
> > from the end to the beginning... sorry :)).
> >
> > Dear all,
> >
> > I am trying to compare two matrices, in order to find in which rows of a
> > matrix A I can find the same values as in matrix B. I am trying to do it
> > for matrices with around 2500 elements, but please find below a toy
> example:
> >
> > A = matrix(1:10,nrow=5)
> > B = A[-c(1,2,3),];
> >
> > So
> > > A
> > [,1] [,2]
> > [1,]16
> > [2,]27
> > [3,]38
> > [4,]49
> > [5,]5 10
> >
> > and
> > > B
> > [,1] [,2]
> > [1,]49
> > [2,]5 10
> >
> > I would like to compare A and B in order to find in which rows of A I can
> > find the rows of B. Something similar to %in% with one dimensional
> arrays.
> > In the example above, the answer should be 4 and 5.
> >
> > I did a function to do it (see it below), it gives me the correct answer
> > for this toy example, but the excess of for-loops makes it extremely slow
> > for larger matrices. I was wondering if there is a better way to do this
> > kind of comparison. Any idea? Sorry if it is a stupid question.
> >
> > matbinmata<-function(B,A){
> > res<-c();
> > rowsB = length(B[,1]);
> > rowsA = length(A[,1]);
> > colsB = length(B[1,]);
> > colsA = length(A[1,]);
> > for (i in 1:rowsB){
> > for (j in 1:colsB){
> > for (k in 1:rowsA){
> > for (l in 1:colsA){
> > if(A[k,l]==B[i,j]){res<-c(res,k);}
> > }
> > }
> > }
> > }
> > return(unique(sort(res)));
> > }
> >
> >
> > Best,
> >
> > Charles
> >
> > On Fri, Oct 31, 2014 at 2:12 PM, Charles Novaes de Santana <
> > charles.sant...@gmail.com> wrote:
> >
> > > A = matrix(1:10,nrow=5)
> > > B = A[-c(1,2,3),];
> > >
> > > So
> > > > A
> > > [,1] [,2]
> > > [1,]16
> > > [2,]27
> > > [3,]38
> > > [4,]49
> > > [5,]5 10
> > >
> > > and
> > > > B
> > > [,1] [,2]
> > > [1,]49
> > > [2,]5 10
> > >
> > > I would like to compare A and B in order to find in which rows of A I
> can
> > > find the rows of B. Something similar to %in% with one dimensional
> arrays.
> > > In the example above, the answer should be 4 and 5.
> > >
> > > I did a function to do it (see it below), it gives me the correct
> answer
> > > for this toy example, but the excess of for-loops makes it extremely
> slow
> > > for larger matrices. I was wondering if there is a better way to do
> this
> > > kind of comparison. Any idea? Sorry if it is a stupid question.
> > >
> > > matbinmata<-function(B,A){
> > > res<-c();
> > > rowsB = length(B[,1]);
> > > rowsA = length(A[,1]);
> > > colsB = length(B[1,]);
> > > colsA = length(A[1,]);
> > > for (i in 1:rowsB){
> > > for (j in 1:colsB){
> > > for (k in 1:rowsA){
> > > for (l in 1:colsA){
> > > if(A[k,l]==B[i,j]){res<-c(res,k);}
> > > }
> > > }
> > > }
> > > }
> > > return(unique(sort(res)));
> > > }
> > >
> > >
> > > Best,
> > >
> > > Charles
> > >
> > >
> > > --
> > > Um axé! :)
> > >
> > > --
> > > Charles Novaes de Santana, PhD
> > > http://www.imedea.uib-csic.es/~charles
> > >
> >
> >
> >
> > --
> > Um axé! :)
> >
> > --
> > Charles Novaes de Santana, PhD
> > http://www.imedea.uib-csic.es/~charles
> >
> > [[alternative HTML version deleted]]
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
>
--
Um axé! :)
--
Charles Novaes de Santana, PhD
http://www.imedea.uib-csic.es/~charles
Re: [R] Comparing matrices in R - matrixB %in% matrixA
Dear Charles,
How about the following?
--- snip -
> AA <- as.list(as.data.frame(t(A)))
> BB <- as.list(as.data.frame(t(B)))
> which(AA %in% BB)
[1] 4 5
--- snip -
This seems reasonably fast. For example:
--- snip -
> A <- matrix(1:1, 1, 10)
> B <- A[1:1000, ]
>
> system.time({
+ AA <- as.list(as.data.frame(t(A)))
+ BB <- as.list(as.data.frame(t(B)))
+ print(sum(AA %in% BB))
+ })
[1] 1000
user system elapsed
0.260.000.26
--- snip -
I hope this helps,
John
John Fox, Professor
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/
On Fri, 31 Oct 2014 14:20:38 +0100
Charles Novaes de Santana wrote:
> My apologies, because I sent the message before finishing it. i am very
> sorry about this. Please find below my message (I use to write the messages
> from the end to the beginning... sorry :)).
>
> Dear all,
>
> I am trying to compare two matrices, in order to find in which rows of a
> matrix A I can find the same values as in matrix B. I am trying to do it
> for matrices with around 2500 elements, but please find below a toy example:
>
> A = matrix(1:10,nrow=5)
> B = A[-c(1,2,3),];
>
> So
> > A
> [,1] [,2]
> [1,]16
> [2,]27
> [3,]38
> [4,]49
> [5,]5 10
>
> and
> > B
> [,1] [,2]
> [1,]49
> [2,]5 10
>
> I would like to compare A and B in order to find in which rows of A I can
> find the rows of B. Something similar to %in% with one dimensional arrays.
> In the example above, the answer should be 4 and 5.
>
> I did a function to do it (see it below), it gives me the correct answer
> for this toy example, but the excess of for-loops makes it extremely slow
> for larger matrices. I was wondering if there is a better way to do this
> kind of comparison. Any idea? Sorry if it is a stupid question.
>
> matbinmata<-function(B,A){
> res<-c();
> rowsB = length(B[,1]);
> rowsA = length(A[,1]);
> colsB = length(B[1,]);
> colsA = length(A[1,]);
> for (i in 1:rowsB){
> for (j in 1:colsB){
> for (k in 1:rowsA){
> for (l in 1:colsA){
> if(A[k,l]==B[i,j]){res<-c(res,k);}
> }
> }
> }
> }
> return(unique(sort(res)));
> }
>
>
> Best,
>
> Charles
>
> On Fri, Oct 31, 2014 at 2:12 PM, Charles Novaes de Santana <
> charles.sant...@gmail.com> wrote:
>
> > A = matrix(1:10,nrow=5)
> > B = A[-c(1,2,3),];
> >
> > So
> > > A
> > [,1] [,2]
> > [1,]16
> > [2,]27
> > [3,]38
> > [4,]49
> > [5,]5 10
> >
> > and
> > > B
> > [,1] [,2]
> > [1,]49
> > [2,]5 10
> >
> > I would like to compare A and B in order to find in which rows of A I can
> > find the rows of B. Something similar to %in% with one dimensional arrays.
> > In the example above, the answer should be 4 and 5.
> >
> > I did a function to do it (see it below), it gives me the correct answer
> > for this toy example, but the excess of for-loops makes it extremely slow
> > for larger matrices. I was wondering if there is a better way to do this
> > kind of comparison. Any idea? Sorry if it is a stupid question.
> >
> > matbinmata<-function(B,A){
> > res<-c();
> > rowsB = length(B[,1]);
> > rowsA = length(A[,1]);
> > colsB = length(B[1,]);
> > colsA = length(A[1,]);
> > for (i in 1:rowsB){
> > for (j in 1:colsB){
> > for (k in 1:rowsA){
> > for (l in 1:colsA){
> > if(A[k,l]==B[i,j]){res<-c(res,k);}
> > }
> > }
> > }
> > }
> > return(unique(sort(res)));
> > }
> >
> >
> > Best,
> >
> > Charles
> >
> >
> > --
> > Um axé! :)
> >
> > --
> > Charles Novaes de Santana, PhD
> > http://www.imedea.uib-csic.es/~charles
> >
>
>
>
> --
> Um axé! :)
>
> --
> Charles Novaes de Santana, PhD
> http://www.imedea.uib-csic.es/~charles
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Comparing matrices in R - matrixB %in% matrixA
My apologies, because I sent the message before finishing it. i am very
sorry about this. Please find below my message (I use to write the messages
from the end to the beginning... sorry :)).
Dear all,
I am trying to compare two matrices, in order to find in which rows of a
matrix A I can find the same values as in matrix B. I am trying to do it
for matrices with around 2500 elements, but please find below a toy example:
A = matrix(1:10,nrow=5)
B = A[-c(1,2,3),];
So
> A
[,1] [,2]
[1,]16
[2,]27
[3,]38
[4,]49
[5,]5 10
and
> B
[,1] [,2]
[1,]49
[2,]5 10
I would like to compare A and B in order to find in which rows of A I can
find the rows of B. Something similar to %in% with one dimensional arrays.
In the example above, the answer should be 4 and 5.
I did a function to do it (see it below), it gives me the correct answer
for this toy example, but the excess of for-loops makes it extremely slow
for larger matrices. I was wondering if there is a better way to do this
kind of comparison. Any idea? Sorry if it is a stupid question.
matbinmata<-function(B,A){
res<-c();
rowsB = length(B[,1]);
rowsA = length(A[,1]);
colsB = length(B[1,]);
colsA = length(A[1,]);
for (i in 1:rowsB){
for (j in 1:colsB){
for (k in 1:rowsA){
for (l in 1:colsA){
if(A[k,l]==B[i,j]){res<-c(res,k);}
}
}
}
}
return(unique(sort(res)));
}
Best,
Charles
On Fri, Oct 31, 2014 at 2:12 PM, Charles Novaes de Santana <
charles.sant...@gmail.com> wrote:
> A = matrix(1:10,nrow=5)
> B = A[-c(1,2,3),];
>
> So
> > A
> [,1] [,2]
> [1,]16
> [2,]27
> [3,]38
> [4,]49
> [5,]5 10
>
> and
> > B
> [,1] [,2]
> [1,]49
> [2,]5 10
>
> I would like to compare A and B in order to find in which rows of A I can
> find the rows of B. Something similar to %in% with one dimensional arrays.
> In the example above, the answer should be 4 and 5.
>
> I did a function to do it (see it below), it gives me the correct answer
> for this toy example, but the excess of for-loops makes it extremely slow
> for larger matrices. I was wondering if there is a better way to do this
> kind of comparison. Any idea? Sorry if it is a stupid question.
>
> matbinmata<-function(B,A){
> res<-c();
> rowsB = length(B[,1]);
> rowsA = length(A[,1]);
> colsB = length(B[1,]);
> colsA = length(A[1,]);
> for (i in 1:rowsB){
> for (j in 1:colsB){
> for (k in 1:rowsA){
> for (l in 1:colsA){
> if(A[k,l]==B[i,j]){res<-c(res,k);}
> }
> }
> }
> }
> return(unique(sort(res)));
> }
>
>
> Best,
>
> Charles
>
>
> --
> Um axé! :)
>
> --
> Charles Novaes de Santana, PhD
> http://www.imedea.uib-csic.es/~charles
>
--
Um axé! :)
--
Charles Novaes de Santana, PhD
http://www.imedea.uib-csic.es/~charles
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Comparing matrices in R - matrixB %in% matrixA
A = matrix(1:10,nrow=5)
B = A[-c(1,2,3),];
So
> A
[,1] [,2]
[1,]16
[2,]27
[3,]38
[4,]49
[5,]5 10
and
> B
[,1] [,2]
[1,]49
[2,]5 10
I would like to compare A and B in order to find in which rows of A I can
find the rows of B. Something similar to %in% with one dimensional arrays.
In the example above, the answer should be 4 and 5.
I did a function to do it (see it below), it gives me the correct answer
for this toy example, but the excess of for-loops makes it extremely slow
for larger matrices. I was wondering if there is a better way to do this
kind of comparison. Any idea? Sorry if it is a stupid question.
matbinmata<-function(B,A){
res<-c();
rowsB = length(B[,1]);
rowsA = length(A[,1]);
colsB = length(B[1,]);
colsA = length(A[1,]);
for (i in 1:rowsB){
for (j in 1:colsB){
for (k in 1:rowsA){
for (l in 1:colsA){
if(A[k,l]==B[i,j]){res<-c(res,k);}
}
}
}
}
return(unique(sort(res)));
}
Best,
Charles
--
Um axé! :)
--
Charles Novaes de Santana, PhD
http://www.imedea.uib-csic.es/~charles
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Comparing matrices
Also look at
?any
and
?all
Very handy functions.
Miguel
On Thu, Mar 11, 2010 at 12:37 PM, Esmail wrote:
> Hello all,
>
> I have two matrices, pop and pop2, each the same number of rows and
> columns that I want to compare for equality. I am concerned about
> efficiency in this operation.
>
> I've tried a few things without success so far. Doing something simple
> like:
>
> if (pop==pop2) { cat('equal') } else { cat('NOT equal') }
>
> results in the warning:
> 1: In if (pop == pop2) { :
> the condition has length > 1 and only the first element will be used
>
> so it seems to look only at the first element.
>
> print(pop==pop2) gives me a listing of TRUE/FALSE values for each
> element.
>
> I am really only looking for a single TRUE or FALSE value to tell me
> if the two populations (ie pop and pop2) are the same or different. In
> my application there will be about 200 columns and 50 rows and this
> comparison will happen very frequently (possibly a few thousand
> times), so I am concerned about efficiency.
>
> I am appending the code I used to generate my matrices and some things
> I tried (mentioned above) and also the output get so far.
>
> FWIW, Linux environment, R 2.10.1.
>
> Thanks,
> Esmail
>
> ps: Am I correct that if I do the assignment
>
>pop2 = pop
>
>I create a totally separate instance/(deep)copy of the
>data? I tried a few tests that seem to confirm this, but I'd
>rather be sure.
>
>
> --- code
>
> # create a binary vector of size "len"
> create_bin_Chromosome <- function(len)
> {
> sample(0:1, len, replace=T)
> }
>
> # create popsize members, each of length len
> create_pop_2 <- function(popsize, len)
> {
> datasize=len*popsize
>
> npop <- matrix(0, popsize, len, byrow=T)
>
> for(i in 1:popsize)
>npop[i,] = create_bin_Chromosome(len)
>
> npop
> }
>
>
> POP_SIZE = 3
> LEN = 8
>
> pop = create_pop_2(POP_SIZE, LEN)
> pop2 = pop
>
> print(pop==pop2)
> if (pop==pop2) { cat('equal\n') } else { cat('NOT equal\n') }
>
> print(pop)
> print(pop2)
>
> pop[2,5] = 99
> pop2[3,3] = 77
>
> print(pop==pop2)
> if (pop==pop2) { cat('equal\n') } else { cat('NOT equal\n') }
>
> print(pop)
> print(pop2)
>
> -- output ---
>
> > source('popc.R')
> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
> [1,] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
> [2,] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
> [3,] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
> equal
> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
> [1,]01010110
> [2,]01100000
> [3,]01001011
> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
> [1,]01010110
> [2,]01100000
> [3,]01001011
> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
> [1,] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
> [2,] TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE
> [3,] TRUE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
> equal
> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
> [1,]01010110
> [2,]0110 99000
> [3,]01001011
> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
> [1,]01010110
> [2,]01100000
> [3,]01 7701011
> Warning messages:
> 1: In if (pop == pop2) { :
> the condition has length > 1 and only the first element will be used
> 2: In if (pop == pop2) { :
> the condition has length > 1 and only the first element will be used
> >
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Comparing matrices
Look at:
identical(pop, pop2) # if you want to do "exact" comparison
Or
all.equal(pop, pop2, tolerance=1.e-10) # if you want to allow for
some small noise
Ravi.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Esmail
Sent: Thursday, March 11, 2010 7:38 AM
To: r-help@r-project.org
Subject: [R] Comparing matrices
Hello all,
I have two matrices, pop and pop2, each the same number of rows and
columns that I want to compare for equality. I am concerned about
efficiency in this operation.
I've tried a few things without success so far. Doing something simple like:
if (pop==pop2) { cat('equal') } else { cat('NOT equal') }
results in the warning:
1: In if (pop == pop2) { :
the condition has length > 1 and only the first element will be used
so it seems to look only at the first element.
print(pop==pop2) gives me a listing of TRUE/FALSE values for each
element.
I am really only looking for a single TRUE or FALSE value to tell me
if the two populations (ie pop and pop2) are the same or different. In
my application there will be about 200 columns and 50 rows and this
comparison will happen very frequently (possibly a few thousand
times), so I am concerned about efficiency.
I am appending the code I used to generate my matrices and some things
I tried (mentioned above) and also the output get so far.
FWIW, Linux environment, R 2.10.1.
Thanks,
Esmail
ps: Am I correct that if I do the assignment
pop2 = pop
I create a totally separate instance/(deep)copy of the
data? I tried a few tests that seem to confirm this, but I'd
rather be sure.
--- code
# create a binary vector of size "len"
create_bin_Chromosome <- function(len)
{
sample(0:1, len, replace=T)
}
# create popsize members, each of length len
create_pop_2 <- function(popsize, len)
{
datasize=len*popsize
npop <- matrix(0, popsize, len, byrow=T)
for(i in 1:popsize)
npop[i,] = create_bin_Chromosome(len)
npop
}
POP_SIZE = 3
LEN = 8
pop = create_pop_2(POP_SIZE, LEN)
pop2 = pop
print(pop==pop2)
if (pop==pop2) { cat('equal\n') } else { cat('NOT equal\n') }
print(pop)
print(pop2)
pop[2,5] = 99
pop2[3,3] = 77
print(pop==pop2)
if (pop==pop2) { cat('equal\n') } else { cat('NOT equal\n') }
print(pop)
print(pop2)
-- output ---
> source('popc.R')
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
[2,] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
[3,] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
equal
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]01010110
[2,]01100000
[3,]01001011
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]01010110
[2,]01100000
[3,]01001011
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
[2,] TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE
[3,] TRUE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
equal
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]01010110
[2,]0110 99000
[3,]01001011
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]01010110
[2,]01100000
[3,]01 7701011
Warning messages:
1: In if (pop == pop2) { :
the condition has length > 1 and only the first element will be used
2: In if (pop == pop2) { :
the condition has length > 1 and only the first element will be used
>
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Comparing matrices
Hello all,
I have two matrices, pop and pop2, each the same number of rows and
columns that I want to compare for equality. I am concerned about
efficiency in this operation.
I've tried a few things without success so far. Doing something simple like:
if (pop==pop2) { cat('equal') } else { cat('NOT equal') }
results in the warning:
1: In if (pop == pop2) { :
the condition has length > 1 and only the first element will be used
so it seems to look only at the first element.
print(pop==pop2) gives me a listing of TRUE/FALSE values for each
element.
I am really only looking for a single TRUE or FALSE value to tell me
if the two populations (ie pop and pop2) are the same or different. In
my application there will be about 200 columns and 50 rows and this
comparison will happen very frequently (possibly a few thousand
times), so I am concerned about efficiency.
I am appending the code I used to generate my matrices and some things
I tried (mentioned above) and also the output get so far.
FWIW, Linux environment, R 2.10.1.
Thanks,
Esmail
ps: Am I correct that if I do the assignment
pop2 = pop
I create a totally separate instance/(deep)copy of the
data? I tried a few tests that seem to confirm this, but I'd
rather be sure.
--- code
# create a binary vector of size "len"
create_bin_Chromosome <- function(len)
{
sample(0:1, len, replace=T)
}
# create popsize members, each of length len
create_pop_2 <- function(popsize, len)
{
datasize=len*popsize
npop <- matrix(0, popsize, len, byrow=T)
for(i in 1:popsize)
npop[i,] = create_bin_Chromosome(len)
npop
}
POP_SIZE = 3
LEN = 8
pop = create_pop_2(POP_SIZE, LEN)
pop2 = pop
print(pop==pop2)
if (pop==pop2) { cat('equal\n') } else { cat('NOT equal\n') }
print(pop)
print(pop2)
pop[2,5] = 99
pop2[3,3] = 77
print(pop==pop2)
if (pop==pop2) { cat('equal\n') } else { cat('NOT equal\n') }
print(pop)
print(pop2)
-- output ---
> source('popc.R')
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
[2,] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
[3,] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
equal
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]01010110
[2,]01100000
[3,]01001011
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]01010110
[2,]01100000
[3,]01001011
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
[2,] TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE
[3,] TRUE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
equal
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]01010110
[2,]0110 99000
[3,]01001011
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]01010110
[2,]01100000
[3,]01 7701011
Warning messages:
1: In if (pop == pop2) { :
the condition has length > 1 and only the first element will be used
2: In if (pop == pop2) { :
the condition has length > 1 and only the first element will be used
>
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] comparing matrices
ONKELINX, Thierry wrote: Have a look at all.equal matA <- matrix(1:4, ncol = 2) matB <- matA all.equal(matA, matB) matB[1,1] <- -10 all.equal(matA, matB) Hi Thierry, Thanks, all.equal does indicate if it's all equal so that works great! Much nicer than my hack - thanks, Esmail __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] comparing matrices
David Winsemius wrote: > identical( matrix((1:4), ncol=2), matrix((1:4), nrow=2)) [1] TRUE > identical( matrix((1:4), ncol=2), matrix((2:5), nrow=2)) [1] FALSE Thanks David, much appreciated. Esmail __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] comparing matrices
baptiste auguie wrote: I'm not sure I'm following you but have you tried, identical(matrix(c(1,1,1,1),ncol=2), matrix(c(1,1,1,1),ncol=2)) ?all.equal ?isTRUE ?identical and possibly the compare package, compare(matrix(c(1,1,1,1),ncol=2),matrix(c(1,1,1,1),ncol=2)) HTH, baptiste Hi Babtiste, Thanks for pointing out the various options that exist. R is a very rich language indeed and it's good to know how to accomplish tasks in various ways. Cheers, Esmail __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] comparing matrices
On Apr 26, 2009, at 1:02 PM, Esmail wrote:
I'm trying to compare two matrices made up of bits.
doing a simple comparison of
matA == matB
> identical( matrix((1:4), ncol=2), matrix((1:4), nrow=2))
[1] TRUE
> identical( matrix((1:4), ncol=2), matrix((2:5), nrow=2))
[1] FALSE
yields this sort of output.
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] FALSE TRUE FALSE TRUE TRUE FALSE
[2,] TRUE TRUE TRUE TRUE TRUE TRUE
[3,] FALSE TRUE FALSE FALSE FALSE TRUE
[4,] FALSE TRUE TRUE FALSE FALSE FALSE
[5,] TRUE TRUE TRUE TRUE FALSE FALSE
[6,] TRUE TRUE TRUE TRUE FALSE FALSE
I really would like just one comprehensive value to say TRUE or FALSE.
This is the hack (rather ugly I think) I put together that works,
but there has to be a nicer way, no?
res=pop[1:ROWS,] == keep[1:ROWS,]
if ((ROWS*COL) == sum(res))
{
cat('they are equal\n')
}else
cat('they are NOT equal\n')
That code is meaningless to us without a definition (in R) of ROWS,
COL, keep, and pop
Thanks!
Esmail
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] comparing matrices
I'm not sure I'm following you but have you tried,
identical(matrix(c(1,1,1,1),ncol=2), matrix(c(1,1,1,1),ncol=2))
?all.equal
?isTRUE
?identical
and possibly the compare package,
compare(matrix(c(1,1,1,1),ncol=2),matrix(c(1,1,1,1),ncol=2))
HTH,
baptiste
On 26 Apr 2009, at 18:02, Esmail wrote:
I'm trying to compare two matrices made up of bits.
doing a simple comparison of
matA == matB
yields this sort of output.
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] FALSE TRUE FALSE TRUE TRUE FALSE
[2,] TRUE TRUE TRUE TRUE TRUE TRUE
[3,] FALSE TRUE FALSE FALSE FALSE TRUE
[4,] FALSE TRUE TRUE FALSE FALSE FALSE
[5,] TRUE TRUE TRUE TRUE FALSE FALSE
[6,] TRUE TRUE TRUE TRUE FALSE FALSE
I really would like just one comprehensive value to say TRUE or FALSE.
This is the hack (rather ugly I think) I put together that works,
but there has to be a nicer way, no?
res=pop[1:ROWS,] == keep[1:ROWS,]
if ((ROWS*COL) == sum(res))
{
cat('they are equal\n')
}else
cat('they are NOT equal\n')
Thanks!
Esmail
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] comparing matrices
Have a look at all.equal
matA <- matrix(1:4, ncol = 2)
matB <- matA
all.equal(matA, matB)
matB[1,1] <- -10
all.equal(matA, matB)
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
thierry.onkel...@inbo.be
www.inbo.be
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Namens Esmail
Verzonden: zondag 26 april 2009 19:03
Aan: R mailing list
Onderwerp: [R] comparing matrices
I'm trying to compare two matrices made up of bits.
doing a simple comparison of
matA == matB
yields this sort of output.
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] FALSE TRUE FALSE TRUE TRUE FALSE
[2,] TRUE TRUE TRUE TRUE TRUE TRUE
[3,] FALSE TRUE FALSE FALSE FALSE TRUE
[4,] FALSE TRUE TRUE FALSE FALSE FALSE
[5,] TRUE TRUE TRUE TRUE FALSE FALSE
[6,] TRUE TRUE TRUE TRUE FALSE FALSE
I really would like just one comprehensive value to say TRUE or FALSE.
This is the hack (rather ugly I think) I put together that works,
but there has to be a nicer way, no?
res=pop[1:ROWS,] == keep[1:ROWS,]
if ((ROWS*COL) == sum(res))
{
cat('they are equal\n')
}else
cat('they are NOT equal\n')
Thanks!
Esmail
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer
en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is
door een geldig ondertekend document. The views expressed in this message
and any annex are purely those of the writer and may not be regarded as stating
an official position of INBO, as long as the message is not confirmed by a duly
signed document.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] comparing matrices
I'm trying to compare two matrices made up of bits.
doing a simple comparison of
matA == matB
yields this sort of output.
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] FALSE TRUE FALSE TRUE TRUE FALSE
[2,] TRUE TRUE TRUE TRUE TRUE TRUE
[3,] FALSE TRUE FALSE FALSE FALSE TRUE
[4,] FALSE TRUE TRUE FALSE FALSE FALSE
[5,] TRUE TRUE TRUE TRUE FALSE FALSE
[6,] TRUE TRUE TRUE TRUE FALSE FALSE
I really would like just one comprehensive value to say TRUE or FALSE.
This is the hack (rather ugly I think) I put together that works,
but there has to be a nicer way, no?
res=pop[1:ROWS,] == keep[1:ROWS,]
if ((ROWS*COL) == sum(res))
{
cat('they are equal\n')
}else
cat('they are NOT equal\n')
Thanks!
Esmail
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] comparing matrices using max or min
you just need pmax() or pmin(), e.g., check this: set.seed(123) M1 <- matrix(rnorm(20), 4, 5) M2 <- matrix(rnorm(20), 4, 5) M3 <- matrix(rnorm(20), 4, 5) M1; M2; M3 pmax(M1, M2, M3) pmin(M1, M2, M3) I hope it helps. Best, Dimitris Diogo André Alagador wrote: Dear all, I have 3 matrices with the same dimension, A,B,C and I would like to produce a matrix D where in each position would retrieve the max(or min) value along A,B,C taken from the same position. I guess that apply functions should fit, but for matrices objects I am not getting it. thanks in advance, Diogo André Alagador [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] comparing matrices using max or min
#how about this A <- rnorm(10) B <- rnorm(10) C <- rnorm(10) D <- data.frame(A,B,C) apply(D, MARGIN=1, FUN=min) On Thu, Nov 6, 2008 at 10:00 AM, Diogo André Alagador <[EMAIL PROTECTED]> wrote: > Dear all, > > I have 3 matrices with the same dimension, A,B,C and I would like to produce > a matrix D where in each position would retrieve the max(or min) value along > A,B,C taken from the same position. > I guess that apply functions should fit, but for matrices objects I am not > getting it. > > thanks in advance, > > Diogo André Alagador > >[[alternative HTML version deleted]] > > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] comparing matrices using max or min
Dear all, I have 3 matrices with the same dimension, A,B,C and I would like to produce a matrix D where in each position would retrieve the max(or min) value along A,B,C taken from the same position. I guess that apply functions should fit, but for matrices objects I am not getting it. thanks in advance, Diogo André Alagador [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
