Re: [R] Using apply with more than one matrix
> Thank you, A.K. I learned from both of your solutions. I find the one that
> uses alply easier to follow and understand intuitively.
Another approach is to only loop over the 3rd dimensional slices of
a1. Your original code, converted to a function so it is easier to
test and think about is
f0 <- function(a1, m1, m2) {
stopifnot(length(dim(a1))==3, length(dim(m1))==2, length(dim(m2))==2,
all(dim(m1)==dim(m2)), all(dim(m1)==dim(a1)[1:2]),
dim(a1)[3] > 1)
condition1 <- !is.na(m1)& m1 > m2
ans <- array(NA_real_, dim(m1)) # initialize
for(i in seq_len(dim(m1)[1])) {
for(j in seq_len(dim(m1)[2])) {
ind.not.na <- which(!is.na(a1[i,j,]))
if(condition1[i,j] && length(ind.not.na) > 0) {
ans[i,j] <- a1[i,j,ind.not.na[1]] + m2[i,j]
}
}
}
ans
}
I believe the following is an equivalent function, but I haven't
checked it much.
Note that the only loop is over the third dimension of a1, which I assume is not
too big. That updating loop could probably be replaced by a call to Reduce.
f1 <- function(a1, m1, m2) {
stopifnot(length(dim(a1))==3, length(dim(m1))==2, length(dim(m2))==2,
all(dim(m1)==dim(m2)), all(dim(m1)==dim(a1)[1:2]),
dim(a1)[3] > 1)
firstGoodInA1 <- array(NA_real_, dim(a1)[1:2])
for(k in seq_len(dim(a1)[3])) {
isStillBad <- is.na(firstGoodInA1)
firstGoodInA1[isStillBad] <- a1[, ,k][isStillBad]
}
condition1 <- !is.na(firstGoodInA1) & !is.na(m1) & (m1 > m2)
ans <- array(NA_real_, dim(m1))
ans[condition1] <- firstGoodInA1[condition1] + m2[condition1]
ans
}
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, May 1, 2014 at 8:30 AM, Waichler, Scott R
wrote:
> Thank you, A.K. I learned from both of your solutions. I find the one that
> uses alply easier to follow and understand intuitively. I guess I'll want to
> learn more about what plyr can do. I've been using R for years but hadn't
> pushed vectorization far enough in my code. Now my computing will be more
> efficient.
>
> Thanks also to David and the others who responded to my question. It all
> helped. --Scott Waichler
>
>> -Original Message-
>> From: arun [mailto:smartpink...@yahoo.com]
>> Sent: Thursday, May 01, 2014 6:24 AM
>> To: R. Help
>> Cc: Waichler, Scott R
>> Subject: Re: [R] Using apply with more than one matrix
>>
>> Hi,
>> Sorry, a typo in the previous function:
>> -
>> if (condition1[i] && !is.na(indx3)) {
>> arr[x1][indx3] + m2[i] ## should be mat2[i]
>> } else NA
>>
>> ---
>>
>> Also, you can try:
>> library(plyr)
>> evaluateNew <- function(arr, mat1, mat2){ if (!all(dim(mat1) ==
>> dim(mat2))) {
>> stop("both matrices should have equal dimensions")
>> }
>> indx1 <- unlist(alply(arr, c(1,2), function(x) {x[!is.na(x)][1]}))
>> condition1 <- !is.na(mat1) & mat1 > mat2 ifelse(condition1, indx1+mat2,
>> NA) } evaluateNew(a1, m1, m2)
>>
>> #[,1] [,2]
>> #[1,] NA 1.66
>> #[2,] 3.14 NA
>> A.K.
>>
>>
>>
>>
>> On Thursday, May 1, 2014 5:49 AM, arun wrote:
>>
>>
>> Hi,
>>
>> You may try:
>> evaluate <- function(arr, mat1, mat2) {
>> if (!all(dim(mat1) == dim(mat2))) {
>> stop("both matrices should have equal dimensions")
>> }
>> indx1 <- as.matrix(do.call(expand.grid, lapply(dim(arr), sequence)))
>> indx2 <- paste0(indx1[, 1], indx1[, 2])
>> condition1 <- !is.na(mat1) & mat1 > mat2
>> ans <- sapply(seq_along(unique(indx2)), function(i) {
>> x1 <- indx1[indx2 %in% unique(indx2)[i], ]
>> indx3 <- which(!is.na(arr[x1]))[1]
>> if (condition1[i] && !is.na(indx3)) {
>> arr[x1][indx3] + m2[i]
>> } else NA
>> })
>> dim(ans) <- dim(mat1)
>> ans
>> }
>> evaluate(a1,m1,m2)
>> # [,1] [,2]
>> #[1,] NA 1.66
>> #[2,] 3.14 NA
>>
>> A.K.
>>
>>
>>
>> On Thursday, May 1, 2014 2:53 AM, "Waichler, Scott R"
>> wrote:
>> > I would ask you to look at this loop-free approach and ask if this is
>> > not equally valid?
>> >
>> > ans <- matrix(NA, ncol=2, nrow=2)
>> > ind.not.na <- which(!is.na(a1))
>> > ans[] <- condition1*a1[,,ind.not.na[1]]+ m2 # two matrices of equal
>> >dimen
Re: [R] Using apply with more than one matrix
Thank you, A.K. I learned from both of your solutions. I find the one that
uses alply easier to follow and understand intuitively. I guess I'll want to
learn more about what plyr can do. I've been using R for years but hadn't
pushed vectorization far enough in my code. Now my computing will be more
efficient.
Thanks also to David and the others who responded to my question. It all
helped. --Scott Waichler
> -Original Message-
> From: arun [mailto:smartpink...@yahoo.com]
> Sent: Thursday, May 01, 2014 6:24 AM
> To: R. Help
> Cc: Waichler, Scott R
> Subject: Re: [R] Using apply with more than one matrix
>
> Hi,
> Sorry, a typo in the previous function:
> -
> if (condition1[i] && !is.na(indx3)) {
> arr[x1][indx3] + m2[i] ## should be mat2[i]
> } else NA
>
> ---
>
> Also, you can try:
> library(plyr)
> evaluateNew <- function(arr, mat1, mat2){ if (!all(dim(mat1) ==
> dim(mat2))) {
> stop("both matrices should have equal dimensions")
> }
> indx1 <- unlist(alply(arr, c(1,2), function(x) {x[!is.na(x)][1]}))
> condition1 <- !is.na(mat1) & mat1 > mat2 ifelse(condition1, indx1+mat2,
> NA) } evaluateNew(a1, m1, m2)
>
> # [,1] [,2]
> #[1,] NA 1.66
> #[2,] 3.14 NA
> A.K.
>
>
>
>
> On Thursday, May 1, 2014 5:49 AM, arun wrote:
>
>
> Hi,
>
> You may try:
> evaluate <- function(arr, mat1, mat2) {
> if (!all(dim(mat1) == dim(mat2))) {
> stop("both matrices should have equal dimensions")
> }
> indx1 <- as.matrix(do.call(expand.grid, lapply(dim(arr), sequence)))
> indx2 <- paste0(indx1[, 1], indx1[, 2])
> condition1 <- !is.na(mat1) & mat1 > mat2
> ans <- sapply(seq_along(unique(indx2)), function(i) {
> x1 <- indx1[indx2 %in% unique(indx2)[i], ]
> indx3 <- which(!is.na(arr[x1]))[1]
> if (condition1[i] && !is.na(indx3)) {
> arr[x1][indx3] + m2[i]
> } else NA
> })
> dim(ans) <- dim(mat1)
> ans
> }
> evaluate(a1,m1,m2)
> # [,1] [,2]
> #[1,] NA 1.66
> #[2,] 3.14 NA
>
> A.K.
>
>
>
> On Thursday, May 1, 2014 2:53 AM, "Waichler, Scott R"
> wrote:
> > I would ask you to look at this loop-free approach and ask if this is
> > not equally valid?
> >
> > ans <- matrix(NA, ncol=2, nrow=2)
> > ind.not.na <- which(!is.na(a1))
> > ans[] <- condition1*a1[,,ind.not.na[1]]+ m2 # two matrices of equal
> >dimensions, one logical.
> > ans
> > [,1] [,2]
> > [1,] NA 1.66
> > [2,] 2.74 NA
>
> Thanks, I am learning something. I didn't know you could multiply a
> logical object by a numerical one. But notice the answer is not the same
> as mine, because I am doing an operation on the vector of values a1[i,j,]
> first.
>
> I tried a modification on sapply below, but it doesn't work because I
> haven't referenced the 3d array a1 properly. So I guess I must try to get
> a 2d result from a1 first, then use that in matrix arithmetic.
>
> Sapply or mapply may work, I haven't used these much and will try to learn
> better how to use them. Your use of sapply looks good; but I'm trying to
> understand if and how I can bring in the operation on a1. This doesn't
> work:
>
> evaluate <- function(idx) {
> ind.not.na <- which(!is.na(a1[idx,])) ])) # doesn't work; improper
> indexing for a1
> if(length(ind.not.na) > 0) {
> return(condition1*(a1[idx,ind.not.na[1]] + m2[idx])) # doesn't work;
> improper indexing for a1
> }
> }
> vec <- sapply(seq(length(m2)), evaluate)
>
> Scott Waichler
>
>
> > -Original Message-
> > From: David Winsemius [mailto:dwinsem...@comcast.net]
> > Sent: Wednesday, April 30, 2014 8:46 PM
> > To: Waichler, Scott R
> > Cc: Bert Gunter; r-help@r-project.org
> > Subject: Re: [R] Using apply with more than one matrix
> >
> >
> > On Apr 30, 2014, at 6:03 PM, Waichler, Scott R wrote:
> >
> > > Here is a working example with no random parts. Thanks for your
> > patience and if I'm still off the mark with my presentation I'll drop
> > the matter.
> > >
> > > v <- c(NA, 1.5, NA, NA,
> > > NA, 1.1, 0.5, NA,
> > > NA, 1.3, 0.4, 0.9)
> > > a1 <- array(v, dim=c(2,2,3))
> > > m1 <- matrix(c(NA, 1.5, 2.1, NA), ncol=2, byrow=T)
> > > m2 <- matrix(c(1.56, 1.64, 1.16, 2.92), ncol=2)
> > >
Re: [R] Using apply with more than one matrix
> > Sapply or mapply may work, I haven't used these much and will try to
> learn better how to use them. Your use of sapply looks good; but I'm
> trying to understand if and how I can bring in the operation on a1. This
> doesn't work:
> >
> > evaluate <- function(idx) {
> > ind.not.na <- which(!is.na(a1[idx,])) ])) # doesn't work; improper
> > indexing for a1
> > if(length(ind.not.na) > 0) {
> >return(condition1*(a1[idx,ind.not.na[1]] + m2[idx])) # doesn't
> > work; improper indexing for a1 } } vec <- sapply(seq(length(m2)),
> > evaluate)
>
> Are we to assume that the length of `which(!is.na(a1[idx,])) ]))` is
> guaranteed to equal the length of the two other matrices? If not then what
> sort of relationships should be assumed?
ind.not.na is just a vector that could be any length. It is a selection on the
vector a1[i,j,]. I want to get the first element of that selection. The key
relationship is that the dimensions of the matrices and the first two
dimensions of the 3d arrays are the same. That is, i and j have the same range
for all of them.
Scott
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Re: [R] Using apply with more than one matrix
Hi,
Sorry, a typo in the previous function:
-
if (condition1[i] && !is.na(indx3)) {
arr[x1][indx3] + m2[i] ## should be mat2[i]
} else NA
---
Also, you can try:
library(plyr)
evaluateNew <- function(arr, mat1, mat2){
if (!all(dim(mat1) == dim(mat2))) {
stop("both matrices should have equal dimensions")
}
indx1 <- unlist(alply(arr, c(1,2), function(x) {x[!is.na(x)][1]}))
condition1 <- !is.na(mat1) & mat1 > mat2
ifelse(condition1, indx1+mat2, NA)
}
evaluateNew(a1, m1, m2)
# [,1] [,2]
#[1,] NA 1.66
#[2,] 3.14 NA
A.K.
On Thursday, May 1, 2014 5:49 AM, arun wrote:
Hi,
You may try:
evaluate <- function(arr, mat1, mat2) {
if (!all(dim(mat1) == dim(mat2))) {
stop("both matrices should have equal dimensions")
}
indx1 <- as.matrix(do.call(expand.grid, lapply(dim(arr), sequence)))
indx2 <- paste0(indx1[, 1], indx1[, 2])
condition1 <- !is.na(mat1) & mat1 > mat2
ans <- sapply(seq_along(unique(indx2)), function(i) {
x1 <- indx1[indx2 %in% unique(indx2)[i], ]
indx3 <- which(!is.na(arr[x1]))[1]
if (condition1[i] && !is.na(indx3)) {
arr[x1][indx3] + m2[i]
} else NA
})
dim(ans) <- dim(mat1)
ans
}
evaluate(a1,m1,m2)
# [,1] [,2]
#[1,] NA 1.66
#[2,] 3.14 NA
A.K.
On Thursday, May 1, 2014 2:53 AM, "Waichler, Scott R"
wrote:
> I would ask you to look at this loop-free approach and ask if this is not
> equally valid?
>
> ans <- matrix(NA, ncol=2, nrow=2)
> ind.not.na <- which(!is.na(a1))
> ans[] <- condition1*a1[,,ind.not.na[1]]+ m2 # two matrices of equal
> dimensions, one logical.
> ans
> [,1] [,2]
> [1,] NA 1.66
> [2,] 2.74 NA
Thanks, I am learning something. I didn't know you could multiply a logical
object by a numerical one. But notice the answer is not the same as mine,
because I am doing an operation on the vector of values a1[i,j,] first.
I tried a modification on sapply below, but it doesn't work because I haven't
referenced the 3d array a1 properly. So I guess I must try to get a 2d result
from a1 first, then use that in matrix arithmetic.
Sapply or mapply may work, I haven't used these much and will try to learn
better how to use them. Your use of sapply looks good; but I'm trying to
understand if and how I can bring in the operation on a1. This doesn't work:
evaluate <- function(idx) {
ind.not.na <- which(!is.na(a1[idx,])) ])) # doesn't work; improper indexing
for a1
if(length(ind.not.na) > 0) {
return(condition1*(a1[idx,ind.not.na[1]] + m2[idx])) # doesn't work;
improper indexing for a1
}
}
vec <- sapply(seq(length(m2)), evaluate)
Scott Waichler
> -Original Message-----
> From: David Winsemius [mailto:dwinsem...@comcast.net]
> Sent: Wednesday, April 30, 2014 8:46 PM
> To: Waichler, Scott R
> Cc: Bert Gunter; r-help@r-project.org
> Subject: Re: [R] Using apply with more than one matrix
>
>
> On Apr 30, 2014, at 6:03 PM, Waichler, Scott R wrote:
>
> > Here is a working example with no random parts. Thanks for your
> patience and if I'm still off the mark with my presentation I'll drop the
> matter.
> >
> > v <- c(NA, 1.5, NA, NA,
> > NA, 1.1, 0.5, NA,
> > NA, 1.3, 0.4, 0.9)
> > a1 <- array(v, dim=c(2,2,3))
> > m1 <- matrix(c(NA, 1.5, 2.1, NA), ncol=2, byrow=T)
> > m2 <- matrix(c(1.56, 1.64, 1.16, 2.92), ncol=2)
> > condition1 <- !is.na(m1)& m1 > m2
> >
> > ans <- matrix(NA, ncol=2, nrow=2) # initialize for(i in 1:2) { for(j
> > in 1:2) {
> > ind.not.na <- which(!is.na(a1[i,j,]))
> > if(condition1[i,j] && length(ind.not.na) > 0) ans[i,j] <-
> > a1[i,j,ind.not.na[1]] + m2[i,j] } } ans
> > [,1] [,2]
> > [1,] NA 1.66
> > [2,] 3.14 NA
>
> I would ask you to look at this loop-free approach and ask if this is not
> equally valid?
>
> ans <- matrix(NA, ncol=2, nrow=2)
> ind.not.na <- which(!is.na(a1))
> ans[] <- condition1*a1[,,ind.not.na[1]]+ m2 # two matrices of equal
> dimensions, one logical.
> ans
> [,1] [,2]
> [1,] NA 1.66
> [2,] 2.74 NA
> >
> > Let me try asking again in words. If I have multiple matrices or slices
> of 3d arrays that are the same dimension, is there a way to pass them all
> to apply, and have apply take care of looping through i,j?
>
> I don't think `apply` is the correct function for this. Either `mapply` or
> basic matrix operation seem more likely to deliver correct results:
>
>
> > I understand that apply has just one input object x. I want
Re: [R] Using apply with more than one matrix
Hi,
You may try:
evaluate <- function(arr, mat1, mat2) {
if (!all(dim(mat1) == dim(mat2))) {
stop("both matrices should have equal dimensions")
}
indx1 <- as.matrix(do.call(expand.grid, lapply(dim(arr), sequence)))
indx2 <- paste0(indx1[, 1], indx1[, 2])
condition1 <- !is.na(mat1) & mat1 > mat2
ans <- sapply(seq_along(unique(indx2)), function(i) {
x1 <- indx1[indx2 %in% unique(indx2)[i], ]
indx3 <- which(!is.na(arr[x1]))[1]
if (condition1[i] && !is.na(indx3)) {
arr[x1][indx3] + m2[i]
} else NA
})
dim(ans) <- dim(mat1)
ans
}
evaluate(a1,m1,m2)
# [,1] [,2]
#[1,] NA 1.66
#[2,] 3.14 NA
A.K.
On Thursday, May 1, 2014 2:53 AM, "Waichler, Scott R"
wrote:
> I would ask you to look at this loop-free approach and ask if this is not
> equally valid?
>
> ans <- matrix(NA, ncol=2, nrow=2)
> ind.not.na <- which(!is.na(a1))
> ans[] <- condition1*a1[,,ind.not.na[1]]+ m2 # two matrices of equal
> dimensions, one logical.
> ans
> [,1] [,2]
> [1,] NA 1.66
> [2,] 2.74 NA
Thanks, I am learning something. I didn't know you could multiply a logical
object by a numerical one. But notice the answer is not the same as mine,
because I am doing an operation on the vector of values a1[i,j,] first.
I tried a modification on sapply below, but it doesn't work because I haven't
referenced the 3d array a1 properly. So I guess I must try to get a 2d result
from a1 first, then use that in matrix arithmetic.
Sapply or mapply may work, I haven't used these much and will try to learn
better how to use them. Your use of sapply looks good; but I'm trying to
understand if and how I can bring in the operation on a1. This doesn't work:
evaluate <- function(idx) {
ind.not.na <- which(!is.na(a1[idx,])) ])) # doesn't work; improper indexing
for a1
if(length(ind.not.na) > 0) {
return(condition1*(a1[idx,ind.not.na[1]] + m2[idx])) # doesn't work;
improper indexing for a1
}
}
vec <- sapply(seq(length(m2)), evaluate)
Scott Waichler
> -Original Message-
> From: David Winsemius [mailto:dwinsem...@comcast.net]
> Sent: Wednesday, April 30, 2014 8:46 PM
> To: Waichler, Scott R
> Cc: Bert Gunter; r-help@r-project.org
> Subject: Re: [R] Using apply with more than one matrix
>
>
> On Apr 30, 2014, at 6:03 PM, Waichler, Scott R wrote:
>
> > Here is a working example with no random parts. Thanks for your
> patience and if I'm still off the mark with my presentation I'll drop the
> matter.
> >
> > v <- c(NA, 1.5, NA, NA,
> > NA, 1.1, 0.5, NA,
> > NA, 1.3, 0.4, 0.9)
> > a1 <- array(v, dim=c(2,2,3))
> > m1 <- matrix(c(NA, 1.5, 2.1, NA), ncol=2, byrow=T)
> > m2 <- matrix(c(1.56, 1.64, 1.16, 2.92), ncol=2)
> > condition1 <- !is.na(m1)& m1 > m2
> >
> > ans <- matrix(NA, ncol=2, nrow=2) # initialize for(i in 1:2) { for(j
> > in 1:2) {
> > ind.not.na <- which(!is.na(a1[i,j,]))
> > if(condition1[i,j] && length(ind.not.na) > 0) ans[i,j] <-
> > a1[i,j,ind.not.na[1]] + m2[i,j] } } ans
> > [,1] [,2]
> > [1,] NA 1.66
> > [2,] 3.14 NA
>
> I would ask you to look at this loop-free approach and ask if this is not
> equally valid?
>
> ans <- matrix(NA, ncol=2, nrow=2)
> ind.not.na <- which(!is.na(a1))
> ans[] <- condition1*a1[,,ind.not.na[1]]+ m2 # two matrices of equal
> dimensions, one logical.
> ans
> [,1] [,2]
> [1,] NA 1.66
> [2,] 2.74 NA
> >
> > Let me try asking again in words. If I have multiple matrices or slices
> of 3d arrays that are the same dimension, is there a way to pass them all
> to apply, and have apply take care of looping through i,j?
>
> I don't think `apply` is the correct function for this. Either `mapply` or
> basic matrix operation seem more likely to deliver correct results:
>
>
> > I understand that apply has just one input object x. I want to work on
> more than one array object at once using a custom function that has this
> characteristic: in order to compute the answer at i,j I need a result
> from higher order array at the same i,j.
>
> If you want to iterate over matrix indices you can either use the vector
> version e.g. m2[3] or the matrix version, m2[2,1[.
>
> vec <- sapply(seq(length(m2) , function(idx) m2[idx]*condition1[idx] )
>
>
>
> > This is what I tried to demonstrate in my example above.
> >
> > Thanks,
> > Scott
>
> David Winsemius
> Alameda, CA, USA
__
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] Using apply with more than one matrix
On Apr 30, 2014, at 9:32 PM, Waichler, Scott R wrote:
>> I would ask you to look at this loop-free approach and ask if this is not
>> equally valid?
>>
>> ans <- matrix(NA, ncol=2, nrow=2)
>> ind.not.na <- which(!is.na(a1))
>> ans[] <- condition1*a1[,,ind.not.na[1]]+ m2 # two matrices of equal
>> dimensions, one logical.
>> ans
>> [,1] [,2]
>> [1,] NA 1.66
>> [2,] 2.74 NA
>
> Thanks, I am learning something. I didn't know you could multiply a logical
> object by a numerical one. But notice the answer is not the same as mine,
> because I am doing an operation on the vector of values a1[i,j,] first.
>
> I tried a modification on sapply below, but it doesn't work because I
> haven't referenced the 3d array a1 properly. So I guess I must try to get a
> 2d result from a1 first, then use that in matrix arithmetic.
>
> Sapply or mapply may work, I haven't used these much and will try to learn
> better how to use them. Your use of sapply looks good; but I'm trying to
> understand if and how I can bring in the operation on a1. This doesn't work:
>
> evaluate <- function(idx) {
> ind.not.na <- which(!is.na(a1[idx,])) ])) # doesn't work; improper indexing
> for a1
> if(length(ind.not.na) > 0) {
>return(condition1*(a1[idx,ind.not.na[1]] + m2[idx])) # doesn't work;
> improper indexing for a1
> }
> }
> vec <- sapply(seq(length(m2)), evaluate)
Are we to assume that the length of `which(!is.na(a1[idx,])) ]))` is guaranteed
to equal the length of the two other matrices? If not then what sort of
relationships should be assumed?
--
David.
>
> Scott Waichler
>
>> -----Original Message-
>> From: David Winsemius [mailto:dwinsem...@comcast.net]
>> Sent: Wednesday, April 30, 2014 8:46 PM
>> To: Waichler, Scott R
>> Cc: Bert Gunter; r-help@r-project.org
>> Subject: Re: [R] Using apply with more than one matrix
>>
>>
>> On Apr 30, 2014, at 6:03 PM, Waichler, Scott R wrote:
>>
>>> Here is a working example with no random parts. Thanks for your
>> patience and if I'm still off the mark with my presentation I'll drop the
>> matter.
>>>
>>> v <- c(NA, 1.5, NA, NA,
>>> NA, 1.1, 0.5, NA,
>>> NA, 1.3, 0.4, 0.9)
>>> a1 <- array(v, dim=c(2,2,3))
>>> m1 <- matrix(c(NA, 1.5, 2.1, NA), ncol=2, byrow=T)
>>> m2 <- matrix(c(1.56, 1.64, 1.16, 2.92), ncol=2)
>>> condition1 <- !is.na(m1)& m1 > m2
>>>
>>> ans <- matrix(NA, ncol=2, nrow=2) # initialize for(i in 1:2) { for(j
>>> in 1:2) {
>>> ind.not.na <- which(!is.na(a1[i,j,]))
>>> if(condition1[i,j] && length(ind.not.na) > 0) ans[i,j] <-
>>> a1[i,j,ind.not.na[1]] + m2[i,j] } } ans
>>>[,1] [,2]
>>> [1,] NA 1.66
>>> [2,] 3.14 NA
>>
>> I would ask you to look at this loop-free approach and ask if this is not
>> equally valid?
>>
>> ans <- matrix(NA, ncol=2, nrow=2)
>> ind.not.na <- which(!is.na(a1))
>> ans[] <- condition1*a1[,,ind.not.na[1]]+ m2 # two matrices of equal
>> dimensions, one logical.
>> ans
>> [,1] [,2]
>> [1,] NA 1.66
>> [2,] 2.74 NA
>>>
>>> Let me try asking again in words. If I have multiple matrices or slices
>> of 3d arrays that are the same dimension, is there a way to pass them all
>> to apply, and have apply take care of looping through i,j?
>>
>> I don't think `apply` is the correct function for this. Either `mapply` or
>> basic matrix operation seem more likely to deliver correct results:
>>
>>
>>> I understand that apply has just one input object x. I want to work on
>> more than one array object at once using a custom function that has this
>> characteristic: in order to compute the answer at i,j I need a result
>> from higher order array at the same i,j.
>>
>> If you want to iterate over matrix indices you can either use the vector
>> version e.g. m2[3] or the matrix version, m2[2,1[.
>>
>> vec <- sapply(seq(length(m2) , function(idx) m2[idx]*condition1[idx] )
>>
>>
>>
>>> This is what I tried to demonstrate in my example above.
>>>
>>> Thanks,
>>> Scott
>>
>> David Winsemius
>> Alameda, CA, USA
>
David Winsemius
Alameda, CA, USA
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Using apply with more than one matrix
> I would ask you to look at this loop-free approach and ask if this is not
> equally valid?
>
> ans <- matrix(NA, ncol=2, nrow=2)
> ind.not.na <- which(!is.na(a1))
> ans[] <- condition1*a1[,,ind.not.na[1]]+ m2 # two matrices of equal
> dimensions, one logical.
> ans
> [,1] [,2]
> [1,] NA 1.66
> [2,] 2.74 NA
Thanks, I am learning something. I didn't know you could multiply a logical
object by a numerical one. But notice the answer is not the same as mine,
because I am doing an operation on the vector of values a1[i,j,] first.
I tried a modification on sapply below, but it doesn't work because I haven't
referenced the 3d array a1 properly. So I guess I must try to get a 2d result
from a1 first, then use that in matrix arithmetic.
Sapply or mapply may work, I haven't used these much and will try to learn
better how to use them. Your use of sapply looks good; but I'm trying to
understand if and how I can bring in the operation on a1. This doesn't work:
evaluate <- function(idx) {
ind.not.na <- which(!is.na(a1[idx,])) ])) # doesn't work; improper indexing
for a1
if(length(ind.not.na) > 0) {
return(condition1*(a1[idx,ind.not.na[1]] + m2[idx])) # doesn't work;
improper indexing for a1
}
}
vec <- sapply(seq(length(m2)), evaluate)
Scott Waichler
> -Original Message-
> From: David Winsemius [mailto:dwinsem...@comcast.net]
> Sent: Wednesday, April 30, 2014 8:46 PM
> To: Waichler, Scott R
> Cc: Bert Gunter; r-help@r-project.org
> Subject: Re: [R] Using apply with more than one matrix
>
>
> On Apr 30, 2014, at 6:03 PM, Waichler, Scott R wrote:
>
> > Here is a working example with no random parts. Thanks for your
> patience and if I'm still off the mark with my presentation I'll drop the
> matter.
> >
> > v <- c(NA, 1.5, NA, NA,
> > NA, 1.1, 0.5, NA,
> > NA, 1.3, 0.4, 0.9)
> > a1 <- array(v, dim=c(2,2,3))
> > m1 <- matrix(c(NA, 1.5, 2.1, NA), ncol=2, byrow=T)
> > m2 <- matrix(c(1.56, 1.64, 1.16, 2.92), ncol=2)
> > condition1 <- !is.na(m1)& m1 > m2
> >
> > ans <- matrix(NA, ncol=2, nrow=2) # initialize for(i in 1:2) { for(j
> > in 1:2) {
> >ind.not.na <- which(!is.na(a1[i,j,]))
> >if(condition1[i,j] && length(ind.not.na) > 0) ans[i,j] <-
> > a1[i,j,ind.not.na[1]] + m2[i,j] } } ans
> > [,1] [,2]
> > [1,] NA 1.66
> > [2,] 3.14 NA
>
> I would ask you to look at this loop-free approach and ask if this is not
> equally valid?
>
> ans <- matrix(NA, ncol=2, nrow=2)
> ind.not.na <- which(!is.na(a1))
> ans[] <- condition1*a1[,,ind.not.na[1]]+ m2 # two matrices of equal
> dimensions, one logical.
> ans
> [,1] [,2]
> [1,] NA 1.66
> [2,] 2.74 NA
> >
> > Let me try asking again in words. If I have multiple matrices or slices
> of 3d arrays that are the same dimension, is there a way to pass them all
> to apply, and have apply take care of looping through i,j?
>
> I don't think `apply` is the correct function for this. Either `mapply` or
> basic matrix operation seem more likely to deliver correct results:
>
>
> > I understand that apply has just one input object x. I want to work on
> more than one array object at once using a custom function that has this
> characteristic: in order to compute the answer at i,j I need a result
> from higher order array at the same i,j.
>
> If you want to iterate over matrix indices you can either use the vector
> version e.g. m2[3] or the matrix version, m2[2,1[.
>
> vec <- sapply(seq(length(m2) , function(idx) m2[idx]*condition1[idx] )
>
>
>
> > This is what I tried to demonstrate in my example above.
> >
> > Thanks,
> > Scott
>
> David Winsemius
> Alameda, CA, USA
__
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] Using apply with more than one matrix
On Apr 30, 2014, at 6:03 PM, Waichler, Scott R wrote:
> Here is a working example with no random parts. Thanks for your patience and
> if I'm still off the mark with my presentation I'll drop the matter.
>
> v <- c(NA, 1.5, NA, NA,
> NA, 1.1, 0.5, NA,
> NA, 1.3, 0.4, 0.9)
> a1 <- array(v, dim=c(2,2,3))
> m1 <- matrix(c(NA, 1.5, 2.1, NA), ncol=2, byrow=T)
> m2 <- matrix(c(1.56, 1.64, 1.16, 2.92), ncol=2)
> condition1 <- !is.na(m1)& m1 > m2
>
> ans <- matrix(NA, ncol=2, nrow=2) # initialize
> for(i in 1:2) {
> for(j in 1:2) {
>ind.not.na <- which(!is.na(a1[i,j,]))
>if(condition1[i,j] && length(ind.not.na) > 0) ans[i,j] <-
> a1[i,j,ind.not.na[1]] + m2[i,j]
> }
> }
> ans
> [,1] [,2]
> [1,] NA 1.66
> [2,] 3.14 NA
I would ask you to look at this loop-free approach and ask if this is not
equally valid?
ans <- matrix(NA, ncol=2, nrow=2)
ind.not.na <- which(!is.na(a1))
ans[] <- condition1*a1[,,ind.not.na[1]]+ m2 # two matrices of equal
dimensions, one logical.
ans
[,1] [,2]
[1,] NA 1.66
[2,] 2.74 NA
>
> Let me try asking again in words. If I have multiple matrices or slices of
> 3d arrays that are the same dimension, is there a way to pass them all to
> apply, and have apply take care of looping through i,j?
I don't think `apply` is the correct function for this. Either `mapply` or
basic matrix operation seem more likely to deliver correct results:
> I understand that apply has just one input object x. I want to work on more
> than one array object at once using a custom function that has this
> characteristic: in order to compute the answer at i,j I need a result from
> higher order array at the same i,j.
If you want to iterate over matrix indices you can either use the vector
version e.g. m2[3] or the matrix version, m2[2,1[.
vec <- sapply(seq(length(m2) , function(idx) m2[idx]*condition1[idx] )
> This is what I tried to demonstrate in my example above.
>
> Thanks,
> Scott
David Winsemius
Alameda, CA, USA
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using apply with more than one matrix
Try
?mapply
Description:
'mapply' is a multivariate version of 'sapply'. 'mapply' applies
'FUN' to the first elements of each ... argument, the second
elements, the third elements, and so on. Arguments are recycled
if necessary.
Rich
On Wed, Apr 30, 2014 at 9:03 PM, Waichler, Scott R
wrote:
> Here is a working example with no random parts. Thanks for your patience and
> if I'm still off the mark with my presentation I'll drop the matter.
>
> v <- c(NA, 1.5, NA, NA,
>NA, 1.1, 0.5, NA,
>NA, 1.3, 0.4, 0.9)
> a1 <- array(v, dim=c(2,2,3))
> m1 <- matrix(c(NA, 1.5, 2.1, NA), ncol=2, byrow=T)
> m2 <- matrix(c(1.56, 1.64, 1.16, 2.92), ncol=2)
> condition1 <- !is.na(m1)& m1 > m2
>
> ans <- matrix(NA, ncol=2, nrow=2) # initialize
> for(i in 1:2) {
> for(j in 1:2) {
> ind.not.na <- which(!is.na(a1[i,j,]))
> if(condition1[i,j] && length(ind.not.na) > 0) ans[i,j] <-
> a1[i,j,ind.not.na[1]] + m2[i,j]
> }
> }
> ans
> [,1] [,2]
> [1,] NA 1.66
> [2,] 3.14 NA
>
> Let me try asking again in words. If I have multiple matrices or slices of
> 3d arrays that are the same dimension, is there a way to pass them all to
> apply, and have apply take care of looping through i,j? I understand that
> apply has just one input object x. I want to work on more than one array
> object at once using a custom function that has this characteristic: in
> order to compute the answer at i,j I need a result from higher order array at
> the same i,j. This is what I tried to demonstrate in my example above.
>
> Thanks,
> Scott
>
> __
> 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] Using apply with more than one matrix
Here is a working example with no random parts. Thanks for your patience and
if I'm still off the mark with my presentation I'll drop the matter.
v <- c(NA, 1.5, NA, NA,
NA, 1.1, 0.5, NA,
NA, 1.3, 0.4, 0.9)
a1 <- array(v, dim=c(2,2,3))
m1 <- matrix(c(NA, 1.5, 2.1, NA), ncol=2, byrow=T)
m2 <- matrix(c(1.56, 1.64, 1.16, 2.92), ncol=2)
condition1 <- !is.na(m1)& m1 > m2
ans <- matrix(NA, ncol=2, nrow=2) # initialize
for(i in 1:2) {
for(j in 1:2) {
ind.not.na <- which(!is.na(a1[i,j,]))
if(condition1[i,j] && length(ind.not.na) > 0) ans[i,j] <-
a1[i,j,ind.not.na[1]] + m2[i,j]
}
}
ans
[,1] [,2]
[1,] NA 1.66
[2,] 3.14 NA
Let me try asking again in words. If I have multiple matrices or slices of 3d
arrays that are the same dimension, is there a way to pass them all to apply,
and have apply take care of looping through i,j? I understand that apply has
just one input object x. I want to work on more than one array object at once
using a custom function that has this characteristic: in order to compute the
answer at i,j I need a result from higher order array at the same i,j. This is
what I tried to demonstrate in my example above.
Thanks,
Scott
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using apply with more than one matrix
On Apr 30, 2014, at 1:21 PM, Waichler, Scott R wrote:
> Ok, here is a toy example. I want to use a custom function that depends on
> more than one matrix/array as input, and I can't figure out how to do that
> with apply.
>
> v <- c(NA, 1.5, NA, NA,
> NA, 1.1, 0.5, NA,
> NA, 1.3, 0.4, 0.9)
> a1 <- array(v, dim=c(2,2,3))
> m1 <- matrix(c(NA, 1.5, 2.1, NA), ncol=2, byrow=T)
> m2 <- matrix(runif(n=4, min=1, max=3), ncol=2)
> condition1 <- ifelse(!is.na(m1) & m1 > m2, T, F)
Just use:
condition1 <- !is.na(m1) & m1 > m2
For the next set of operation you should heed Jim Holtman's tagline:
"What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it."
>
> ans <- matrix(NA, ncol=2, nrow=2) # initialize
> for(i in 1:2) {
> for(j in 1:2) {
>ind.not.na <- which(!is.na(a1[i,j,]))
>if(condition1[i,j] && length(ind.not.na) > 0) {
> ans[i,j] <- a1[i,j,ind.not.na[1]] + m2[i,j]
>}
> }
> }
At least in the realizatiuon I used all of the entries for conditon1 were
FALSE, so a better test might be to specify set.seed() to give a reproducible
example.
--
David.
>
> Scott Waichler
>
>
>
>> -Original Message-
>> From: Bert Gunter [mailto:gunter.ber...@gene.com]
>> Sent: Wednesday, April 30, 2014 12:18 PM
>> To: Waichler, Scott R
>> Cc: r-help@r-project.org
>> Subject: Re: [R] Using apply with more than one matrix
>>
>> Scott:
>>
>> Your problem specification is rather vague: What do you mean by "use"?
>>
>> In general, matrices are merely vectors with a dim attribute, so if you
>> can do what you want with them as vectors, then that will work for them as
>> matrices. For example:
>>
>>> m1<- matrix(1:6, nr=3)
>>> m2 <- matrix(11:16, nr=2)
>>> m2*m2
>> [,1] [,2] [,3]
>> [1,] 121 169 225
>> [2,] 144 196 256
>>
>> If this does not meet your needs, you will have to follow the posting
>> guide and provide both code and a minimal reproducible example.
>>
>> Cheers,
>> Bert
>>
>>
>> Bert Gunter
>> Genentech Nonclinical Biostatistics
>> (650) 467-7374
>>
>> "Data is not information. Information is not knowledge. And knowledge is
>> certainly not wisdom."
>> H. Gilbert Welch
>>
>>
>>
>>
>> On Wed, Apr 30, 2014 at 10:54 AM, Waichler, Scott R
>> wrote:
>>> Hi,
>>>
>>> I want to apply a custom function to all the elements of one matrix.
>> The function uses the value in the same i,j in a second matrix. How can I
>> use apply() or similar function to avoid nested loops in i and j?
>>>
>>> Thanks,
>>> Scott Waichler
>>> Pacific Northwest National Laboratory
>>> Richland, WA, USA
>>>
>>> __
>>> 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.
David Winsemius
Alameda, CA, USA
__
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] Using apply with more than one matrix
Ok, here is a toy example. I want to use a custom function that depends on
more than one matrix/array as input, and I can't figure out how to do that with
apply.
v <- c(NA, 1.5, NA, NA,
NA, 1.1, 0.5, NA,
NA, 1.3, 0.4, 0.9)
a1 <- array(v, dim=c(2,2,3))
m1 <- matrix(c(NA, 1.5, 2.1, NA), ncol=2, byrow=T)
m2 <- matrix(runif(n=4, min=1, max=3), ncol=2)
condition1 <- ifelse(!is.na(m1) & m1 > m2, T, F)
ans <- matrix(NA, ncol=2, nrow=2) # initialize
for(i in 1:2) {
for(j in 1:2) {
ind.not.na <- which(!is.na(a1[i,j,]))
if(condition1[i,j] && length(ind.not.na) > 0) {
ans[i,j] <- a1[i,j,ind.not.na[1]] + m2[i,j]
}
}
}
Scott Waichler
> -Original Message-
> From: Bert Gunter [mailto:gunter.ber...@gene.com]
> Sent: Wednesday, April 30, 2014 12:18 PM
> To: Waichler, Scott R
> Cc: r-help@r-project.org
> Subject: Re: [R] Using apply with more than one matrix
>
> Scott:
>
> Your problem specification is rather vague: What do you mean by "use"?
>
> In general, matrices are merely vectors with a dim attribute, so if you
> can do what you want with them as vectors, then that will work for them as
> matrices. For example:
>
> > m1<- matrix(1:6, nr=3)
> > m2 <- matrix(11:16, nr=2)
> > m2*m2
> [,1] [,2] [,3]
> [1,] 121 169 225
> [2,] 144 196 256
>
> If this does not meet your needs, you will have to follow the posting
> guide and provide both code and a minimal reproducible example.
>
> Cheers,
> Bert
>
>
> Bert Gunter
> Genentech Nonclinical Biostatistics
> (650) 467-7374
>
> "Data is not information. Information is not knowledge. And knowledge is
> certainly not wisdom."
> H. Gilbert Welch
>
>
>
>
> On Wed, Apr 30, 2014 at 10:54 AM, Waichler, Scott R
> wrote:
> > Hi,
> >
> > I want to apply a custom function to all the elements of one matrix.
> The function uses the value in the same i,j in a second matrix. How can I
> use apply() or similar function to avoid nested loops in i and j?
> >
> > Thanks,
> > Scott Waichler
> > Pacific Northwest National Laboratory
> > Richland, WA, USA
> >
> > __
> > 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] Using apply with more than one matrix
Scott: Your problem specification is rather vague: What do you mean by "use"? In general, matrices are merely vectors with a dim attribute, so if you can do what you want with them as vectors, then that will work for them as matrices. For example: > m1<- matrix(1:6, nr=3) > m2 <- matrix(11:16, nr=2) > m2*m2 [,1] [,2] [,3] [1,] 121 169 225 [2,] 144 196 256 If this does not meet your needs, you will have to follow the posting guide and provide both code and a minimal reproducible example. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 "Data is not information. Information is not knowledge. And knowledge is certainly not wisdom." H. Gilbert Welch On Wed, Apr 30, 2014 at 10:54 AM, Waichler, Scott R wrote: > Hi, > > I want to apply a custom function to all the elements of one matrix. The > function uses the value in the same i,j in a second matrix. How can I use > apply() or similar function to avoid nested loops in i and j? > > Thanks, > Scott Waichler > Pacific Northwest National Laboratory > Richland, WA, USA > > __ > 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] Using apply with more than one matrix
Hi Scott, You could set up a three-dimensional array and then use apply on the desired dimension (MARGIN in apply language). HTH! Many thanks, Ranjan On Wed, 30 Apr 2014 17:54:48 + "Waichler, Scott R" wrote: > Hi, > > I want to apply a custom function to all the elements of one matrix. The > function uses the value in the same i,j in a second matrix. How can I use > apply() or similar function to avoid nested loops in i and j? > > Thanks, > Scott Waichler > Pacific Northwest National Laboratory > Richland, WA, USA > > __ > 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. > -- Important Notice: This mailbox is ignored: e-mails are set to be deleted on receipt. Please respond to the mailing list if appropriate. For those needing to send personal or professional e-mail, please use appropriate addresses. FREE 3D EARTH SCREENSAVER - Watch the Earth right on your desktop! __ 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] Using apply with more than one matrix
Hi, I want to apply a custom function to all the elements of one matrix. The function uses the value in the same i,j in a second matrix. How can I use apply() or similar function to avoid nested loops in i and j? Thanks, Scott Waichler Pacific Northwest National Laboratory Richland, WA, USA __ 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.
