[R] Using apply with more than one matrix

[David Winsemius]

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:dwinsemius at comcast.net]
>> Sent: Wednesday, April 30, 2014 8:46 PM
>> To: Waichler, Scott R
>> Cc: Bert Gunter; r-help at 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