R: Creating a matrix with loops/apply (Original code Fortran) - r

Question

Any help appreciated, I have been struggling with this problem far too long today, and I hope a fresh pair of eyes and set of braincells can help. Suggestions on how to make the code more efficient will also be greatly appreciated.
I am in the process of rewriting a program from Fortran into R. The eventual matrix, once all the data comes in, will be bigger that 1000x1000.
The first element of the code looked like this:
allocate (S(nrecords))
do i=1,nrecords
S(i)=ZZ(i,i)
end do
which in R simply became this: S<-diag(ZZ) **nrecords in the example data = 10
The example dataset I am using consists of a 10x10 matrix ZZ:
167315 136626 138035 150376 137080 136561 139467 137161 151010 140947
136626 171188 139660 138286 138161 138709 139713 138422 138138 140265
138035 139660 170362 138202 138643 138168 140629 139121 137675 139288
150376 138286 138202 167354 138025 138029 140168 137797 144110 139955
137080 138161 138643 138025 168606 144637 140715 138636 142043 141936
136561 138709 138168 138029 144637 167756 140256 138348 140914 152011
139467 139713 140629 140168 140715 140256 172119 141704 140553 140769
137161 138422 139121 137797 138636 138348 141704 169635 137902 138752
151010 138138 137675 144110 142043 140914 140553 137902 169823 142444
140947 140265 139288 139955 141936 152011 140769 138752 142444 173183
so S is a vector containing the diagonal values.
I am stuck in translating this Fortran element though:
allocate(D(nrecords,nrecords))
sumD=0
do i=1,nrecords
do j=1,nrecords
D(i,j)=S(i)+S(j)-2*ZZ(i,j)
sumD=sumD+D(i,j)
end do
end do
deallocate(ZZ)
sumD=sumD/(nrecords*nrecords)
I know that at the end of the day I am supposed to end up with another 10x10 matrix, where D1,1 will equal to 0, and D1,2 will be 65251. But between reading-up on for-loops, apply(), sapply() and tapply() I am rather lost and confused.
This is another element that has already been translated, and I wanted to base the fortran translation on this, but I think I have been staring at it too long, and I strongly suspect that there is a more efficient answer:
n <-6
sumA <- 0
for (i in 1:n) {
for (j in 1:n) {
sumA <- sumA+A[i,j]
}
}
sumA2 <- 0
for (i in 1:n) {
for (j in 1:n) {
sumA2 <- sumA2+A[i,j]^2
}
}
with the corresponding fortran:
sumA2=0.0;sumA=0.0
do i=1,nrecords
do j=1,nrecords
if(A(i,j) > 0.0) then
sumA2=sumA2+(A(i,j)*A(i,j))
sumA=sumA+A(i,j)
end if
end do
end do
sumMMA=0.0;sumZZ=0.0
do i=1,nrecords
do j=1,nrecords
sumMMA=sumMMA+(ZZ(i,j)*A(i,j))
sumZZ=sumZZ+ZZ(i,j) !this will not work using the sum(ZZ) function
end do
end do
Matrix A is simply
1 0 0 0 0 0 0 0 0 0
0 0.75 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0.5 0 0 0 0
0 0 0 0 0 0 0.75 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 1
Thanks in advance!

Answer 1

The purpose of the apply functions is to improve readability. If you don't understand them you don't need to use them. They are more or less wrappers for for loops. In your case, you can almost translate your code verbatim.
R
nrecords <- 10
ZZ <- as.matrix(read.table(header=F, text='
167315 136626 138035 150376 137080 136561 139467 137161 151010 140947
136626 171188 139660 138286 138161 138709 139713 138422 138138 140265
138035 139660 170362 138202 138643 138168 140629 139121 137675 139288
150376 138286 138202 167354 138025 138029 140168 137797 144110 139955
137080 138161 138643 138025 168606 144637 140715 138636 142043 141936
136561 138709 138168 138029 144637 167756 140256 138348 140914 152011
139467 139713 140629 140168 140715 140256 172119 141704 140553 140769
137161 138422 139121 137797 138636 138348 141704 169635 137902 138752
151010 138138 137675 144110 142043 140914 140553 137902 169823 142444
140947 140265 139288 139955 141936 152011 140769 138752 142444 173183
'))
S <- diag(ZZ)
Fortran
allocate(D(nrecords,nrecords))
sumD=0
do i=1,nrecords
do j=1,nrecords
D(i,j)=S(i)+S(j)-2*ZZ(i,j)
sumD=sumD+D(i,j)
end do
end do
deallocate(ZZ)
sumD=sumD/(nrecords*nrecords)
R
D <- matrix(0, nrecords, nrecords)
sumD = 0
for(i in 1:nrecords){
for(j in 1:nrecords){
D[i,j] = S[i] + S[j] - 2*ZZ[i,j]
sumD = sumD + D[i,j]
}
}
sumD = sumD/(nrecords*nrecords)
Fortran
do i=1,nrecords
do j=1,nrecords
if(A(i,j) > 0.0) then
sumA2=sumA2+(A(i,j)*A(i,j))
sumA=sumA+A(i,j)
end if
end do
end do
sumMMA=0.0;sumZZ=0.0
do i=1,nrecords
do j=1,nrecords
sumMMA=sumMMA+(ZZ(i,j)*A(i,j))
sumZZ=sumZZ+ZZ(i,j) !this will not work using the sum(ZZ) function
end do
end do
R
A <- matrix(0, nrecords, nrecords)
diag(A) <- c(1,.75,1,1,1,.5,.75,1,1,1)
sumA2 = 0
sumA = 0
for(i in 1:nrecords){
for(j in 1:nrecords){
if(A[i,j] > 0){
sumA2=sumA2+(A[i,j]*A[i,j])
sumA = sumA+A[i,j]
}
}
}
sumMMA=0
sumZZ=0
for(i in 1:nrecords){
for(j in 1:nrecords){
sumMMa=sumMMA+(ZZ[i,j]*A[i,j])
sumZZ=sumZZ+ZZ[i,j]
}
}