Is there a way in Fortran to access many elements of an array without using a loop?
For example given array of 100 elements
real(100) :: a
can I do something like this to access elements 1,4,7,54,81 that do not follow a regular step?
a(1,4,7,54,81)= 3.21423
you could use a vector subscript: a( (/1,4,7,54,81/) )= 3.21423
As noted before, an array may be used as the indexes of an array. This is a so-called vector subscript.
A([1,4,7,54,81]) = 3.21423
sets the elements given to that value. (This is the same as the earlier answer but using the Fortran 2003+/modern array constructor notation.)
The array can be any rank-1 array, such as a variable or expression:
integer :: idx(5)=[1,4,7,54,81]
A(idx) = 3.21423
A(idx-1+1) = 3.21423
Of course, vector subscripts are of use in other settings, such as referencing:
print *, A(idx)
call sub(A(idx))
A(idx) = A(idx+1) + 5
However, array sections with vector subscripts are subject to various restrictions, such as:
In the third case, if the same index appears more than once in the subscript we can't define it. So
print *, A([1,5,1])
is allowed, but
A([1,5,1]) = 3.
is not.
(/1,5,1/) rather than [1,5,1].RESHAPE and WHERE are worth looking at. If you are determining which elements to 'pull out' then maybe ALLOCATE a new variable and stuff the elements of A into B. Maybe something like this:
REAL, DIMENSION(100) :: A
LOGICAL, DIMENSION(100) :: A_Mask
INTEGER :: SizeB
REAL, DIMENSION(:), ALLOCATABLE :: B
!...
A_Mask = .FALSE.
WHERE(A > 1.0)
A_Mask = .TRUE.
ENDWHERE
SizeB = SUM(A_Mask)
!... then allocate B and fill it.
COUNT true values in a logical array expression. I don't think the code in this answer addresses the question in any useful way.
pack function !