How to find statistical mode in Fortran - fortran

Question

I'm trying to write a program to find the mean, median, mode of an integer array but am having some complications in finding the mode. The following is the code that I've written so far.
First, the program will prompt user to enter a value for the number of integers that will be entered followed by request to enter that number of integers. The integers are then sorted in ascending order and the mean and median are found.
The problem I am having is when I try to get the mode. I am able to count the number of occurrence of a repetitive value. By finding the value with highest occurrence, we'll be able to find Mode. But I am unsure how to do this. Is there any intrinsic function in Fortran to calculate number of occurrence of input values and the value with highest occurrence?
PROGRAM STATISTICS
!Created by : Rethnaraj Rambabu
IMPLICIT NONE
REAL, DIMENSION(:), ALLOCATABLE:: VAL
REAL TEMP, MEDIAN
REAL EVEN, MEAN, SUM, FMODE
INTEGER N, I,J
WRITE(*,*)' WHAT IS THE VALUE FOR N? '
READ(*,*) N
ALLOCATE(VAL(N))
WRITE(*,*) 'ENTER THE NUMBERS'
OPEN(1,FILE='FILE.TXT')
READ(1,*)(VAL(I),I=1,N)
CLOSE(1)
WRITE(*,*) VAL
!/---FOR SORTING----/!
DO I=1,N-1
DO J=1,N-1
IF(VAL(J) > VAL(J+1)) THEN
TEMP=VAL(J)
VAL(J)=VAL(J+1)
VAL(J+1)=TEMP
END IF
END DO
END DO
WRITE(*,*) VAL
!/-----MEDIAN----/!
IF ((N/2*2) /= N) THEN
MEDIAN=VAL((N+1)/2)
ELSE IF ((N/2*2) == N) THEN
EVEN= (VAL(N/2)+VAL((N+2)/2))
MEDIAN=EVEN/2
END IF
WRITE(*,*)'MEDIAN=', MEDIAN
!/----MEAN----/
SUM=0
DO I=1,N
SUM=SUM+VAL(I)
END DO
MEAN=SUM/N
WRITE(*,*)'MEAN=', MEAN
!/------MODE----/
FMODE=1
DO I=1,N-1
IF (VAL(I) == VAL(I+1)) THEN
FMODE=FMODE+1
END IF
END DO
WRITE(*,*)FMODE
END PROGRAM
The FILE.TXT contains
10 8 1 9 8 9 9 7 5 9 3 5 6

Answer 1

But, how to do that? Or is there any intrinsic function in Fortran to calculate number of occurrence of input values and the value with highest occurrence.
No, there is not. You'll have to calculate the mode by hand.
The following code should work (on a sorted array):
FMODE = VAL(1)
COUNT = 1
CURRENTCOUNT = 1
DO I = 2, N
! We are going through the loop looking for values == VAL(I-1)...
IF (VAL(I) == VAL(I-1)) THEN
! We spotted another VAL(I-1), so increment the count.
CURRENTCOUNT = CURRENTCOUNT + 1
ELSE
! There are no more VAL(I-1)
IF (CURRENTCOUNT > COUNT) THEN
! There were more elements of value VAL(I-1) than of value FMODE
COUNT = CURRENTCOUNT
FMODE = VAL(I-1)
END IF
! Next we are looking for values == VAL(I), so far we have spotted one...
CURRENTCOUNT = 1
END
END DO
IF (CURRENTCOUNT > COUNT) THEN
! This means there are more elements of value VAL(N) than of value FMODE.
FMODE = VAL(N)
END IF
Explanation:
We keep the best-so-far mode in the FMODE variable, and the count of the FMODE in the COUNT variable. As we step through the array we count the number of hits that are equal to what we are looking at now, in the CURRENTCOUNT variable.
If the next item we look at is equal to the previous, we simply increment the CURRENTCOUNT. If it's different, then we need to reset the CURRENTCOUNT, because we will now count the number of duplications of the next element.
Before we reset the CURRENTCOUNT we check if it's bigger than the previous best result, and if it is, we overwrite the previous best result (the FMODE and COUNT variables) with the new best results (whatever is at VAL(I) and CURRENTCOUNT), before we continue.
This reset doesn't happen at the end of the loop, so I inserted another check at the end in case the most frequent element happens to be the final element of the loop. In that case we overwrite FMODE, like we would have done in the loop.

Answer 2

It is a bit lengthy, you could probably get rid of the optional argument, but there is an example provided here. They use the quick sort algorithm as implemented here.
Alternatively, you could use
integer function mode(arr) result(m)
implicit none
integer, dimension(:), intent(in) :: arr
! Local variables
integer, dimension(:), allocatable :: counts
integer :: i, astat
character(len=128) :: error_str
! Initialise array to count occurrences of each value.
allocate(counts(minval(arr):maxval(arr)), stat=astat, errmsg=error_str)
if (astat/=0) then
print'("Allocation of counts array failed.")'
print*, error_str
end if
counts = 0
! Loop over inputted array, counting occurrence of each value.
do i=1,size(arr)
counts(arr(i)) = counts(arr(i)) + 1
end do
! Finally, find the mode
m = minloc(abs(counts - maxval(counts)),1)
end function mode
This doesn't require any sorting.