从 Python 的 fortran 获取两个值？

Question

是否可以从 fortran 中获取两个值？ 例如，我想从一个矩阵中得到最高分和这个坐标

# python code 
import numpy as np

matrix = np.array([[1, 10, 3, 4, 9],
                  [2, 1, 0, 9, 13], 
                  [3, 5, 10, 18, 3]])

max_score= 0
column_coord = 0
row_coord = 0

for i in range(len(matrix[:,0])):
    for j in range(len(matrix[0,:])):
        if matrix[i, j] >= max_score:
            # getting max_score, column, row coordinate
            max_score= matrix[i, j]
            column_coord = i
            row_coord = j
print(max_score, column_coord, row_coord)

这段代码可以正常工作，但是如果矩阵变大，会花费很多时间 找到我想要的值。
所以，我决定使用 f2py 来进行更快的计算，这就是 fortran 代码。 cc 是列长，rr 是行长。

subroutine findthemax(cc, rr, matrix, max_score, col_coord, row_coord)
integer, intent(in) :: cc, rr 
integer, intent(in) :: matrix(0:cc, 0:rr)
integer, intent(out) :: max_score, col_coord, row_coord

max_score = 0
col_coord = 0
row_coord = 0

do i = 1, cc
    do j = 1, rr
        if (matrix(i, j).GE.max_score) then
            max_score = matrix(i, j)
            col_coord = i
            row_coord = j
        end if        
    end do
end do
return
end subroutine

我想得到max_score、col_coord、row_coord，所以我导入了findthemax模块
（名为 findthemax.f90）是我用 f2py 改造的。

import numpy as np

matrix = np.array([[1, 10, 3, 4, 9],
                  [2, 1, 0, 9, 13], 
                  [3, 5, 10, 18, 3]])

cc = len(matrix[:,0])
rr = len(matrix[0,:])

max_score, column_coord, row_coord = findthemax.findthemax(cc, rr, matrix)

我不知道为什么这不起作用，那是因为我实际上并不知道
如何使用 fortran 和 f2py 返回两个以上的值。有人可以告诉我如何
从 fortran 中获取多个值？

Answer 1

在 fortran 方面你有

subroutine findthemax(cc, rr, matrix, max_score, col_coord, row_coord)
integer, intent(in) :: cc, rr 
integer, intent(in) :: matrix(0:cc, 0:rr)
integer, intent(out) :: max_score, col_coord, row_coord

max_score = 0
col_coord = 0
row_coord = 0

do i = 1, cc    !Does this need a 0, cc ??
    do j = 1, rr   !Does this need a 0, rr ??
        if (matrix(i, j).GE.max_score) then
            max_score = matrix(i, j)
            col_coord = i
            row_coord = j
        end if        
    end do
end do
return
end subroutine

你可能想要这样的东西：

subroutine findthemax(cc, rr, matrix, max_score, col_coord, row_coord)
integer, intent(in) :: cc, rr 
integer, intent(in) :: matrix(cc+1, rr+1)  ! not zero ??
integer, intent(out) :: max_score, col_coord, row_coord

max_score = 0
col_coord = 0
row_coord = 0

Outter_Loop: do j = 1, UBOUND(MATRIX, DIM=2)
  Inner_Loop: do i = 1, UBOUND(MATRIX, DIM=1)
        if (matrix(i, j) >= max_score) then
            max_score = matrix(i, j)
            col_coord = i
            row_coord = j
        end if        
    end do Inner_Loop
end do Outter_Loop
return

end subroutine

在 fortran 中使用内部函数会更优雅：

subroutine findthemax(cc, rr, matrix, threeple)
IMPLICIT NONE
integer, intent(in)  :: cc, rr 
integer, intent(in)  :: matrix(cc+1,rr+1)  ! not 0:rr, or its it 1:rr+1 ???
integer, intent(OUT)  :: Threeple(3)
integer, DIMENSION(2) :: At_Min

! YOU MAY NEED TRANSPOSE - but just intentionally reverse rows and columns if python is row major??

Threeple(1) = MAXVAL(matrix)

At_Min = MAXLOC(Matrix)
Threeple(2) = At_Min(2) ! Or is it (1)??
Threeple(3) = At_Min(1) ! Or is it (2)??

return
end subroutine findthemax

Answer 2

您可以通过以下方式使用具有多个返回值的fortran子例程。

Fortran 代码findthemax.f90

subroutine findthemax(matrix, max_score, row, col)
  integer, intent(in)  :: matrix(:,:)
  integer, intent(out) :: max_score, row, col

  integer :: ind(2)

  ind       = maxloc(matrix)
  row       = ind(1)
  col       = ind(2)
  max_score = matrix(row,col)
end subroutine

通过 f2py 编译

$ f2py -c findthemax.f90 -m findthemax

将其导入您的 python 代码

import findthemax
import numpy as np

matrix = np.array([[1, 10,  3,  4,  9],
                   [2,  1,  0,  9, 13],
                   [3,  5, 10, 18,  3]])

(max_score, row, col) = findthemax.findthemax(matrix)

print(max_score)        # 18
print(row)              #  3
print(col)              #  4

Answer 3

您对最大值及其索引的实现是慢，因为它使用python 的循环标准。 Numpy 拥有大量快速/优化的内置函数，大多数甚至在底层使用 fortran 代码。

对于您的问题，您应该查看numpy.argmax。

示例代码

import numpy as np

matrix = np.array([[1, 10, 3, 4, 9],
                  [2, 1, 0, 9, 13], 
                  [3, 5, 10, 18, 3]])

# get flat index of maximum
flat = np.argmax(matrix)

# flat index to row/col indices
(row, col) = np.unravel_index(flat, matrix.shape)

# max value
mymax = matrix[row,col]