在 Fortran 中存储具有多维索引的变量
Storing a Variable with a Multi-Dimensional Index in Fortran
问题
考虑以下代码：
program example
implicit none
integer, parameter :: n_coeffs = 1000
integer, parameter :: n_indices = 5
integer :: i
real(8), dimension(n_coeffs) :: coeff
integer, dimension(n_coeffs,n_indices) :: index
do i = 1, n_coeffs
coeff(i) = real(i*3,8)
index(i,:) = [2,4,8,16,32]*i
end do
end
For any 5 dimensional index I need to obtain the associated coefficient, without knowing or calculating i. For instance, given [2,4,8,16,32] I need to obtain 3.0 without computing i.
Is there a reasonable solution, perhaps using sparse matrices, that would work for n_indices in the order of 100 (though n_coeffs still in the order of 1000)?
一个糟糕的解决方案
一个解决方案是定义一个 5 维数组，如
real(8), dimension(2000,4000,8000,16000,32000) :: coeff2
do i = 1, ncoeffs
coeff2(index(i,1),index(i,2),index(i,3),index(i,4),index(i,5)) = coeff(i)
end do
然后，要获得与 [2,4,8,16,32] 关联的系数，请调用
coeff2(2,4,8,16,32)
但是，除了非常浪费内存之外，鉴于数组的 7 维限制，此解决方案不允许将 n_indices 设置为大于 7 的数字。
OBS：这个问题是 的衍生问题。我试图更准确地提出这个问题，但在第一次尝试中失败了，@Rodrigo_Rodrigues.
的回答让我受益匪浅。
实际代码
如果它有帮助，这里是我试图解决的实际问题的代码。它是一种用于逼近函数的自适应稀疏网格方法。主要目标是尽可能快地进行插值：
MODULE MOD_PARAMETERS
IMPLICIT NONE
SAVE
INTEGER, PARAMETER :: d = 2 ! number of dimensions
INTEGER, PARAMETER :: L_0 = 4 ! after this adaptive grid kicks in, for L <= L_0 usual sparse grid
INTEGER, PARAMETER :: L_max = 9 ! maximum level
INTEGER, PARAMETER :: bound = 0 ! 0 -> for f = 0 at boundary
! 1 -> adding grid points at boundary
! 2 -> extrapolating close to boundary
INTEGER, PARAMETER :: max_error = 1
INTEGER, PARAMETER :: L2_error = 1
INTEGER, PARAMETER :: testing_sample = 1000000
REAL(8), PARAMETER :: eps = 0.01D0 ! epsilon for adaptive grid
END MODULE MOD_PARAMETERS
PROGRAM MAIN
USE MOD_PARAMETERS
IMPLICIT NONE
INTEGER, DIMENSION(d,d) :: ident
REAL(8), DIMENSION(d) :: xd
INTEGER, DIMENSION(2*d) :: temp
INTEGER, DIMENSION(:,:), ALLOCATABLE :: grid_index, temp_grid_index, grid_index_new, J_index
REAL(8), DIMENSION(:), ALLOCATABLE :: coeff, temp_coeff, J_coeff
REAL(8) :: temp_min, temp_max, V, T, B, F, x1
INTEGER :: k, k_1, k_2, h, i, j, L, n, dd, L1, L2, dsize, count, first, repeated, add, ind
INTEGER :: time1, time2, clock_rate, clock_max
REAL(8), DIMENSION(L_max,L_max,2**(L_max),2**(L_max)) :: coeff_grid
INTEGER, DIMENSION(d) :: level, LL, ii
REAL(8), DIMENSION(testing_sample,d) :: x_rand
REAL(8), DIMENSION(testing_sample) :: interp1, interp2
! ============================================================================
! EXECUTABLE
! ============================================================================
ident = 0
DO i = 1,d
ident(i,i) = 1
ENDDO
! Initial grid point
dsize = 1
ALLOCATE(grid_index(dsize,2*d),grid_index_new(dsize,2*d))
grid_index(1,:) = 1
grid_index_new = grid_index
ALLOCATE(coeff(dsize))
xd = (/ 0.5D0, 0.5D0 /)
CALL FF(xd,coeff(1))
CALL FF(xd,coeff_grid(1,1,1,1))
L = 1
n = SIZE(grid_index_new,1)
ALLOCATE(J_index(n*2*d,2*d))
ALLOCATE(J_coeff(n*2*d))
CALL SYSTEM_CLOCK (time1,clock_rate,clock_max)
DO WHILE (L .LT. L_max)
L = L+1
n = SIZE(grid_index_new,1)
count = 0
first = 1
DEALLOCATE(J_index,J_coeff)
ALLOCATE(J_index(n*2*d,2*d))
ALLOCATE(J_coeff(n*2*d))
J_index = 0
J_coeff = 0.0D0
DO k = 1,n
DO i = 1,d
DO j = 1,2
IF ((bound .EQ. 0) .OR. (bound .EQ. 2)) THEN
temp = grid_index_new(k,:)+(/ident(i,:),ident(i,:)*(grid_index_new(k,d+i)-(-1)**j)/)
ELSEIF (bound .EQ. 1) THEN
IF (grid_index_new(k,i) .EQ. 1) THEN
temp = grid_index_new(k,:)+(/ident(i,:),ident(i,:)*(-(-1)**j)/)
ELSE
temp = grid_index_new(k,:)+(/ident(i,:),ident(i,:)*(grid_index_new(k,d+i)-(-1)**j)/)
ENDIF
ENDIF
CALL XX(d,temp(1:d),temp(d+1:2*d),xd)
temp_min = MINVAL(xd)
temp_max = MAXVAL(xd)
IF ((temp_min .GE. 0.0D0) .AND. (temp_max .LE. 1.0D0)) THEN
IF (first .EQ. 1) THEN
first = 0
count = count+1
J_index(count,:) = temp
V = 0.0D0
DO k_1 = 1,SIZE(grid_index,1)
T = 1.0D0
DO k_2 = 1,d
CALL XX(1,temp(k_2),temp(d+k_2),x1)
CALL BASE(x1,grid_index(k_1,k_2),grid_index(k_1,k_2+d),B)
T = T*B
ENDDO
V = V+coeff(k_1)*T
ENDDO
CALL FF(xd,F)
J_coeff(count) = F-V
ELSE
repeated = 0
DO h = 1,count
IF (SUM(ABS(J_index(h,:)-temp)) .EQ. 0) THEN
repeated = 1
ENDIF
ENDDO
IF (repeated .EQ. 0) THEN
count = count+1
J_index(count,:) = temp
V = 0.0D0
DO k_1 = 1,SIZE(grid_index,1)
T = 1.0D0
DO k_2 = 1,d
CALL XX(1,temp(k_2),temp(d+k_2),x1)
CALL BASE(x1,grid_index(k_1,k_2),grid_index(k_1,k_2+d),B)
T = T*B
ENDDO
V = V+coeff(k_1)*T
ENDDO
CALL FF(xd,F)
J_coeff(count) = F-V
ENDIF
ENDIF
ENDIF
ENDDO
ENDDO
ENDDO
ALLOCATE(temp_grid_index(dsize,2*d))
ALLOCATE(temp_coeff(dsize))
temp_grid_index = grid_index
temp_coeff = coeff
DEALLOCATE(grid_index,coeff)
ALLOCATE(grid_index(dsize+count,2*d))
ALLOCATE(coeff(dsize+count))
grid_index(1:dsize,:) = temp_grid_index
coeff(1:dsize) = temp_coeff
DEALLOCATE(temp_grid_index,temp_coeff)
grid_index(dsize+1:dsize+count,:) = J_index(1:count,:)
coeff(dsize+1:dsize+count) = J_coeff(1:count)
dsize = dsize + count
DO i = 1,count
coeff_grid(J_index(i,1),J_index(i,2),J_index(i,3),J_index(i,4)) = J_coeff(i)
ENDDO
IF (L .LE. L_0) THEN
DEALLOCATE(grid_index_new)
ALLOCATE(grid_index_new(count,2*d))
grid_index_new = J_index(1:count,:)
ELSE
add = 0
DO h = 1,count
IF (ABS(J_coeff(h)) .GT. eps) THEN
add = add + 1
J_index(add,:) = J_index(h,:)
ENDIF
ENDDO
DEALLOCATE(grid_index_new)
ALLOCATE(grid_index_new(add,2*d))
grid_index_new = J_index(1:add,:)
ENDIF
ENDDO
CALL SYSTEM_CLOCK (time2,clock_rate,clock_max)
PRINT *, 'Elapsed real time1 = ', DBLE(time2-time1)/DBLE(clock_rate)
PRINT *, 'Grid Points = ', SIZE(grid_index,1)
! ============================================================================
! Compute interpolated values:
! ============================================================================
CALL RANDOM_NUMBER(x_rand)
CALL SYSTEM_CLOCK (time1,clock_rate,clock_max)
DO i = 1,testing_sample
V = 0.0D0
DO L1=1,L_max
DO L2=1,L_max
IF (L1+L2 .LE. L_max+1) THEN
level = (/L1,L2/)
T = 1.0D0
DO dd = 1,d
T = T*(1.0D0-ABS(x_rand(i,dd)/2.0D0**(-DBLE(level(dd)))-DBLE(2*FLOOR(x_rand(i,dd)*2.0D0**DBLE(level(dd)-1))+1)))
ENDDO
V = V + coeff_grid(L1,L2,2*FLOOR(x_rand(i,1)*2.0D0**DBLE(L1-1))+1,2*FLOOR(x_rand(i,2)*2.0D0**DBLE(L2-1))+1)*T
ENDIF
ENDDO
ENDDO
interp2(i) = V
ENDDO
CALL SYSTEM_CLOCK (time2,clock_rate,clock_max)
PRINT *, 'Elapsed real time2 = ', DBLE(time2-time1)/DBLE(clock_rate)
END PROGRAM
For any 5 dimensional index I need to obtain the associated
coefficient, without knowing or calculating i. For instance, given
[2,4,8,16,32] I need to obtain 3.0 without computing i.
function findloc_vector(matrix, vector) result(out)
integer, intent(in) :: matrix(:, :)
integer, intent(in) :: vector(size(matrix, dim=2))
integer :: out, i
do i = 1, size(matrix, dim=1)
if (all(matrix(i, :) == vector)) then
out = i
return
end if
end do
stop "No match for this vector"
end
这就是您的使用方式：
print*, coeff(findloc_vector(index, [2,4,8,16,32])) ! outputs 3.0
我必须承认我不愿意 post 这个代码，因为尽管这个 回答了你的问题 ，但老实说，我认为这是 not 你想要的真的 want/need，但你没有提供足够的信息让我知道你真的 do want/need.
编辑（根据 OP 的实际代码）：
如果我正确地解密了你的代码（并考虑到你在上一个问题中所说的话），你就是在声明：
REAL(8), DIMENSION(L_max,L_max,2**(L_max),2**(L_max)) :: coeff_grid
（其中 L_max = 9，所以 size(coeff_grid) = 21233664 =~160MB）然后填充它：
DO i = 1,count
coeff_grid(J_index(i,1),J_index(i,2),J_index(i,3),J_index(i,4)) = J_coeff(i)
ENDDO
（其中 count 是 1000 的数量级，即它的元素的 0.005%），所以这样你就可以通过它的 4 个索引和数组符号来获取值。
请不要那样做。在这种情况下，您也不需要稀疏矩阵。您提出的新方法要好得多：将索引存储在较小数组的每一行中，并通过这些索引在其自身数组中的相应位置获取系数数组。这样速度更快（避免大量分配）并且内存效率更高。
PS：您是否必须坚持使用 Fortran 90？它是一个非常旧的标准版本，您使用的编译器很可能实现了更新的版本。您可以使用内在 move_alloc（为了减少数组副本）、内在模块 iso_fortran_env 中的种类常量（为了可移植性）、[], >, <, <=,... 符号（为了便于阅读)...
问题
考虑以下代码：
program example
implicit none
integer, parameter :: n_coeffs = 1000
integer, parameter :: n_indices = 5
integer :: i
real(8), dimension(n_coeffs) :: coeff
integer, dimension(n_coeffs,n_indices) :: index
do i = 1, n_coeffs
coeff(i) = real(i*3,8)
index(i,:) = [2,4,8,16,32]*i
end do
end
For any 5 dimensional index I need to obtain the associated coefficient, without knowing or calculating
i. For instance, given[2,4,8,16,32]I need to obtain3.0without computingi.Is there a reasonable solution, perhaps using sparse matrices, that would work for
n_indicesin the order of 100 (thoughn_coeffsstill in the order of 1000)?
一个糟糕的解决方案
一个解决方案是定义一个 5 维数组，如
real(8), dimension(2000,4000,8000,16000,32000) :: coeff2
do i = 1, ncoeffs
coeff2(index(i,1),index(i,2),index(i,3),index(i,4),index(i,5)) = coeff(i)
end do
然后，要获得与 [2,4,8,16,32] 关联的系数，请调用
coeff2(2,4,8,16,32)
但是，除了非常浪费内存之外，鉴于数组的 7 维限制，此解决方案不允许将 n_indices 设置为大于 7 的数字。
OBS：这个问题是
实际代码
如果它有帮助，这里是我试图解决的实际问题的代码。它是一种用于逼近函数的自适应稀疏网格方法。主要目标是尽可能快地进行插值：
MODULE MOD_PARAMETERS
IMPLICIT NONE
SAVE
INTEGER, PARAMETER :: d = 2 ! number of dimensions
INTEGER, PARAMETER :: L_0 = 4 ! after this adaptive grid kicks in, for L <= L_0 usual sparse grid
INTEGER, PARAMETER :: L_max = 9 ! maximum level
INTEGER, PARAMETER :: bound = 0 ! 0 -> for f = 0 at boundary
! 1 -> adding grid points at boundary
! 2 -> extrapolating close to boundary
INTEGER, PARAMETER :: max_error = 1
INTEGER, PARAMETER :: L2_error = 1
INTEGER, PARAMETER :: testing_sample = 1000000
REAL(8), PARAMETER :: eps = 0.01D0 ! epsilon for adaptive grid
END MODULE MOD_PARAMETERS
PROGRAM MAIN
USE MOD_PARAMETERS
IMPLICIT NONE
INTEGER, DIMENSION(d,d) :: ident
REAL(8), DIMENSION(d) :: xd
INTEGER, DIMENSION(2*d) :: temp
INTEGER, DIMENSION(:,:), ALLOCATABLE :: grid_index, temp_grid_index, grid_index_new, J_index
REAL(8), DIMENSION(:), ALLOCATABLE :: coeff, temp_coeff, J_coeff
REAL(8) :: temp_min, temp_max, V, T, B, F, x1
INTEGER :: k, k_1, k_2, h, i, j, L, n, dd, L1, L2, dsize, count, first, repeated, add, ind
INTEGER :: time1, time2, clock_rate, clock_max
REAL(8), DIMENSION(L_max,L_max,2**(L_max),2**(L_max)) :: coeff_grid
INTEGER, DIMENSION(d) :: level, LL, ii
REAL(8), DIMENSION(testing_sample,d) :: x_rand
REAL(8), DIMENSION(testing_sample) :: interp1, interp2
! ============================================================================
! EXECUTABLE
! ============================================================================
ident = 0
DO i = 1,d
ident(i,i) = 1
ENDDO
! Initial grid point
dsize = 1
ALLOCATE(grid_index(dsize,2*d),grid_index_new(dsize,2*d))
grid_index(1,:) = 1
grid_index_new = grid_index
ALLOCATE(coeff(dsize))
xd = (/ 0.5D0, 0.5D0 /)
CALL FF(xd,coeff(1))
CALL FF(xd,coeff_grid(1,1,1,1))
L = 1
n = SIZE(grid_index_new,1)
ALLOCATE(J_index(n*2*d,2*d))
ALLOCATE(J_coeff(n*2*d))
CALL SYSTEM_CLOCK (time1,clock_rate,clock_max)
DO WHILE (L .LT. L_max)
L = L+1
n = SIZE(grid_index_new,1)
count = 0
first = 1
DEALLOCATE(J_index,J_coeff)
ALLOCATE(J_index(n*2*d,2*d))
ALLOCATE(J_coeff(n*2*d))
J_index = 0
J_coeff = 0.0D0
DO k = 1,n
DO i = 1,d
DO j = 1,2
IF ((bound .EQ. 0) .OR. (bound .EQ. 2)) THEN
temp = grid_index_new(k,:)+(/ident(i,:),ident(i,:)*(grid_index_new(k,d+i)-(-1)**j)/)
ELSEIF (bound .EQ. 1) THEN
IF (grid_index_new(k,i) .EQ. 1) THEN
temp = grid_index_new(k,:)+(/ident(i,:),ident(i,:)*(-(-1)**j)/)
ELSE
temp = grid_index_new(k,:)+(/ident(i,:),ident(i,:)*(grid_index_new(k,d+i)-(-1)**j)/)
ENDIF
ENDIF
CALL XX(d,temp(1:d),temp(d+1:2*d),xd)
temp_min = MINVAL(xd)
temp_max = MAXVAL(xd)
IF ((temp_min .GE. 0.0D0) .AND. (temp_max .LE. 1.0D0)) THEN
IF (first .EQ. 1) THEN
first = 0
count = count+1
J_index(count,:) = temp
V = 0.0D0
DO k_1 = 1,SIZE(grid_index,1)
T = 1.0D0
DO k_2 = 1,d
CALL XX(1,temp(k_2),temp(d+k_2),x1)
CALL BASE(x1,grid_index(k_1,k_2),grid_index(k_1,k_2+d),B)
T = T*B
ENDDO
V = V+coeff(k_1)*T
ENDDO
CALL FF(xd,F)
J_coeff(count) = F-V
ELSE
repeated = 0
DO h = 1,count
IF (SUM(ABS(J_index(h,:)-temp)) .EQ. 0) THEN
repeated = 1
ENDIF
ENDDO
IF (repeated .EQ. 0) THEN
count = count+1
J_index(count,:) = temp
V = 0.0D0
DO k_1 = 1,SIZE(grid_index,1)
T = 1.0D0
DO k_2 = 1,d
CALL XX(1,temp(k_2),temp(d+k_2),x1)
CALL BASE(x1,grid_index(k_1,k_2),grid_index(k_1,k_2+d),B)
T = T*B
ENDDO
V = V+coeff(k_1)*T
ENDDO
CALL FF(xd,F)
J_coeff(count) = F-V
ENDIF
ENDIF
ENDIF
ENDDO
ENDDO
ENDDO
ALLOCATE(temp_grid_index(dsize,2*d))
ALLOCATE(temp_coeff(dsize))
temp_grid_index = grid_index
temp_coeff = coeff
DEALLOCATE(grid_index,coeff)
ALLOCATE(grid_index(dsize+count,2*d))
ALLOCATE(coeff(dsize+count))
grid_index(1:dsize,:) = temp_grid_index
coeff(1:dsize) = temp_coeff
DEALLOCATE(temp_grid_index,temp_coeff)
grid_index(dsize+1:dsize+count,:) = J_index(1:count,:)
coeff(dsize+1:dsize+count) = J_coeff(1:count)
dsize = dsize + count
DO i = 1,count
coeff_grid(J_index(i,1),J_index(i,2),J_index(i,3),J_index(i,4)) = J_coeff(i)
ENDDO
IF (L .LE. L_0) THEN
DEALLOCATE(grid_index_new)
ALLOCATE(grid_index_new(count,2*d))
grid_index_new = J_index(1:count,:)
ELSE
add = 0
DO h = 1,count
IF (ABS(J_coeff(h)) .GT. eps) THEN
add = add + 1
J_index(add,:) = J_index(h,:)
ENDIF
ENDDO
DEALLOCATE(grid_index_new)
ALLOCATE(grid_index_new(add,2*d))
grid_index_new = J_index(1:add,:)
ENDIF
ENDDO
CALL SYSTEM_CLOCK (time2,clock_rate,clock_max)
PRINT *, 'Elapsed real time1 = ', DBLE(time2-time1)/DBLE(clock_rate)
PRINT *, 'Grid Points = ', SIZE(grid_index,1)
! ============================================================================
! Compute interpolated values:
! ============================================================================
CALL RANDOM_NUMBER(x_rand)
CALL SYSTEM_CLOCK (time1,clock_rate,clock_max)
DO i = 1,testing_sample
V = 0.0D0
DO L1=1,L_max
DO L2=1,L_max
IF (L1+L2 .LE. L_max+1) THEN
level = (/L1,L2/)
T = 1.0D0
DO dd = 1,d
T = T*(1.0D0-ABS(x_rand(i,dd)/2.0D0**(-DBLE(level(dd)))-DBLE(2*FLOOR(x_rand(i,dd)*2.0D0**DBLE(level(dd)-1))+1)))
ENDDO
V = V + coeff_grid(L1,L2,2*FLOOR(x_rand(i,1)*2.0D0**DBLE(L1-1))+1,2*FLOOR(x_rand(i,2)*2.0D0**DBLE(L2-1))+1)*T
ENDIF
ENDDO
ENDDO
interp2(i) = V
ENDDO
CALL SYSTEM_CLOCK (time2,clock_rate,clock_max)
PRINT *, 'Elapsed real time2 = ', DBLE(time2-time1)/DBLE(clock_rate)
END PROGRAM
For any 5 dimensional index I need to obtain the associated coefficient, without knowing or calculating
i. For instance, given[2,4,8,16,32]I need to obtain3.0without computingi.
function findloc_vector(matrix, vector) result(out)
integer, intent(in) :: matrix(:, :)
integer, intent(in) :: vector(size(matrix, dim=2))
integer :: out, i
do i = 1, size(matrix, dim=1)
if (all(matrix(i, :) == vector)) then
out = i
return
end if
end do
stop "No match for this vector"
end
这就是您的使用方式：
print*, coeff(findloc_vector(index, [2,4,8,16,32])) ! outputs 3.0
我必须承认我不愿意 post 这个代码，因为尽管这个 回答了你的问题 ，但老实说，我认为这是 not 你想要的真的 want/need，但你没有提供足够的信息让我知道你真的 do want/need.
编辑（根据 OP 的实际代码）：
如果我正确地解密了你的代码（并考虑到你在上一个问题中所说的话），你就是在声明：
REAL(8), DIMENSION(L_max,L_max,2**(L_max),2**(L_max)) :: coeff_grid
（其中 L_max = 9，所以 size(coeff_grid) = 21233664 =~160MB）然后填充它：
DO i = 1,count
coeff_grid(J_index(i,1),J_index(i,2),J_index(i,3),J_index(i,4)) = J_coeff(i)
ENDDO
（其中 count 是 1000 的数量级，即它的元素的 0.005%），所以这样你就可以通过它的 4 个索引和数组符号来获取值。
请不要那样做。在这种情况下，您也不需要稀疏矩阵。您提出的新方法要好得多：将索引存储在较小数组的每一行中，并通过这些索引在其自身数组中的相应位置获取系数数组。这样速度更快（避免大量分配）并且内存效率更高。
PS：您是否必须坚持使用 Fortran 90？它是一个非常旧的标准版本，您使用的编译器很可能实现了更新的版本。您可以使用内在 move_alloc（为了减少数组副本）、内在模块 iso_fortran_env 中的种类常量（为了可移植性）、[], >, <, <=,... 符号（为了便于阅读)...