如何获得先前未知的数组作为 Fortran 中函数的输出 [英] How to get priorly-unknown array as the output of a function in Fortran

问题描述

在 Python 中:

def select(x):y = []对于 x 中的 e:如果 e!=0:y.append(e)返回 y

作为:

<前>x = [1,0,2,0,0,3]选择(x)[1,2,3]

翻译成Fortran:

function select(x,n) result(y)隐式无整数:: x(n),n,i,j,y(?)j = 0做我= 1，n如果 (x(i)/=0) 那么j = j+1y(j) = x(i)万一端部结束函数

问题在 Fortran 中:

1. 如何声明y(?)?
2. 如何声明 x 的预定义值
3. 如何避免维度信息 n

如果定义为 y(n) 则为 1，则输出为:

<前>x = (/1,0,2,0,0,3/)打印*,选择(x,6)1,2,3,0,0,0

这是不想要的！
!-------------------------------
评论:
1- 所有给出的答案在这篇文章中都很有用.特别是 M.S.B 和 eryksun's.
2- 我尝试针对我的问题调整想法并使用 F2Py 进行编译，但没有成功.我已经使用 GFortran 调试过它们，并且都成功了.它可能是 F2Py 中的一个错误，或者我不知道正确使用它的东西.我会尝试在另一篇文章中讨论这个问题.

更新:可以在 此处.

解决方案

我希望一个真正的 Fortran 程序员出现，但在没有更好的建议的情况下，我只会指定 x(:)，使用临时数组temp(size(x))，并使输出y allocable.然后在第一遍之后，allocate(y(j)) 并从临时数组中复制值.但是我不能强调我不是 Fortran 程序员，所以我不能说该语言是否有可增长的数组，或者是否存在用于后者的库.

程序测试隐式无整数:: x(10) = (/1,0,2,0,3,0,4,0,5,0/)打印(10I2.1)"，选择(x)包含函数选择(x) 结果(y)隐式无整数，意图(输入):: x(:)整数:: i, j, temp(size(x))整数，可分配:: y(:)j = 0做我= 1，大小(x)如果 (x(i)/= 0) 那么j = j + 1温度(j) = x(i)万一端部分配(y(j))y = 温度(:j)结束函数选择结束程序测试

<小时>

根据 M.S.B. 的回答，这里是函数的修订版，它会随着过度分配而增长 temp y.和之前一样，它最后将结果复制到 y. 事实证明，我没有必要以最终大小显式分配一个新数组.相反，它可以通过分配自动完成.

 函数选择(x) 结果(y)隐式无整数，意图(输入):: x(:)整数:: i, j, dsize整数，可分配:: temp(:), y(:)dsize = 0;分配(y(0))j = 0做我= 1，大小(x)如果 (x(i)/= 0) 那么j = j + 1if (j >= dsize) then !grow y using tempdsize = j + j/8 + 8分配(临时(dsize))温度(:大小(y))= y调用 move_alloc(temp, y) !temp 被释放万一y(j) = x(i)万一端部y = y(:j)结束函数选择

In Python:

def select(x):
    y = []
    for e in x:
        if e!=0:
            y.append(e)
    return y

that works as:

x = [1,0,2,0,0,3]
select(x)
[1,2,3]

to be translated into Fortran:

function select(x,n) result(y)
    implicit none
    integer:: x(n),n,i,j,y(?)
    j = 0
    do i=1,n
        if (x(i)/=0) then
            j = j+1
            y(j) = x(i)
        endif
    enddo
end function

The questions are in Fortran:

1. how to declare y(?)?
2. how to declare predefined values for x
3. how to avoid dimension info n

for 1 if it is defined as y(n) the output will be:

x = (/1,0,2,0,0,3/)
print *,select(x,6)
1,2,3,0,0,0

which is not desired!
!-------------------------------
Comments:
1- All given answers are useful in this post. Specially M.S.B and eryksun's.
2- I tried to adapt the ideas for my problem and compile with F2Py however it was not successful. I had already debugged them using GFortran and all were successful. It might be a bug in F2Py or something that I don't know about using it properly. I will try to cover this issue in another post.

Update: A linked question could be found at here.

解决方案

I hope a real Fortran programmer comes along, but in the absence of better advice, I would only specify the shape and not the size of x(:), use a temporary array temp(size(x)), and make the output y allocatable. Then after the first pass, allocate(y(j)) and copy the values from the temporary array. But I can't stress enough that I'm not a Fortran programmer, so I can't say if the language has a growable array or if a library exists for the latter.

program test
    implicit none
    integer:: x(10) = (/1,0,2,0,3,0,4,0,5,0/)
    print "(10I2.1)", select(x)

contains

    function select(x) result(y)
        implicit none
        integer, intent(in):: x(:) 
        integer:: i, j, temp(size(x))
        integer, allocatable:: y(:)

        j = 0
        do i = 1, size(x)
            if (x(i) /= 0) then
                j = j + 1
                temp(j) = x(i)
            endif
        enddo

        allocate(y(j))
        y = temp(:j)
    end function select

end program test

---

Edit:

Based on M.S.B.'s answer, here's a revised version of the function that grows temp y with over-allocation. As before it copies the result to y at the end. It turns out i's not necessary to explicitly allocate a new array at the final size. Instead it can be done automatically with assignment.

    function select(x) result(y)
        implicit none
        integer, intent(in):: x(:) 
        integer:: i, j, dsize
        integer, allocatable:: temp(:), y(:)

        dsize = 0; allocate(y(0))

        j = 0
        do i = 1, size(x)
            if (x(i) /= 0) then
                j = j + 1

                if (j >= dsize) then         !grow y using temp
                    dsize = j + j / 8 + 8 
                    allocate(temp(dsize))
                    temp(:size(y)) = y
                    call move_alloc(temp, y) !temp gets deallocated
                endif

                y(j) = x(i)
            endif
        enddo
        y = y(:j)
    end function select

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