在Fortran中，如何从一个数组中删除第N个元素？ [英] In Fortran, how do I remove Nth element from an array?

问题描述

例如我有阵列（/ 1,3,4,5,7,9,11 /），我怎么删除其3元？我找不到它确实是一个阵列的功能，也没有我发现一个循环一个优雅的解决方案，因为我不知道该怎么追加到一个数组（这意味着，添加元素的previous定义的元素旁边。 ）

Eg I have array (/1,3,4,5,7,9,11/), how do I remove its 3rd element? I couldn't find an array function which does that, nor I find a loop an elegant solution, since I don't know how to append to an array (this means, add an element next to the previous defined element.)

我想从一个数组中删除所有偶数元素......我知道，只有一个。

I want to remove all even elements from an array... I know there is only one.

我可以找到使用它的索引 MINLOC ，但我不知道如何从数组中删除一个元素。

I can find its index using MINLOC, but I don't know how to remove an element from array.

推荐答案

让

a = (/1,3,4,5,7,9,11/)

然后

pack(a,mod(a,2)/=0)

将返回的奇数元素。这是不太一样去掉了第三个元素，但你的问题表明，去除偶数元素（s）是你真正想要做的事情。

will return the odd elements of a. This isn't quite the same as removing the 3rd element, but your question suggests that removing the even element(s) is really what you want to do.

如果您声明

integer, dimension(:), allocatable :: oddones

然后

oddones = pack(a,mod(a,2)/=0)

将离开 oddones 包含的奇数元素。你需要一个向上最新的编译器使用这种自动分配。

will leave oddones containing the odd elements of a. You'll need an up-to-date compiler to use this automatic allocation.

请注意，在Fortran语言，如任何理智的语言中，数组是固定大小的，所以移除元素是不是真的支持。但是，如果 A 本身是可分配那么你可以使用 A 上的 LHS 的前任pression的。让我们把它留到哲学家与否 A 保持此操作下是相同的。

Note that in Fortran, as in any sane language, arrays are of fixed size so removing an element isn't really supported. However, if a itself were allocatable then you could use a on the lhs of the expression. Let's leave it to the philosophers whether or not a remains the same under this operation.

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