如何检查 Fortran 数组是否包含值?

问题描述

我已经看到这要求使用其他语言，但刚刚发现 Fortran 可以很好地处理数组，我认为可能有一种简单的方法可以在没有循环的情况下做到这一点.

I've seen this asked for other languages, but having just found out how nicely Fortran can handle arrays, I thought there might be an easy way to do this without loops.

目前我正在搜索一个 3D 数组，查看最近的邻居"以查看它们是否包含字母n"，并且每当它找到这个值时，我希望它执行一些 clusterLabel 分配(这不相关对于这个问题)

Currently I'm searching over a 3D array looking at 'nearest neighbours' to see if they contain the letter 'n', and whenever it finds this value, I want it to perform some clusterLabel assignment (which isn't relevant for this question)

我想使用 if(lastNeighArray.eq."n") then...但出于显而易见的原因，它不喜欢根据值检查数组.它也不喜欢我使用 lastNeighArray(:)，即使我希望它一次检查每个元素.where(lastNeighArray.eq."n") 不起作用，因为我在 where 循环中有一个 case 语句，并且出现错误 WHERE 语句和构造不能嵌套.代码>

I wanted to use if(lastNeighArray.eq."n") then...<rest of code>but for obvious reasons it doesn't like checking an array against a value. Neither does it like me using lastNeighArray(:), even though I'd like it to check each of the elements one at a time. where(lastNeighArray.eq."n") doesn't work as I have a case statement inside the where loop and I get the error WHERE statements and constructs must not be nested.

所以我有点卡住了.我真正想要的是 when(lastNeighArray.eq."n") 之类的东西，但它不存在.

So I'm a little stuck. What I really want is something like when(lastNeighArray.eq."n") but that doesn't exist.

我也看过 any 和 forall 但它们似乎不是正确的选择.

I've also looked at any and forall but they don't seem like the right choice.

推荐答案

ANY 实际上应该是正确的选择

ANY should actually be the right choice

if ( ANY( lastNeighArray=="n" ) ) then

如果您希望整个数组包含该值，也可以使用 ALL.

there is also ALL if you wanted the whole array to contain that value.

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