In Fortran, how do I remove Nth element from an array?

Question

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.) I want to remove all even elements from an array... I know there is only one. I can find its index using MINLOC, but I don't know how to remove an element from array.

Accepted Answer

Let

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

then

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

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.

If you declare

```
integer, dimension(:), allocatable :: oddones
```

then

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

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

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.

In Fortran, how do I remove Nth element from an array?

Solution

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