std way to remove multiple indices from an array at once

[Steven Schveighoffer]

On 12/20/17 7:52 PM, Nicholas Wilson wrote:
> On Thursday, 21 December 2017 at 00:23:08 UTC, Steven Schveighoffer wrote:
>> On 12/20/17 6:01 PM, aliak wrote:
>>> Hi, is there a way to remove a number of elements from an array by a 
>>> range of indices in the standard library somewhere?
>>>
>>> I wrote one (code below), but I'm wondering if there's a better way?
>>>
>>> Also, can the below be made more efficient?
>>>
>>> auto without(T, R)(T[] array, R indices) if (isForwardRange!R && 
>>> isIntegral!(ElementType!R) && !isInfinite!R) {
>>>      T[] newArray;
>>>      ElementType!R start = 0;
>>>      foreach (i; indices) {
>>>          newArray ~= array[start .. i];
>>>          start = i + 1;
>>>      }
>>>      newArray ~= array[start .. $];
>>>      return newArray;
>>> }
>>>
>>> // Usage
>>> long[] aa = [1, 2, 3, 4]
>>> aa = aa.without([1, 3])
>>>
>>> Thanks!
>>
>> I'm assuming here indices is sorted? Because it appears you expect 
>> that in your code. However, I'm going to assume it isn't sorted at first.
>>
>> Now:
>>
>> import std.range;
>> import std.algorithm;
>>
>> auto indices = [1,2,3,4];
>> aa = aa.enumerate.filter!(a => !indicies.canFind(a[0])).map(a => 
>> a[1]).array;
>>
>> Now, this is going to be O(n^2), but if indices is sorted, you could 
>> use binary search:
>>
>> auto sortedIdxs = indices.assumeSorted; // also could be = indices.sort()
>>
>> arr = arr.enumerate.filter!(a => !sortedIdxs.contains(a[0])).map(a => 
>> a[1]).array;
>>
>> Complexity would be O(n lg(n))
>>
>> It's not going to be as good as hand-written code, complexity wise, 
>> but it's definitely shorter to write :)
>>
> 
> isn't that n log(m), where n is length of array and m is length of indices?

Strictly speaking, yes :) But lg(n) and lg(m) are going to be pretty 
insignificantly different.

> If indices is sorted with no duplicates and random access then you can 
> do it in linear time.
> 
> int i;
> int ii;
> int[] indicies = ...;
> arr = arr.filter!((T t)
> {
>      scope(exit) i++;
>      if (i == indicies[ii])
>      {
>          ii++;
>          return false;
>      }
>      return true;
> }).array;
> 

Very nice! as aliak mentioned, however, you have a bug, as ii might 
extend beyond the size of indicies. So should be if(ii < indices.length 
&& indicies[ii] == i).

We are always focused on using a chained one-liner, but your lambda 
stretches that ;)

Here's a similar solution with an actual range:

https://run.dlang.io/is/gR3CjF

Note, all done lazily. However, the indices must be sorted/unique.

-Steve