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
