On Thursday, 26 January 2017 at 08:22:09 UTC, albert-j wrote:
What is the D idiom for removing array elements that are
present in another array?
Is this the right/fastest way?
int[] a = [1, 2, 3, 4, 5, 6, 7, 4];
int[] b = [3, 4, 6];
auto c = a.remove!(x => b.canFind(x));
assert(c == [1, 2, 5, 7]);
import std.stdio, std.algorithm, std.range, std.array;
int[] a = [1, 2, 3, 4, 5, 6, 7, 4];
int[] b = [3, 4, 6];
auto sortedB = sort(b.dup);
auto c = a
. filter!(i => !sortedB.contains(i))
. array
;
assert(c == [1, 2, 5, 7]);
If arrays get large, this should be more efficient since it
performs O(n * n.log) instead of O(n * n).
On the other hand you could also just use assumeSorted if b is
already sorted, as in this case. But if you do, I still recommend
adding an assert to check the range truly is sorted when
debugging.
It can be made to perform O(n) by sorting both a and b, but then
you need to figure a way to return a back to normal order after
filtering. I tried, and found it to be so hard I didn't bother.
