Programming-Idioms

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Idiom #218 List intersection

Create list c containing all unique elements that are contained in both lists a and b.
c should not contain any duplicates, even if a and b do.
The order of c doesn't matter.

using System.Linq;
using System.Collections.Generic;
c = a.Intersect(b).ToList();
seta := make(map[T]bool, len(a))
for _, x := range a {
	seta[x] = true
}
setb := make(map[T]bool, len(a))
for _, y := range b {
	setb[y] = true
}

var c []T
for x := range seta {
	if setb[x] {
		c = append(c, x)
	}
}
import Data.List (intersect)
a `intersect` b
const c = [...new Set(a)].filter(e => b.includes(e));
Classes
for elem in a do
  if (b.indexof(elem) >= 0) and (c.indexof(elem) = -1) then
    c.add(elem);
c = list(set(a).intersection(b))
c = list(set(a) & set(b))
c = a & b
use std::collections::HashSet;
let unique_a = a.iter().collect::<HashSet<_>>();
let unique_b = b.iter().collect::<HashSet<_>>();

let c = unique_a.intersection(&unique_b).collect<Vec<_>>();
use std::collections::HashSet;
let set_a: HashSet<_> = a.into_iter().collect();
let set_b: HashSet<_> = b.into_iter().collect();
let c = set_a.intersection(&set_b);
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Related idioms

c = a.Intersect(b).ToList();
seta := make(map[T]bool, len(a))
for _, x := range a {
	seta[x] = true
}
setb := make(map[T]bool, len(a))
for _, y := range b {
	setb[y] = true
}

var c []T
for x := range seta {
	if setb[x] {
		c = append(c, x)
	}
}
a `intersect` b
const c = [...new Set(a)].filter(e => b.includes(e));
for elem in a do
  if (b.indexof(elem) >= 0) and (c.indexof(elem) = -1) then
    c.add(elem);
c = list(set(a).intersection(b))
c = list(set(a) & set(b))
c = a & b
let unique_a = a.iter().collect::<HashSet<_>>();
let unique_b = b.iter().collect::<HashSet<_>>();

let c = unique_a.intersection(&unique_b).collect<Vec<_>>();
let set_a: HashSet<_> = a.into_iter().collect();
let set_b: HashSet<_> = b.into_iter().collect();
let c = set_a.intersection(&set_b);