arrIntersection

Returns the values that are present in every one of the given arrays.

The result is unique and keeps the order of the first array, so arrIntersection([2, 1, 2], [2]) returns [2]. Calling it with a single array is therefore the same as removing its duplicates, and calling it with no array at all returns an empty array.

Values are compared by value, not by reference, so nested arrays and objects are matched as well: arrIntersection([[1], [2]], [[2], [3]]) returns [[2]]. Types are not coerced — 1 and '1' are different values.

The arrays are passed as separate arguments.

The arrays are passed as a single list of lists.

Parameters

Name	Type	Required	Default
arrays	any[][]List<List<dynamic>>list[list[Any]]	●	–
The arrays to intersect. None of them is modified.

Returns

any[]

List<dynamic>

list[Any]

Examples

arrIntersection([2, 1], [2, 3]); // Returns [2]
arrIntersection([1, 2, 3], [2, 3, 4], [3, 2]); // Returns [2, 3]
arrIntersection([2, 1, 2], [2]); // Returns [2]
arrIntersection([1, 2], [3]); // Returns []
arrIntersection([[1], [2]], [[2], [3]]); // Returns [[2]]

arrIntersection([[2, 1], [2, 3]]); // Returns [2]
arrIntersection([[1, 2, 3], [2, 3, 4], [3, 2]]); // Returns [2, 3]
arrIntersection([[2, 1, 2], [2]]); // Returns [2]
arrIntersection([[1, 2], [3]]); // Returns []
arrIntersection([[[1], [2]], [[2], [3]]]); // Returns [[2]]

arrIntersection([2, 1], [2, 3])  # Returns [2]
arrIntersection([1, 2, 3], [2, 3, 4], [3, 2])  # Returns [2, 3]
arrIntersection([2, 1, 2], [2])  # Returns [2]
arrIntersection([1, 2], [3])  # Returns []
arrIntersection([[1], [2]], [[2], [3]])  # Returns [[2]]