numpy.isin

numpy.isin(element, test_elements, assume_unique=False, invert=False)

Calculates element in test_elements, broadcasting over element only. Returns a boolean array of the same shape as element that is True where an element of element is in test_elements and False otherwise.

Parameters

elementarray_like

Input array.

test_elementsarray_like

The values against which to test each value of element. This argument is flattened if it is an array or array_like. See notes for behavior with non-array-like parameters.

assume_uniquebool, optional

If True, the input arrays are both assumed to be unique, which can speed up the calculation. Default is False.

invertbool, optional

If True, the values in the returned array are inverted, as if calculating element not in test_elements. Default is False. np.isin(a, b, invert=True) is equivalent to (but faster than) np.invert(np.isin(a, b)).

Returns

isinndarray, bool

Has the same shape as element. The values element[isin] are in test_elements.

See also

in1d

Flattened version of this function.

numpy.lib.arraysetops

Module with a number of other functions for performing set operations on arrays.

Notes

isin is an element-wise function version of the python keyword in. isin(a, b) is roughly equivalent to np.array([item in b for item in a]) if a and b are 1-D sequences.

element and test_elements are converted to arrays if they are not already. If test_elements is a set (or other non-sequence collection) it will be converted to an object array with one element, rather than an array of the values contained in test_elements. This is a consequence of the array constructor’s way of handling non-sequence collections. Converting the set to a list usually gives the desired behavior.

New in version 1.13.0.

Examples

>>> element = 2*np.arange(4).reshape((2, 2))
>>> element
array([[0, 2],
       [4, 6]])
>>> test_elements = [1, 2, 4, 8]
>>> mask = np.isin(element, test_elements)
>>> mask
array([[False,  True],
       [ True, False]])
>>> element[mask]
array([2, 4])

The indices of the matched values can be obtained with nonzero:

>>> np.nonzero(mask)
(array([0, 1]), array([1, 0]))

The test can also be inverted:

>>> mask = np.isin(element, test_elements, invert=True)
>>> mask
array([[ True, False],
       [False,  True]])
>>> element[mask]
array([0, 6])

Because of how array handles sets, the following does not work as expected:

>>> test_set = {1, 2, 4, 8}
>>> np.isin(element, test_set)
array([[False, False],
       [False, False]])

Casting the set to a list gives the expected result:

>>> np.isin(element, list(test_set))
array([[False,  True],
       [ True, False]])

numpy.intersect1d numpy.setdiff1d