numpy.isin#

numpy.isin(element, test_elements, assume_unique=False, invert=False)[source]#

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