numpy.isin
-
numpy.isin(element, test_elements, assume_unique=False, invert=False)
[source] -
Calculates
element in test_elements
, broadcasting overelement
only. Returns a boolean array of the same shape aselement
that is True where an element ofelement
is intest_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 valueselement[isin]
are intest_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 keywordin
.isin(a, b)
is roughly equivalent tonp.array([item in b for item in a])
ifa
andb
are 1-D sequences.element
andtest_elements
are converted to arrays if they are not already. Iftest_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 intest_elements
. This is a consequence of thearray
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]])
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https://numpy.org/doc/1.21/reference/generated/numpy.isin.html