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: element : array_like
Input array.
test_elements : array_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_unique : bool, optional
If True, the input arrays are both assumed to be unique, which can speed up the calculation. Default is False.
invert : bool, 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: isin : ndarray, 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 tonp.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]) >>> 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]])