mask = np.isin(A, B)
idx=np.nonzere(mask) # return index of element A which is in element B
-
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.
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