numpy.in1d(ar1, ar2, assume_unique=False, invert=False)[source]

Test whether each element of a 1-D array is also present in a second array.

Returns a boolean array the same length as ar1 that is True where an element of ar1 is in ar2 and False otherwise.

We recommend using isin instead of in1d for new code.


ar1 : (M,) array_like

Input array.

ar2 : array_like

The values against which to test each value of ar1.

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 (that is, False where an element of ar1 is in ar2 and True otherwise). Default is False. np.in1d(a, b, invert=True) is equivalent to (but is faster than) np.invert(in1d(a, b)).

New in version 1.8.0.


in1d : (M,) ndarray, bool

The values ar1[in1d] are in ar2.

See also

Version of this function that preserves the shape of ar1.
Module with a number of other functions for performing set operations on arrays.


in1d can be considered as an element-wise function version of the python keyword in, for 1-D sequences. in1d(a, b) is roughly equivalent to np.array([item in b for item in a]). However, this idea fails if ar2 is a set, or similar (non-sequence) container: As ar2 is converted to an array, in those cases asarray(ar2) is an object array rather than the expected array of contained values.

New in version 1.4.0.


>>> test = np.array([0, 1, 2, 5, 0])
>>> states = [0, 2]
>>> mask = np.in1d(test, states)
>>> mask
array([ True, False,  True, False,  True], dtype=bool)
>>> test[mask]
array([0, 2, 0])
>>> mask = np.in1d(test, states, invert=True)
>>> mask
array([False,  True, False,  True, False], dtype=bool)
>>> test[mask]
array([1, 5])

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