numpy.in1d¶

numpy.
in1d
(ar1, ar2, assume_unique=False, invert=False)[source]¶ Test whether each element of a 1D 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 ofin1d
for new code.Parameters: 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.
Returns: in1d : (M,) ndarray, bool
The values ar1[in1d] are in ar2.
See also
isin
 Version of this function that preserves the shape of ar1.
numpy.lib.arraysetops
 Module with a number of other functions for performing set operations on arrays.
Notes
in1d
can be considered as an elementwise function version of the python keyword in, for 1D sequences.in1d(a, b)
is roughly equivalent tonp.array([item in b for item in a])
. However, this idea fails if ar2 is a set, or similar (nonsequence) container: Asar2
is converted to an array, in those casesasarray(ar2)
is an object array rather than the expected array of contained values.New in version 1.4.0.
Examples
>>> test = np.array([0, 1, 2, 5, 0]) >>> states = [0, 2] >>> mask = np.in1d(test, states) >>> mask array([ True, False, True, False, True]) >>> test[mask] array([0, 2, 0]) >>> mask = np.in1d(test, states, invert=True) >>> mask array([False, True, False, True, False]) >>> test[mask] array([1, 5])