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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
ar1that is True where an element ofar1is inar2and False otherwise.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
ar1is inar2and 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 inar2.See also
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numpy.lib.arraysetops - Module with a number of other functions for performing set operations on arrays.
Notes
in1dcan be considered as an element-wise function version of the python keywordin, for 1-D sequences.in1d(a, b)is roughly equivalent tonp.array([item in b for item in a]). However, this idea fails ifar2is a set, or similar (non-sequence) container: Asar2is 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], 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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numpy.in1d()
2017-01-10 18:14:25
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