Set routines

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numpy.intersect1d

# numpy.in1d¶

`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.

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. 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 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.

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])
```