#### Previous topic

numpy.unique

#### Next topic

numpy.intersect1d

# numpy.in1d

`numpy.``in1d`(ar1, ar2, assume_unique=False, invert=False)

  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.

  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

  `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])`.

  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])