# 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.

  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.

  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 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])

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