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# numpy.any

`numpy.``any`(a, axis=None, out=None, keepdims=False)

  Test whether any array element along a given axis evaluates to True.

  Returns single boolean unless axis is not `None`

  Parameters : | a : array_like Input array or object that can be converted to an array. axis : None or int or tuple of ints, optional Axis or axes along which a logical OR reduction is performed. The default (axis = None) is perform a logical OR over all the dimensions of the input array. axis may be negative, in which case it counts from the last to the first axis. New in version 1.7.0. If this is a tuple of ints, a reduction is performed on multiple axes, instead of a single axis or all the axes as before. out : ndarray, optional Alternate output array in which to place the result. It must have the same shape as the expected output and its type is preserved (e.g., if it is of type float, then it will remain so, returning 1.0 for True and 0.0 for False, regardless of the type of a). See `doc.ufuncs` (Section “Output arguments”) for details. keepdims : bool, optional If this is set to True, the axes which are reduced are left in the result as dimensions with size one. With this option, the result will broadcast correctly against the original arr.
  Returns : | any : bool or ndarray A new boolean or `ndarray` is returned unless out is specified, in which case a reference to out is returned.

  See also

  `ndarray.any`
    equivalent method
  `all`
    Test whether all elements along a given axis evaluate to True.

  Notes

  Not a Number (NaN), positive infinity and negative infinity evaluate to True because these are not equal to zero.

  Examples

      >>> np.any([[True, False], [True, True]])
      True

      >>> np.any([[True, False], [False, False]], axis=0)
      array([ True, False], dtype=bool)

      >>> np.any([-1, 0, 5])
      True

      >>> np.any(np.nan)
      True

      >>> o=np.array([False])
      >>> z=np.any([-1, 4, 5], out=o)
      >>> z, o
      (array([ True], dtype=bool), array([ True], dtype=bool))
      >>> # Check now that z is a reference to o
      >>> z is o
      True
      >>> id(z), id(o) # identity of z and o
      (191614240, 191614240)