numpy.all¶

numpy.
all
(a, axis=None, out=None, keepdims=<no value>)[source]¶ Test whether all array elements along a given axis evaluate to True.
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 AND reduction is performed. The default (axis = None) is to perform a logical AND 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
dtype(out)
is float, the result will consist of 0.0’s and 1.0’s). Seedoc.ufuncs
(Section “Output arguments”) for more 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 input array.
If the default value is passed, then keepdims will not be passed through to the
all
method of subclasses ofndarray
, however any nondefault value will be. If the subclass’ method does not implement keepdims any exceptions will be raised.
Returns:  all : ndarray, bool
A new boolean or array is returned unless out is specified, in which case a reference to out is returned.
See also
ndarray.all
 equivalent method
any
 Test whether any element along a given axis evaluates to True.
Notes
Not a Number (NaN), positive infinity and negative infinity evaluate to True because these are not equal to zero.
Examples
>>> np.all([[True,False],[True,True]]) False
>>> np.all([[True,False],[True,True]], axis=0) array([ True, False])
>>> np.all([1, 4, 5]) True
>>> np.all([1.0, np.nan]) True
>>> o=np.array(False) >>> z=np.all([1, 4, 5], out=o) >>> id(z), id(o), z (28293632, 28293632, array(True)) # may vary