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Test whether all array elements along a given axis evaluate to True.
Input array or object that can be converted to an array.
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. 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.
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 0.0s and 1.0s). See Output type determination
for more details.
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 sub-classes of
ndarray, however any non-default value will be. If the
sub-class method does not implement keepdims any
exceptions will be raised.
Elements to include in checking for all True values.
See reduce for details.
New in version 1.20.0.
A new boolean or array is returned unless out is specified, in which case a reference to out is returned.
See also
ndarray.allequivalent method
anyTest 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.
Changed in version 2.0: Before NumPy 2.0, all did not return booleans for object dtype
input arrays.
This behavior is still available via np.logical_and.reduce.
Examples
>>> import numpy as np
>>> 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
>>> np.all([[True, True], [False, True]], where=[[True], [False]])
True
>>> o=np.array(False)
>>> z=np.all([-1, 4, 5], out=o)
>>> id(z), id(o), z
(28293632, 28293632, array(True)) # may vary
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