Computes the sum of elements across dimensions of a tensor.
meridian
.
backend
.
reduce_sum
(
input_tensor
,
axis
=
None
,
keepdims
=
False
,
name
=
None
)
This is the reduction operation for the elementwise tf.math.add
op.
Reduces input_tensor
along the dimensions given in axis
.
Unless keepdims
is true, the rank of the tensor is reduced by 1 for each
of the entries in axis
, which must be unique. If keepdims
is true, the
reduced dimensions are retained with length 1.
If axis
is None, all dimensions are reduced, and a
tensor with a single element is returned.
For example
x has a shape of (2, 3) (two rows and three columns):
x = tf.constant([[1, 1, 1], [1, 1, 1]]) x.numpy() array([[1, 1, 1], [1, 1, 1]], dtype=int32)
sum all the elements
1 + 1 + 1 + 1 + 1+ 1 = 6
tf.reduce_sum(x).numpy().item() 6
reduce along the first dimension
the result is [1, 1, 1] + [1, 1, 1] = [2, 2, 2]
tf.reduce_sum(x, 0).numpy() array([2, 2, 2], dtype=int32)
reduce along the second dimension
the result is [1, 1] + [1, 1] + [1, 1] = [3, 3]
tf.reduce_sum(x, 1).numpy() array([3, 3], dtype=int32)
keep the original dimensions
tf.reduce_sum(x, 1, keepdims=True).numpy() array([[3], [3]], dtype=int32)
reduce along both dimensions
the result is 1 + 1 + 1 + 1 + 1 + 1 = 6
or, equivalently, reduce along rows, then reduce the resultant array
[1, 1, 1] + [1, 1, 1] = [2, 2, 2]
2 + 2 + 2 = 6
tf.reduce_sum(x, [0, 1]).numpy().item() 6 ```
Args
None
(the default), reduces all
dimensions. Must be in the range [-rank(input_tensor),
rank(input_tensor)]
.
Returns
numpy compatibility
Equivalent to np.sum apart the fact that numpy upcast uint8 and int32 to int64 while tensorflow returns the same dtype as the input.
