深入浅出TensorFlow2函数——tf.math.reduce_sum

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计算张量各维度上元素的总和。

语法

tf.math.reduce_sum(
    input_tensor, axis=None, keepdims=False, name=None
)

参数

• input_tensor：[Tensor] 待求和的多维Tensor。
• axis：求和运算的维度。如果为None，则计算所有元素的和并返回包含单个元素的Tensor变量，否则必须在 [ − rank ( x ) , rank ( x ) ] [-\text{rank}(x), \text{rank}(x)] [−rank(x),rank(x)]范围内。如果 axis [ i ] < 0 \text{axis}[i]<0 axis[i]<0，则维度将变为 rank + axis [ i ] \text{rank} + \text{axis}[i] rank+axis[i]，默认值为None。
• keepdim：[bool] 是否在输出Tensor中保留减小的维度。如keepdim=True，否则结果张量的维度将比输入张量小，默认值为False。
• name：[可选, str] 操作的名称，默认值为None。

返回值

与

input_tensor

具有相同

dtype

的求和后的

tensor

。

实例

>>> # 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()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()6

函数实现

@tf_export("math.reduce_sum","reduce_sum", v1=[])
@dispatch.add_dispatch_support
def reduce_sum(input_tensor, axis=None, keepdims=False, name=None):"""Computes the sum of elements across dimensions of a tensor.
  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 1for 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()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()6
  Args:
    input_tensor: The tensor to reduce. Should have numeric type.
    axis: The dimensions to reduce. If `None` (the default), reduces all
      dimensions. Must be in the range `[-rank(input_tensor),rank(input_tensor)]`.
    keepdims: If true, retains reduced dimensions with length 1.
    name: A name for the operation(optional).
  Returns:
    The reduced tensor, of the same dtype as the input_tensor.
  @compatibility(numpy)
  Equivalent to np.sum apart the fact that numpy upcast uint8 and int32 to
  int64 while tensorflow returns the same dtype as the input.
  @end_compatibility
  """

  returnreduce_sum_with_dims(input_tensor, axis, keepdims, name,_ReductionDims(input_tensor, axis))