TensorFlow中矩阵和向量的高效逐元素乘法 [英] Efficient element-wise multiplication of a matrix and a vector in TensorFlow

问题描述

乘以2维张量(矩阵)的最有效方法是什么:

What would be the most efficient way to multiply (element-wise) a 2D tensor (matrix):

x11 x12 .. x1N
...
xM1 xM2 .. xMN

通过垂直向量:

w1
...
wN

获得一个新矩阵:

x11*w1 x12*w2 ... x1N*wN
...
xM1*w1 xM2*w2 ... xMN*wN

为了提供一些背景信息，我们有一个批处理中的M个数据样本可以并行处理，并且每个N元素样本都必须乘以存储在变量中的权重w才能最终选择最大Xij*wj每行i.

To give some context, we have M data samples in a batch that can be processed in parallel, and each N-element sample must be multiplied by weights w stored in a variable to eventually pick the largest Xij*wj for each row i.

推荐答案

执行此操作的最简单代码取决于

The simplest code to do this relies on the broadcasting behavior of tf.multiply()^*, which is based on numpy's broadcasting behavior:

x = tf.constant(5.0, shape=[5, 6])
w = tf.constant([0.0, 1.0, 2.0, 3.0, 4.0, 5.0])
xw = tf.multiply(x, w)
max_in_rows = tf.reduce_max(xw, 1)

sess = tf.Session()
print sess.run(xw)
# ==> [[0.0, 5.0, 10.0, 15.0, 20.0, 25.0],
#      [0.0, 5.0, 10.0, 15.0, 20.0, 25.0],
#      [0.0, 5.0, 10.0, 15.0, 20.0, 25.0],
#      [0.0, 5.0, 10.0, 15.0, 20.0, 25.0],
#      [0.0, 5.0, 10.0, 15.0, 20.0, 25.0]]

print sess.run(max_in_rows)
# ==> [25.0, 25.0, 25.0, 25.0, 25.0]

^* 在旧版本的TensorFlow中，tf.multiply()被称为tf.mul().您还可以使用*运算符(即xw = x * w)执行相同的操作.

^* In older versions of TensorFlow, tf.multiply() was called tf.mul(). You can also use the * operator (i.e. xw = x * w) to perform the same operation.

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