如何将参数传递给Tensorflow中tf.cond内的函数？

Question

我有以下简单的占位符:

x = tf.placeholder(tf.float32, shape=[1])
y = tf.placeholder(tf.float32, shape=[1])
z = tf.placeholder(tf.float32, shape=[1])

有两个功能fn1,fn2定义如下:

def fn1(a, b): 
    return tf.mul(a, b)
def fn2(a, b): 
    return tf.add(a, b)

现在我想根据pred条件计算结果:

pred = tf.placeholder(tf.bool, shape=[1])
result = tf.cond(pred, fn1(x,y), fn2(y,z))

但它给我一个错误的说法fn1 and fn2 must be callable.

我怎么写fn1,fn2以便他们可以在运行时接收参数？我想打电话给以下人士:

sess.run(result, feed_dict={x:1,y:2,z:3,pred:True})

Answer 1

您可以使用lambda将参数传递给函数,代码如下所示.

x = tf.placeholder(tf.float32)
y = tf.placeholder(tf.float32)
z = tf.placeholder(tf.float32)

def fn1(a, b):
  return tf.mul(a, b)

def fn2(a, b):
  return tf.add(a, b)

pred = tf.placeholder(tf.bool)
result = tf.cond(pred, lambda: fn1(x, y), lambda: fn2(y, z))

然后你可以称之为:

with tf.Session() as sess:
  print sess.run(result, feed_dict={x: 1, y: 2, z: 3, pred: True})
  # The result is 2.0

Answer 2

最简单的方法是在调用中定义函数：

result = tf.cond(pred, lambda: tf.mul(a, b), lambda: tf.add(a, b))