tensorflow Tf.cond提供意外输出

Question

我似乎对tf.cond的工作方式产生了误解。在tensorflow 文档中，给出了以下示例：

z = tf.multiply(a, b)
result = tf.cond(x < y, lambda: tf.add(x, z), lambda: tf.square(y))

示例的结果是，如果x<y是True，则为tf.add(x,z) tf.square(y) tf.square(y)。

按照这个示例，我尝试用tf.cond构建一个小示例，其结果与文档中提到的不同。

在我的例子中，deterministic_action = 4，random_action = 11，chose_random=False。stochastic_action应该是4，而不是1。1的价值从何而来？

#!/usr/bin/env python3

import tensorflow as tf
import numpy as np

with tf.Graph().as_default():
    with tf.device('/cpu:0'):
        stochastic_ph = tf.placeholder(tf.bool, (), name="stochastic")
        eps = tf.get_variable("eps", (), initializer=tf.constant_initializer(0))
        with tf.variable_scope('test_cond') as sc:
            deterministic_action = tf.random_uniform([], minval=0, maxval=15, dtype=tf.int64, seed=0) # 4
            random_action = tf.random_uniform([], minval=0, maxval=15, dtype=tf.int64, seed=1) # 11
            chose_random = tf.random_uniform([], minval=0, maxval=1, dtype=tf.float32) < eps # False because eps = 0
            stochastic_action = tf.cond(chose_random, lambda: random_action, lambda: deterministic_action) # S_action should be 4 but it is 1
            #output_action = tf.cond(stochastic_ph, lambda: stochastic_action, lambda: deterministic_action)


    init = tf.global_variables_initializer()
    sess = tf.Session()
    sess.run(init, feed_dict={stochastic_ph: True})
    print ("s_ph = ", stochastic_ph)
    d_action = sess.run(deterministic_action)
    print ("det_action= ", d_action)
    r_action = sess.run(random_action)
    print ("rand_action= ", r_action)
    e = sess.run(eps)
    c_action = sess.run(chose_random)
    print ("chose_rand= ", c_action)
    s_action = sess.run(stochastic_action)
    print ("s_action= ", s_action)
    #output = sess.run(output_action)

这是输出：

python random_vec.py
2018-10-31 09:46:15.028376: I tensorflow/core/platform/cpu_feature_guard.cc:140] Your CPU supports instructions that this TensorFlow binary was not compiled to use: AVX2 FMA
s_ph =  Tensor("stochastic:0", shape=(), dtype=bool, device=/device:CPU:0)
det_action=  4
rand_action=  11
chose_rand=  False
s_action=  1

Answer 1

这是因为您再次在新的sess.run中进行评估。由于您正在为deterministic_action生成一个随机数，结果是4之后的下一个随机数，即1。这是您的代码的结果，在最后一步提取deterministic_action的值时。

修改：

print ("s_ph = ", stochastic_ph)
d_action = sess.run(deterministic_action)
print ("det_action= ", d_action)
r_action = sess.run(random_action)
print ("rand_action= ", r_action)
e = sess.run(eps)
c_action = sess.run(chose_random)
print ("chose_rand= ", c_action)
s_action, d_action = sess.run([stochastic_action, deterministic_action])
print ("s_action= ", s_action)
print ("det_action= ", d_action)

结果：

s_ph =  Tensor("stochastic:0", shape=(), dtype=bool, device=/device:CPU:0)
det_action=  4
rand_action=  11
chose_rand=  False
s_action=  1
det_action=  1

现在您所需要做的就是在一个sess.run中运行所有的东西。

d_action, r_action, e,  c_action, s_action = sess.run([deterministic_action, random_action, eps, chose_random, stochastic_action])
print ("det_action= ", d_action)
print ("rand_action= ", r_action)
print ("chose_rand= ", c_action)
print ("s_action= ", s_action)

结果：

s_ph =  Tensor("stochastic:0", shape=(), dtype=bool, device=/device:CPU:0)
det_action=  4
rand_action=  11
chose_rand=  False
s_action=  4

更新：

当设置种子时，我不清楚为什么random_uniform会生成不同的值。这是因为代码使用它初始化变量的同一个会话对象运行。使用新的session对象修改代码，会发生这样的情况：

print ("s_ph = ", stochastic_ph)
d_action = sess.run(deterministic_action)
print ("det_action= ", d_action)
sess.close()
sess = tf.Session()
sess.run(init, feed_dict={stochastic_ph: True})
s_action = sess.run(stochastic_action)
print ("s_action= ", s_action)

结果：

s_ph =  Tensor("stochastic:0", shape=(), dtype=bool, device=/device:CPU:0)
det_action=  4
s_action=  4