Question

我正在寻找与numpy.random.choice(range(3),replacement=False,size=2,p=[0.1,0.2,0.7])类似的东西在TensorFlow。

与它最接近的Op似乎是tf.multinomial(tf.log(p))，它将记录作为输入，但无法替换而无法进行采样。还有其他方法可以在TensorFlow中对非均匀分布进行采样吗？

感谢。

Answer 1

You could just use tf.py_func to wrap numpy.random.choice and make it available as a TensorFlow op:

a = tf.placeholder(tf.float32)
size = tf.placeholder(tf.int32)
replace = tf.placeholder(tf.bool)
p = tf.placeholder(tf.float32)

y = tf.py_func(np.random.choice, [a, size, replace, p], tf.float32)

with tf.Session() as sess:
    print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))

You can specify the numpy seed as usual:

np.random.seed(1)
print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))
print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))
print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))
np.random.seed(1)
print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))
print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))
print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))
np.random.seed(1)
print(sess.run(y, {a: range(3), size: 2, replace:False, p:[0.1,0.2,0.7]}))

would print:

[ 2.  0.]
[ 2.  1.]
[ 0.  1.]
[ 2.  0.]
[ 2.  1.]
[ 0.  1.]
[ 2.  0.]

Answer 2

是的，有。有关一些背景信息，请参见here和here。解决方案是：

z = -tf.log(-tf.log(tf.random_uniform(tf.shape(p),0,1))) 
_, indices = tf.nn.top_k(tf.log(p) + z, size)

在TensorFlow中无需替换给定的非均匀分布即可进行抽样

2 个答案: