在TensorFlow中无需替换给定的非均匀分布即可进行抽样

Question

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

最接近Op它似乎是tf.multinomial(tf.log(p))将logits作为输入,但它不能在没有替换的情况下进行采样.在TensorFlow中,是否还有其他方法可以从非均匀分布中进行采样？

谢谢.

Answer 1

您只需使用tf.py_func包装numpy.random.choice并将其作为 TensorFlow 操作即可使用：

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]}))

您可以像往常一样指定 numpy 种子：

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]}))

会打印：

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