不，但是您可以使用tf.multinomial获得相同的结果：

elems = tf.convert_to_tensor([1,2,3,5])
samples = tf.multinomial(tf.log([[1, 0, 0.3, 0.6]]), 1) # note log-prob
elems[tf.cast(samples[0][0], tf.int32)].eval()
Out: 1
elems[tf.cast(samples[0][0], tf.int32)].eval()
Out: 5

[0][0]部分在这里，因为multinomial期望该批次中每个元素的行都具有未归一化的对数概率，并且对于样本数量还具有另一个维度。

如果要从n维张量中随机采样行，而不是从一维张量中采样随机元素，则可以组合tf.multinomial和tf.gather 。

def _random_choice(inputs, n_samples):
    """
    With replacement.
    Params:
      inputs (Tensor): Shape [n_states, n_features]
      n_samples (int): The number of random samples to take.
    Returns:
      sampled_inputs (Tensor): Shape [n_samples, n_features]
    """
    # (1, n_states) since multinomial requires 2D logits.
    uniform_log_prob = tf.expand_dims(tf.zeros(tf.shape(inputs)[0]), 0)

    ind = tf.multinomial(uniform_log_prob, n_samples)
    ind = tf.squeeze(ind, 0, name="random_choice_ind")  # (n_samples,)

    return tf.gather(inputs, ind, name="random_choice")