tensorflow 创建不同长度的掩码

Question

我在 tensorflow 中有一个长度的张量，假设它看起来像这样：

[4, 3, 5, 2]

我希望创建一个 1 和 0 的掩码，其 1 的数量对应于该张量的条目，用 0 填充到总长度 8。即我想创建这个张量：

[[1,1,1,1,0,0,0,0],
 [1,1,1,0,0,0,0,0],
 [1,1,1,1,1,0,0,0],
 [1,1,0,0,0,0,0,0]
]

我该怎么做？

Answer 1

这可以通过多种方式实现 TensorFlow transformations:

# Make a 4 x 8 matrix where each row contains the length repeated 8 times.
lengths = [4, 3, 5, 2]
lengths_transposed = tf.expand_dims(lengths, 1)

# Make a 4 x 8 matrix where each row contains [0, 1, ..., 7]
range = tf.range(0, 8, 1)
range_row = tf.expand_dims(range, 0)

# Use the logical operations to create a mask
mask = tf.less(range_row, lengths_transposed)

# Use the select operation to select between 1 or 0 for each value.
result = tf.select(mask, tf.ones([4, 8]), tf.zeros([4, 8]))

Answer 2

我得到的版本比以前的答案要短一些。不确定它是否更有效

 def mask(self, seq_length, max_seq_length):
    return tf.map_fn(
        lambda x: tf.pad(tf.ones([x], dtype=tf.int32), [[0, max_seq_length - x]]),
        seq_length)

Answer 3

现在可以通过 tf.sequence_mask 实现。更多详情 here.

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