用一个张量的值作为另一个张量的形状？

Question

我的代码中有以下两行：

numSequences = tf.placeholder(tf.float32, shape=())
...
prediction = tf.reshape(predictionFlat, [numSequences, sequenceLength, vocabSize])

在定义prediction张量的形状时，是否可以从numSequences张量中提取标量值作为值使用？在

编辑

下面是我的更多代码：

^{pr2}$

编辑2

我正在尝试做类似的事情，我需要我的sequenceLength变量（张量形状的参数）是占位符而不是固定值。我的实现方法与对numSequences相同，但是我得到了如下所示的错误。我不明白这与我最初询问的前一个numSequences实现有何不同。在

代码：

numSequences = tf.placeholder(tf.int32, shape=())
seqLength = tf.placeholder(tf.int32, shape=())
x = tf.placeholder(tf.float32, [None, seqLength, vocabSize])
y = tf.placeholder(tf.float32, [None, seqLength, vocabSize])
xFlat = tf.contrib.layers.flatten(x)                                                # [batchSize, sequenceLength*vocabSize]

W = tf.Variable(tf.random_normal([hiddenDimension, seqLength, vocabSize]))
b = tf.Variable(tf.random_normal([1, seqLength, vocabSize]))
WFlat = tf.contrib.layers.flatten(W)                                                # [hiddenDimension, sequenceLength*vocabSize]
bFlat = tf.contrib.layers.flatten(b)                                                # [1, sequenceLength*vocabSize]

cell = rnn.BasicLSTMCell(hiddenDimension, forget_bias=forgetRate)
outputs, states = tf.nn.static_rnn(cell, [xFlat], dtype=tf.float32)                 # outputs    = [[batchSize, hiddenDimension]]
predictionFlat = tf.add(tf.matmul(outputs[0], WFlat), bFlat)                        # outputs[0] = [batchSize, hiddenDimension]
prediction = tf.reshape(predictionFlat, [numSequences, seqLength, vocabSize])

错误：

    x = tf.placeholder(tf.float32, [None, seqLength, vocabSize])
  File "/usr/lib/python3.6/site-packages/tensorflow/python/ops/array_ops.py", line 1530, in placeholder
    return gen_array_ops._placeholder(dtype=dtype, shape=shape, name=name)
  File "/usr/lib/python3.6/site-packages/tensorflow/python/ops/gen_array_ops.py", line 1954, in _placeholder
    name=name)
  File "/usr/lib/python3.6/site-packages/tensorflow/python/framework/op_def_library.py", line 705, in apply_op
    attr_value.shape.CopyFrom(_MakeShape(value, key))
  File "/usr/lib/python3.6/site-packages/tensorflow/python/framework/op_def_library.py", line 198, in _MakeShape
    return tensor_shape.as_shape(v).as_proto()
  File "/usr/lib/python3.6/site-packages/tensorflow/python/framework/tensor_shape.py", line 798, in as_shape
    return TensorShape(shape)
  File "/usr/lib/python3.6/site-packages/tensorflow/python/framework/tensor_shape.py", line 434, in __init__
    self._dims = [as_dimension(d) for d in dims_iter]
  File "/usr/lib/python3.6/site-packages/tensorflow/python/framework/tensor_shape.py", line 434, in <listcomp>
    self._dims = [as_dimension(d) for d in dims_iter]
  File "/usr/lib/python3.6/site-packages/tensorflow/python/framework/tensor_shape.py", line 376, in as_dimension
    return Dimension(value)
  File "/usr/lib/python3.6/site-packages/tensorflow/python/framework/tensor_shape.py", line 32, in __init__
    self._value = int(value)
TypeError: int() argument must be a string, a bytes-like object or a number, not 'Tensor'

Answer 1

是的，张量形状通常可以是张量本身，但是它们必须是整数类型。在

import tensorflow as tf

x = tf.constant([2, 3], dtype=tf.int32)
y = tf.zeros((x[0], x[1], 5))

sess = tf.InteractiveSession()
print(y.eval().shape)
# (2, 3, 5)

编辑

更接近你的例子

^{pr2}$

当然，整形应该保留y中元素的总数，因此它的用处有限。在

编辑2

不能用另一个placehoder参数化一个占位符的形状。因为同时提供了占位符，所以没有意义。提供一个形状未知的占位符，然后像第一个示例中那样提供参数化的重塑操作。在