Python Tensorflow 重塑张量
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Tensorflow reshape tensor
提问by Kendall Weihe
I have a prediction tensor (the actual network)
我有一个预测张量（实际网络）
(Pdb) pred
<tf.Tensor 'transpose_1:0' shape=(?, 200, 200) dtype=float32>
and a y tensor
和张量
y = tf.placeholder("float", [None, n_steps, n_classes])
(Pdb) y
<tf.Tensor 'Placeholder_1:0' shape=(?, 200, 200) dtype=float32>
I want to feed it into
我想喂它
f.reduce_mean(tf.nn.softmax_cross_entropy_with_logits(pred, y))
f.reduce_mean(tf.nn.softmax_cross_entropy_with_logits(pred, y))
However, it requires the dimensions to be [batch_size, num_classes]
但是，它要求尺寸为 [batch_size, num_classes]
So I want to reshape both predand yso that they look like this
所以我想重塑两者pred，y让它们看起来像这样
<tf.Tensor 'transpose_1:0' shape=(?, 40000) dtype=float32>
But when I run reshapeI get
但是当我跑步时reshape我得到
(Pdb) tf.reshape(pred, [40000])
<tf.Tensor 'Reshape_1:0' shape=(40000,) dtype=float32>
instead of (?,40000)how can I maintain that Nonedimension? (the batch size dimension)
而不是(?,40000)我如何保持那个None维度？（批量大小维度）
I've also posted all of the relevant code...
我还发布了所有相关代码...
# tf Graph input
x = tf.placeholder("float", [None, n_steps, n_input])
y = tf.placeholder("float", [None, n_steps, n_classes])
# Define weights
weights = {
'hidden': tf.Variable(tf.random_normal([n_hidden, n_classes]), dtype="float32"),
'out': tf.Variable(tf.random_normal([n_hidden, n_classes]), dtype="float32")
}
biases = {
'hidden': tf.Variable(tf.random_normal([n_hidden]), dtype="float32"),
'out': tf.Variable(tf.random_normal([n_classes]), dtype="float32")
}
def RNN(x, weights, biases):
# Prepare data shape to match `rnn` function requirements
# Current data input shape: (batch_size, n_steps, n_input)
# Permuting batch_size and n_steps
x = tf.transpose(x, [1, 0, 2])
# Reshaping to (n_steps*batch_size, n_input)
x = tf.reshape(x, [-1, n_input])
# Split to get a list of 'n_steps' tensors of shape (batch_size, n_hidden)
# This input shape is required by `rnn` function
x = tf.split(0, n_steps, x)
# Define a lstm cell with tensorflow
lstm_cell = rnn_cell.BasicLSTMCell(n_hidden, forget_bias=1.0, state_is_tuple=True)
outputs, states = rnn.rnn(lstm_cell, x, dtype=tf.float32)
output_matrix = []
for i in xrange(n_steps):
temp = tf.matmul(outputs[i], weights['out']) + biases['out']
# temp = tf.matmul(weights['hidden'], outputs[i]) + biases['hidden']
output_matrix.append(temp)
pdb.set_trace()
return output_matrix
pred = RNN(x, weights, biases)
# temp = RNN(x)
# pdb.set_trace()
# pred = tf.shape(temp)
pred = tf.pack(tf.transpose(pred, [1,0,2]))
cost = tf.reduce_mean(tf.nn.softmax_cross_entropy_with_logits(pred, y))
回答by weitang114
I'm the author of one of the answers of the other question in Yaroslav's comment. You can use -1 for the None dimension.
我是 Yaroslav 评论中另一个问题的答案之一的作者。您可以将 -1 用于 None 维度。
You can do it easily with tf.reshape() without knowing the batch size.
您可以使用 tf.reshape() 轻松完成，而无需知道批量大小。
x = tf.placeholder(tf.float32, shape=[None, 9,2])
shape = x.get_shape().as_list() # a list: [None, 9, 2]
dim = numpy.prod(shape[1:]) # dim = prod(9,2) = 18
x2 = tf.reshape(x, [-1, dim]) # -1 means "all"
The -1 in the last line means the whole column no matter what the batchsize is in the runtime. You can see it in tf.reshape().
最后一行中的 -1 表示整列，无论运行时的批次大小如何。你可以在 tf.reshape() 中看到它。
