我就废话不多说,咱直接看代码吧! tf.transpose transpose( a, perm=None, name='transpose') Defined in tensorflow/python/ops/array_ops.py. See the guides: Math Matrix Math Functions, Tensor Transformations Slicing and Joining T
我就废话不多说,咱直接看代码吧!
tf.transpose
transpose( a, perm=None, name='transpose' )
Defined in tensorflow/python/ops/array_ops.py.
See the guides: Math > Matrix Math Functions, Tensor Transformations > Slicing and Joining
Transposes a. Permutes the dimensions according to perm.
The returned tensor's dimension i will correspond to the input dimension perm[i]. If perm is not given, it is set to (n-1…0), where n is the rank of the input tensor. Hence by default, this operation performs a regular matrix transpose on 2-D input Tensors.
For example:
x = tf.constant([[1, 2, 3], [4, 5, 6]]) tf.transpose(x) # [[1, 4] # [2, 5] # [3, 6]]
tf.transpose(x, perm=[1, 0]) # [[1, 4] # [2, 5] # [3, 6]]
# 'perm' is more useful for n-dimensional tensors, for n > 2 x = tf.constant([[[ 1, 2, 3], [ 4, 5, 6]], [[ 7, 8, 9], [10, 11, 12]]]) # Take the transpose of the matrices in dimension-0 tf.transpose(x, perm=[0, 2, 1]) # [[[1, 4], # [2, 5], # [3, 6]], # [[7, 10], # [8, 11], # [9, 12]]]
a的转置是根据 perm 的设定值来进行的。
返回数组的 dimension(尺寸、维度) i与输入的 perm[i]的维度相一致。如果未给定perm,默认设置为 (n-1…0),这里的 n 值是输入变量的 rank 。因此默认情况下,这个操作执行了一个正规(regular)的2维矩形的转置
例如:
x = [[1 2 3] [4 5 6]] tf.transpose(x) ==> [[1 4] [2 5] [3 6]] tf.transpose(x) 等价于: tf.transpose(x perm=[1, 0]) ==> [[1 4] [2 5] [3 6]]
a=tf.constant([[[1,2,3],[4,5,6]],[[7,8,9],[10,11,12]]]) array([[[ 1, 2, 3], [ 4, 5, 6]], [[ 7, 8, 9], [10, 11, 12]]]) x=tf.transpose(a,[1,0,2]) array([[[ 1, 2, 3], [ 7, 8, 9]], [[ 4, 5, 6], [10, 11, 12]]]) x=tf.transpose(a,[0,2,1]) array([[[ 1, 4], [ 2, 5], [ 3, 6]], [[ 7, 10], [ 8, 11], [ 9, 12]]]) x=tf.transpose(a,[2,1,0]) array([[[ 1, 7], [ 4, 10]], [[ 2, 8], [ 5, 11]], [[ 3, 9], [ 6, 12]]]) array([[[ 1, 7], [ 4, 10]], [[ 2, 8], [ 5, 11]], [[ 3, 9], [ 6, 12]]]) x=tf.transpose(a,[1,2,0]) array([[[ 1, 7], [ 2, 8], [ 3, 9]], [[ 4, 10], [ 5, 11], [ 6, 12]]])
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