TensorFlow的reshape操作 tf.reshape的实现
初学tensorflow,如果写的不对的,请更正,谢谢!
tf.reshape(tensor, shape, name=None)
函数的作用是将tensor变换为参数shape的形式。
其中shape为一个列表形式,特殊的一点是列表中可以存在-1。-1代表的含义是不用我们自己指定这一维的大小,函数会自动计算,但列表中只能存在一个-1。(当然如果存在多个-1,就是一个存在多解的方程了)
好了我想说的重点还有一个就是根据shape如何变换矩阵。其实简单的想就是,
reshape(t, shape) => reshape(t, [-1]) => reshape(t, shape)
首先将矩阵t变为一维矩阵,然后再对矩阵的形式更改就可以了。
官方的例子:
# tensor 't' is [1, 2, 3, 4, 5, 6, 7, 8, 9]
# tensor 't' has shape [9]
reshape(t, [3, 3]) ==> [[1, 2, 3],
[4, 5, 6],
[7, 8, 9]]
# tensor 't' is [[[1, 1], [2, 2]],
# [[3, 3], [4, 4]]]
# tensor 't' has shape [2, 2, 2]
reshape(t, [2, 4]) ==> [[1, 1, 2, 2],
[3, 3, 4, 4]]
# tensor 't' is [[[1, 1, 1],
# [2, 2, 2]],
# [[3, 3, 3],
# [4, 4, 4]],
# [[5, 5, 5],
# [6, 6, 6]]]
# tensor 't' has shape [3, 2, 3]
# pass '[-1]' to flatten 't'
reshape(t, [-1]) ==> [1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6]
# -1 can also be used to infer the shape
# -1 is inferred to be 9:
reshape(t, [2, -1]) ==> [[1, 1, 1, 2, 2, 2, 3, 3, 3],
[4, 4, 4, 5, 5, 5, 6, 6, 6]]
# -1 is inferred to be 2:
reshape(t, [-1, 9]) ==> [[1, 1, 1, 2, 2, 2, 3, 3, 3],
[4, 4, 4, 5, 5, 5, 6, 6, 6]]
# -1 is inferred to be 3:
reshape(t, [ 2, -1, 3]) ==> [[[1, 1, 1],
[2, 2, 2],
[3, 3, 3]],
[[4, 4, 4],
[5, 5, 5],
[6, 6, 6]]]
# tensor 't' is [7]
# shape `[]` reshapes to a scalar
reshape(t, []) ==> 7
在举几个例子或许就清楚了,有一个数组z,它的shape属性是(4, 4)
z = np.array([[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16]])
z.shape
(4, 4)
z.reshape(-1)
z.reshape(-1)
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16])
z.reshape(-1, 1)
也就是说,先前我们不知道z的shape属性是多少,但是想让z变成只有一列,行数不知道多少,通过`z.reshape(-1,1)`,Numpy自动计算出有12行,新的数组shape属性为(16, 1),与原来的(4, 4)配套。
z.reshape(-1,1)
array([[ 1],
[ 2],
[ 3],
[ 4],
[ 5],
[ 6],
[ 7],
[ 8],
[ 9],
[10],
[11],
[12],
[13],
[14],
[15],
[16]])
z.reshape(-1, 2)
newshape等于-1,列数等于2,行数未知,reshape后的shape等于(8, 2)
z.reshape(-1, 2)
array([[ 1, 2],
[ 3, 4],
[ 5, 6],
[ 7, 8],
[ 9, 10],
[11, 12],
[13, 14],
[15, 16]])
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