Question

我在Python中使用np.dstack来处理3D中的堆栈数组，但形状看起来像这样

a = np.sin(np.cumsum((np.random.normal(scale=0.1, size=(len(x), 12))), axis=0))
b = np.dstack((nlead1,nlead2,nlead3,nlead4,nlead5,nlead6,nlead7,nlead8,nlead9,nlead10,nlead11,nlead12))

print a
print a.shape
print b 
print b.shape

输出看起来像

[[ 0.16172117 -0.06113749 -0.05279262 ..., -0.02493445  0.14569041
  -0.20455989]
 [ 0.12982533 -0.09941658  0.06052276 ...,  0.08941575  0.23417498
  -0.15513552]
 [ 0.21850132 -0.16716535  0.048252   ...,  0.00501475  0.16843843
  -0.15638653]
 ..., 
 [ 0.6436674   0.04253489 -0.58825104 ..., -0.95674562  0.36004254
   0.41308711]
 [ 0.6193386  -0.00224555 -0.56129994 ..., -0.90884016  0.31269762
   0.26443869]
 [ 0.63100264 -0.03888419 -0.61395087 ..., -0.92910613  0.42215252
   0.1646122 ]]
(5500, 12)
[[[ -39   47   86 ..., -100  -77   -3]
  [ -34   50   85 ..., -101  -75   -1]
  [ -36   49   86 ..., -103  -77   -4]
  ..., 
  [   0    0    0 ...,    0    0    0]
  [   0    0    0 ...,    0    0    0]
  [   0    0    0 ...,    0    0    0]]]
(1, 5500, 12)

我想让b的形状像a一样，你能给我一些建议非常感谢你。

Answer 1

您可以squeeze数组，删除单项维度：

b = np.squeeze(b)

虽然，只是连接并采用转置效率更高：

print(timeit.timeit(lambda: np.concatenate(arrays,axis=0).T, number=100))
print(timeit.timeit(lambda: np.squeeze(np.dstack(arrays)), number=100))

# 0.024745813333333335
# 0.07166506666666667

如何以2个维度深度顺序堆叠数组

1 个答案: