Numpy重塑1d到2d阵列,有1列

Question

在numpy结果数组的维度中,运行时会有所不同.1d数组和具有1列的2d数组之间经常存在混淆.在一种情况下,我可以遍历列,在另一种情况下,我不能.

你如何优雅地解决这个问题？为了避免使用if语句检查维度来乱丢我的代码,我使用此函数:

def reshape_to_vect(ar):
    if len(ar.shape) == 1:
      return ar.reshape(ar.shape[0],1)
    return ar

然而,这感觉不优雅且昂贵.有更好的解决方案吗？

Answer 1

最简单的方法:

ar.reshape(-1, 1)

Answer 2

你可以做 -

ar.reshape(ar.shape[0],-1)

第二个输入reshape:-1处理第二轴的元素数量.因此,对于2D输入案例,它没有变化.对于1D输入案例,它创建一个2D数组,其中所有元素都被"推"到第一个轴,因为 ar.shape[0]它是元素的总数.

样品运行

1D案例:

In [87]: ar
Out[87]: array([ 0.80203158,  0.25762844,  0.67039516,  0.31021513,  0.80701097])

In [88]: ar.reshape(ar.shape[0],-1)
Out[88]: 
array([[ 0.80203158],
       [ 0.25762844],
       [ 0.67039516],
       [ 0.31021513],
       [ 0.80701097]])

2D案例:

In [82]: ar
Out[82]: 
array([[ 0.37684126,  0.16973899,  0.82157815,  0.38958523],
       [ 0.39728524,  0.03952238,  0.04153052,  0.82009233],
       [ 0.38748174,  0.51377738,  0.40365096,  0.74823535]])

In [83]: ar.reshape(ar.shape[0],-1)
Out[83]: 
array([[ 0.37684126,  0.16973899,  0.82157815,  0.38958523],
       [ 0.39728524,  0.03952238,  0.04153052,  0.82009233],
       [ 0.38748174,  0.51377738,  0.40365096,  0.74823535]])

Answer 3

为了避免首先需要重新整形，如果您使用列表或“运行”切片对行/列进行切片，您将得到一个具有一行/列的二维数组

import numpy as np
x = np.array(np.random.normal(size=(4,4)))
print x, '\n'

Result:
[[ 0.01360395  1.12130368  0.95429414  0.56827029]
 [-0.66592215  1.04852182  0.20588886  0.37623406]
 [ 0.9440652   0.69157556  0.8252977  -0.53993904]
 [ 0.6437994   0.32704783  0.52523173  0.8320762 ]] 

y = x[:,[0]]
print y, 'col vector \n'
Result:
[[ 0.01360395]
 [-0.66592215]
 [ 0.9440652 ]
 [ 0.6437994 ]] col vector 


y = x[[0],:]
print y, 'row vector \n'

Result:
[[ 0.01360395  1.12130368  0.95429414  0.56827029]] row vector 

# Slice with "running" index on a column
y = x[:,0:1]
print y, '\n'

Result:
[[ 0.01360395]
 [-0.66592215]
 [ 0.9440652 ]
 [ 0.6437994 ]] 

相反，如果您使用单个数字来选择行/列，则会产生一维数组，这是问题的根本原因：

y = x[:,0]
print y, '\n'

Result:
[ 0.01360395 -0.66592215  0.9440652   0.6437994 ]