小编典典

numpy切片数组而不复制它

python

我的矩阵中有大量数据x,我需要分析一些子矩阵。

我正在使用以下代码来选择子矩阵:

>>> import numpy as np
>>> x = np.random.normal(0,1,(20,2))
>>> x
array([[-1.03266826,  0.04646684],
       [ 0.05898304,  0.31834926],
       [-0.1916809 , -0.97929025],
       [-0.48837085, -0.62295003],
       [-0.50731017,  0.50305894],
       [ 0.06457385, -0.10670002],
       [-0.72573604,  1.10026385],
       [-0.90893845,  0.99827162],
       [ 0.20714399, -0.56965615],
       [ 0.8041371 ,  0.21910274],
       [-0.65882317,  0.2657183 ],
       [-1.1214074 , -0.39886425],
       [ 0.0784783 , -0.21630006],
       [-0.91802557, -0.20178683],
       [ 0.88268539, -0.66470235],
       [-0.03652459,  1.49798484],
       [ 1.76329838, -0.26554555],
       [-0.97546845, -2.41823586],
       [ 0.32335103, -1.35091711],
       [-0.12981597,  0.27591674]])
>>> index = x[:,1] > 0
>>> index
array([ True,  True, False, False,  True, False,  True,  True, False,
        True,  True, False, False, False, False,  True, False, False,
       False,  True], dtype=bool)
>>> x1 = x[index, :] #x1 is a copy of the submatrix
>>> x1
array([[-1.03266826,  0.04646684],
       [ 0.05898304,  0.31834926],
       [-0.50731017,  0.50305894],
       [-0.72573604,  1.10026385],
       [-0.90893845,  0.99827162],
       [ 0.8041371 ,  0.21910274],
       [-0.65882317,  0.2657183 ],
       [-0.03652459,  1.49798484],
       [-0.12981597,  0.27591674]])
>>> x1[0,0] = 1000
>>> x1
array([[  1.00000000e+03,   4.64668400e-02],
       [  5.89830401e-02,   3.18349259e-01],
       [ -5.07310170e-01,   5.03058935e-01],
       [ -7.25736045e-01,   1.10026385e+00],
       [ -9.08938455e-01,   9.98271624e-01],
       [  8.04137104e-01,   2.19102741e-01],
       [ -6.58823174e-01,   2.65718300e-01],
       [ -3.65245877e-02,   1.49798484e+00],
       [ -1.29815968e-01,   2.75916735e-01]])
>>> x
array([[-1.03266826,  0.04646684],
       [ 0.05898304,  0.31834926],
       [-0.1916809 , -0.97929025],
       [-0.48837085, -0.62295003],
       [-0.50731017,  0.50305894],
       [ 0.06457385, -0.10670002],
       [-0.72573604,  1.10026385],
       [-0.90893845,  0.99827162],
       [ 0.20714399, -0.56965615],
       [ 0.8041371 ,  0.21910274],
       [-0.65882317,  0.2657183 ],
       [-1.1214074 , -0.39886425],
       [ 0.0784783 , -0.21630006],
       [-0.91802557, -0.20178683],
       [ 0.88268539, -0.66470235],
       [-0.03652459,  1.49798484],
       [ 1.76329838, -0.26554555],
       [-0.97546845, -2.41823586],
       [ 0.32335103, -1.35091711],
       [-0.12981597,  0.27591674]])
>>>

但我希望x1只是一个指针或类似的东西。每当需要子矩阵时,复制数据对我来说都是太昂贵了。我怎样才能做到这一点?

编辑:显然没有与numpy数组的任何解决方案。从这个角度来看,熊猫数据框会更好吗?


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2021-01-20

共1个答案

小编典典

属性中x汇总了有关阵列的信息.__array_interface__

In [433]: x.__array_interface__
Out[433]: 
{'descr': [('', '<f8')],
 'strides': None,
 'data': (171396104, False),
 'typestr': '<f8',
 'version': 3,
 'shape': (20, 2)}

它具有数组shapestrides(这里是默认值)和指向数据缓冲区的指针。Aview可以指向相同的数据缓冲区(可能更远),并具有自己的shapestrides

但是用布尔值建立索引不能用这几个数字来概括。它要么必须一直携带整个index阵列,要么从x数据缓冲区中复制选定的项目。
numpy选择复制。您可以选择何时应用index,立即应用还是在调用堆栈的更下游应用。

2021-01-20