python 等价于 Matlab 展平

Question

我正在尝试将此代码从 matlab 转换为 python:

Matlab

 lP =lP(:)/length(lP(:));
  lM=lM(:)/length(lP(:));

我的尝试：

lP=ravel(lP) / length(ravel(lP))
lM=ravel(lM) / length(ravel(lP))

不幸的是，扁平化过程似乎以不兼容的方式结束： Matlab 结束于： lP =

                     0
                     0
      282.158675652645
      245.206579787908
      180.857334994335
      145.295985608783
      98.6111323759356
      78.6541285064253
                     0
                     0
      235.988440426672
      252.705377114354
      149.700677619887
      149.441877249856
      94.6327116403152
      84.5507642298126
       201.33034579544
        168.5504445176
      112.020019512079
      100.029337003081
      81.9696997469859
      77.2791884039185
      53.8857894853351
      47.1052112310254
      201.005392514542
      170.334647413678
      102.794197425736
      117.624998880039
      63.5474961850914
      85.6855606877716
      40.2032122529948
      55.9023649432995
      113.662165613052
      105.200545764806
       66.249378944203
      60.9266562306452
      46.1411501598512
      44.4884016760164
      31.2403038422145
      31.8768645697115
      96.6943687173171
      94.3667944720464
      63.9090790984522
      63.4837480098802
       43.801313157197
      47.5765809365761
      27.4423759004399
      31.6859217744796
      74.7724121990454
       66.780700287442
      35.6152335123858
      37.2132749966295
      27.2233609912034
      28.2363756837094
      18.7629642336031
       19.217036195859
      55.1857268319806
      49.2057110093965
      35.3827383680906
      33.9961168439708
      26.0418645945658
      26.2135571149649
      17.8586181284183
      18.6567862728906

而 python 以 lP=:

结尾

    [[  0.        ],
   [  0.        ],
   [282.15867565],
   [245.20657979],
   [ 35.61523351],
   [ 37.213275  ],
   [ 74.7724122 ],
   [ 66.78070029],
   [  0.        ],
   [  0.        ],
   [235.98844043],
   [252.70537711],
   [ 35.38273837],
   [ 33.99611684],
   [ 55.18572683],
   [ 49.20571101],
   [201.3303458 ],
   [168.55044452],
   [112.02001951],
   [100.029337  ],
   [ 66.24937894],
   [ 60.92665623],
   [113.66216561],
   [105.20054576],
   [201.00539251],
   [170.33464741],
   [102.79419743],
   [117.62499888],
   [ 63.9090791 ],
   [ 63.48374801],
   [ 96.69436872],
   [ 94.36679447],
   [ 53.88578949],
   [ 47.10521123],
   [ 81.96969975],
   [ 77.2791884 ],
   [ 46.14115016],
   [ 44.48840168],
   [ 31.24030384],
   [ 31.87686457],
   [ 40.20321225],
   [ 55.90236494],
   [ 63.54749619],
   [ 85.68556069],
   [ 43.80131316],
   [ 47.57658094],
   [ 27.4423759 ],
   [ 31.68592177],
   [ 98.61113238],
   [ 78.65412851],
   [180.85733499],
   [145.29598561],
   [ 27.22336099],
   [ 28.23637568],
   [ 18.76296423],
   [ 19.2170362 ],
   [ 94.63271164],
   [ 84.55076423],
   [149.70067762],
   [149.44187725],
   [ 26.04186459],
   [ 26.21355711],
   [ 17.85861813],
   [ 18.65678627]]

似乎 python 对块展平做了不同的选择，有谁知道我怎样才能用 python 达到与 matlab 相同的结果？

Answer 1

最有可能的问题是数组排序。 Numpy 默认使用行优先（C 风格）排序，而 MATLAB 仅支持列优先（Fortran 风格）。所以这应该有效：

lP=ravel(lP, 'F') / length(ravel(lP))
lM=ravel(lM, 'F') / length(ravel(lP))

然而，numpy 将很多东西直接附加到数组，所以这个语法更清晰：

lP = lP.ravel('F') / length(lP.ravel())
lM = lM.ravel('F') / length(lP.ravel())

而展平数组的长度正好是.size，所以可以进一步简化为：

lP = lP.ravel('F') / lP.size
lM = lM.ravel('F') / lP.size

python 等价于 Matlab 展平

python equivalence to Matlab flattening

python

arrays

matlab

matrix

flatten