python等價于Matlab展平

我正在嘗試將此代碼從 matlab 轉換為 python：

MATLAB

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

我的嘗試：

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

不幸的是，扁平化程序似乎以一種不兼容的方式結束：Matlab 以：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=結尾：

    [[  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對塊展平做了不同的選擇，有沒有人看到我怎樣才能用python達到matlab的相同結果？

uj5u.com熱心網友回復：

最可能的問題是陣列排序。Numpy 默認使用行優先（C 風格）排序，而 MATLAB 僅支持列優先（Fortran 風格）。所以這應該有效：

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

然而，numpy 將很多東西直接附加到陣列，所以這個語法更清晰：

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

而扁平化陣列的長度就是它的.size，所以可以進一步簡化為：

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

標籤：Python 数组 MATLAB 矩阵 压平

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