python - 如何反转 python 中 numpy float 列表的列表中的每个元素？

Question

这个问题已经有答案了:

奥 git _a (3 个回答)

奥 git _a (5 个回答)

已关闭 4 年前。

我有一个列表的列表，我想在其中对每个元素 x 执行 1/x 。

我试过了

[1/x for x in list]

我该如何继续？我尝试以乘法为例

[x*10 for x in list]

但这只是将行数增加了 10，并没有乘以实际值。我做错了什么？

数据

[[8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.098671875, 16.09259521484375, 24.296372680664064],
 [8.049411010742187, 15.993977661132812, 24.147247314453125],
 [7.6774951171875, 15.250462036132813, 23.08390319824219],
 [7.450130615234375, 14.684713745117188, 22.35098876953125],
 [6.54051025390625, 12.876199340820312, 19.765568237304688],
 [5.39010986328125, 10.824963989257812, 16.51649169921875],
 [4.979031677246094, 9.665416870117188, 14.883976440429688],
 [4.6164859008789065, 8.65063751220703, 13.361966552734373],
 [3.9221954345703125, 7.188418579101563, 11.591380615234375],
 [3.4298834228515624, 6.534327087402344, 10.445729370117188],
 [2.9979116821289065, 5.92670654296875, 9.218810424804687],
 [2.7159127807617187, 5.452456359863281, 8.101337585449219],
 [2.216434326171875, 4.654409942626954, 6.799687652587892],
 [2.111146240234375, 4.231194915771484, 6.159953155517577],
 [1.6629426574707031, 3.4381407165527342, 5.140887298583984],
 [1.2951142883300781, 2.8314198303222655, 4.534166412353516],
 [1.1690933990478516, 2.3221934509277347, 3.6382754516601565],
 [1.1690933990478516, 2.1871324157714844, 3.230146026611328],
 [0.8830496215820313, 1.901088638305664, 2.645755157470703],
 [0.7231619262695312, 1.4308435058593751, 2.175510025024414],
 [0.2225925636291504, 0.7668316841125488, 1.5114982032775877],
 [0.2225925636291504, 0.4066648483276367, 0.9712825393676758],
 [0.2225925636291504, 0.2225925636291504, 0.2225925636291504],
 [0.2225925636291504, 0.2225925636291504, 0.2225925636291504],
 [0.0, 0.0, 0.0]]

Answer 1

由于这个问题被标记为numpy，我想我建议使用一个专门用于计算1/x的numpy函数:numpy.reciprocal

>>> a
array([[1., 2.],
       [3., 4.]])

>>> np.reciprocal(a)
array([[1.        , 0.5       ],
       [0.33333333, 0.25      ]])

非常简单，如文档中所述:

Calculates 1/x

但您还应该注意:

Note: This function is not designed to work with integers.

For integer arguments with absolute value larger than 1 the result is always zero because of the way Python handles integer division. For integer zero the result is an overflow.