Python 3.x 对列表中的子列表进行排序，然后将其另存为子列表 b=np.数组（[1，-20.559，-5.4188，-4.0916]，[1，-22，-5.9432，-4.0916]，[1，-20.559，-5.4188， -4.0916],[1, -20.559, -5.4188, -4.0916],[2, -20.559, -5.4188, -4.0916],[1, -20.559, -5.4188, -4.0916],[3, -20.559, -5.4188, -4.0916], [2, -20.559, -5.4188, -4.0916],[1, -20.559, -5.4188, -4.0916],[5, -20.559, -5.4188, -4.0916],[1, -20.559, -5.4188, -4.0916],[3, -20.559, -5.4188, -4.0916],[7, -20.559, -5.4188, -4.0916],[1, -20.559, -5.4188, -4.0916],[7, -20.559, -5.4188, -4.0916], [5, -20.559, -5.4188, -4.0916],[2, -20.559, -5.4188, -4.0916],]) list1=b.tolist（） len（列表1） myList=[] mynew=[] 对于范围（8）内的j： 对于范围内的i（len（列表1））： 如果列表1[i][0]==j： myList.append（列表1[i][0:4]） mynew.append（myList） 打印（myList）

我的输出应该如下所示：

[[[1.0, -20.559, -5.4188, -4.0916], [1.0, -22.0, -5.9432, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916]], [[2.0, -20.559, -5.4188, -4.0916], [2.0, -20.559, -5.4188, -4.0916], [2.0, -20.559, -5.4188, -4.0916]], [[3.0, -20.559, -5.4188, -4.0916], [3.0, -20.559, -5.4188, -4.0916]], [[5.0, -20.559, -5.4188, -4.0916], [5.0, -20.559, -5.4188, -4.0916]], [[7.0, -20.559, -5.4188, -4.0916], [7.0, -20.559, -5.4188, -4.0916]]]

[[1.0, -20.559, -5.4188, -4.0916], [1.0, -22.0, -5.9432, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [2.0, -20.559, -5.4188, -4.0916], [2.0, -20.559, -5.4188, -4.0916], [2.0, -20.559, -5.4188, -4.0916], [3.0, -20.559, -5.4188, -4.0916], [3.0, -20.559, -5.4188, -4.0916], [5.0, -20.559, -5.4188, -4.0916], [5.0, -20.559, -5.4188, -4.0916], [7.0, -20.559, -5.4188, -4.0916], [7.0, -20.559, -5.4188, -4.0916]]

我得到的是：

[[[1.0, -20.559, -5.4188, -4.0916], [1.0, -22.0, -5.9432, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916]], [[2.0, -20.559, -5.4188, -4.0916], [2.0, -20.559, -5.4188, -4.0916], [2.0, -20.559, -5.4188, -4.0916]], [[3.0, -20.559, -5.4188, -4.0916], [3.0, -20.559, -5.4188, -4.0916]], [[5.0, -20.559, -5.4188, -4.0916], [5.0, -20.559, -5.4188, -4.0916]], [[7.0, -20.559, -5.4188, -4.0916], [7.0, -20.559, -5.4188, -4.0916]]]

[[1.0, -20.559, -5.4188, -4.0916], [1.0, -22.0, -5.9432, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [1.0, -20.559, -5.4188, -4.0916], [2.0, -20.559, -5.4188, -4.0916], [2.0, -20.559, -5.4188, -4.0916], [2.0, -20.559, -5.4188, -4.0916], [3.0, -20.559, -5.4188, -4.0916], [3.0, -20.559, -5.4188, -4.0916], [5.0, -20.559, -5.4188, -4.0916], [5.0, -20.559, -5.4188, -4.0916], [7.0, -20.559, -5.4188, -4.0916], [7.0, -20.559, -5.4188, -4.0916]]

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任何帮助都将不胜感激。

您需要一个临时

jList

来存储第一个元素相同的所有列表

范围（8）内j的

：
jList=[]
对于范围内的i（len（列表1））：
如果列表1[i][0]==j：
jList.append（列表1[i][：]）
如果len（jList）：
myList.append（jList）
从pprint导入pprint
pprint（myList）
"""
[[[1.0, -20.559, -5.4188, -4.0916],
[1.0, -22.0, -5.9432, -4.0916],
[1.0, -20.559, -5.4188, -4.0916],
[1.0, -20.559, -5.4188, -4.0916],
[1.0, -20.559, -5.4188, -4.0916],
[1.0, -20.559, -5.4188, -4.0916],
[1.0, -20.559, -5.4188, -4.0916],
[1.0, -20.559, -5.4188, -4.0916]],
[[2.0, -20.559, -5.4188, -4.0916],
[2.0, -20.559, -5.4188, -4.0916],
[2.0, -20.559, -5.4188, -4.0916]],
[[3.0, -20.559, -5.4188, -4.0916], [3.0, -20.559, -5.4188, -4.0916]],
[[5.0, -20.559, -5.4188, -4.0916], [5.0, -20.559, -5.4188, -4.0916]],
[[7.0, -20.559, -5.4188, -4.0916], [7.0, -20.559, -5.4188, -4.0916]]]
"""

但是你也可以用熊猫来做

df=pd.DataFrame（b）
结果=[]
对于df.groupby（0）中的（方法，组）：
result.append（group.values.tolist（））
pprint（结果）

尽管将所有相同的第一个元素分组到一个列表中，您也可以通过在

groupby

df.sort_值（[0,1,2,3]，轴=1，升序=[True，True，True，True]，inplace=True）

您好，将列表分成相等的部分有帮助吗？没有，因为我不知道尺寸。我将在一个大列表中使用该代码，其中我不确定第一列中的值的范围。虽然我正在使用范围（8），但我会将其更改为一些。谢谢