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python - 订购 list

转载 作者:行者123 更新时间:2023-11-28 20:59:16 31 4
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我有一个这样的列表,

M=[[75], [95, 64], [17, 47, 82], [18, 35, 87, 10], [20, 4, 82, 47, 65], [19, 
1, 23, 75, 3, 34], [88, 2, 77, 73, 7, 63, 67], [99, 65, 4, 28, 6, 16, 70,
92], [41, 41, 26, 56, 83, 40, 80, 70, 33], [41, 48, 72, 33, 47, 32, 37, 16,
94, 29], [53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14], [70, 11, 33, 28, 77,
73, 17, 78, 39, 68, 17, 57], [91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27,
29, 48], [63, 66, 4, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31], [4, 62,
98, 27, 23, 9, 70, 98, 73, 93, 38, 53, 60, 4, 23]]

例如,我想按列表中每个元素的“零”位置上的最大数字对其进行排序,

M=[[75], [95, 64], [82, 47, 17], [87, 35, 18, 10].....]

我尝试使用 key ,但效果不佳我也不知道为什么效果不佳..这是 key

def Len(elem):
for i in range(16):
y=len(K[i])
return elem[y-1]
y=sorted(K,key=Len)
print(y)

可能是我没看懂按键功能吧。谢谢

最佳答案

只需使用sorted(或list.sort)而不使用key。他们已经按字典顺序排序。

>>> sorted(M, reverse=True)
[[99, 65, 4, 28, 6, 16, 70, 92], [95, 64], [91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48], [88, 2, 77, 73, 7, 63, 67], [75], [70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57], [63, 66, 4, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31], [53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14], [41, 48, 72, 33, 47, 32, 37, 16, 94, 29], [41, 41, 26, 56, 83, 40, 80, 70, 33], [20, 4, 82, 47, 65], [19, 1, 23, 75, 3, 34], [18, 35, 87, 10], [17, 47, 82], [4, 62, 98, 27, 23, 9, 70, 98, 73, 93, 38, 53, 60, 4, 23]]

编辑:

单独排序列表:

>>> [sorted(sublist, reverse=True) for sublist in M]
[[75], [95, 64], [82, 47, 17], [87, 35, 18, 10], [82, 65, 47, 20, 4], [75, 34, 23, 19, 3, 1], [88, 77, 73, 67, 63, 7, 2], [99, 92, 70, 65, 28, 16, 6, 4], [83, 80, 70, 56, 41, 41, 40, 33, 26], [94, 72, 48, 47, 41, 37, 33, 32, 29, 16], [97, 91, 71, 65, 53, 52, 51, 44, 43, 25, 14], [78, 77, 73, 70, 68, 57, 39, 33, 28, 17, 17, 11], [91, 91, 71, 58, 52, 50, 48, 43, 38, 29, 27, 17, 14], [89, 87, 73, 69, 68, 67, 66, 63, 53, 40, 31, 30, 16, 4], [98, 98, 93, 73, 70, 62, 60, 53, 38, 27, 23, 23, 9, 4, 4]]

关于python - 订购 list ,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49870913/

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