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This topic created in 2089 days ago, the information mentioned may be changed or developed.
data_list = [{1: "1"}, {3: "3"}, {4: "4"}, {10: "10"}, {7: "7"}, {6: "6"}]，
类似这样的数据，怎么用 sort 去排序，sort 里的 key 关键字参数要怎么写
类似这样的数据，怎么用 sort 去排序，sort 里的 key 关键字参数要怎么写
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1
Akikiki Aug 19, 2020
data_list.sort(key=lambda x: x.keys()[0])
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casatAway Aug 19, 2020
就假设你的字典都是一个 key，sorted(data_list, key=lambda k: list(k.keys())[0])
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4
duyuyouci OP @h272377502 对呀，转化一下类型就好了，厉害
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6
mahonejolla Aug 19, 2020
data_list = [{1: "1"}, {3: "3"}, {4: "4"}, {10: "10"}, {7: "7"}, {6: "6"}]
kk = data_list.sort(key=lambda x: list(x.keys())[0]) print(data_list) # [{1: '1'}, {3: '3'}, {4: '4'}, {6: '6'}, {7: '7'}, {10: '10'}] kk = sorted(data_list, key=lambda k: list(k.keys())[0]) print(kk) # [{1: '1'}, {3: '3'}, {4: '4'}, {6: '6'}, {7: '7'}, {10: '10'}] |
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lithbitren Aug 21, 2020
data_list.sort(key=lambda x: next(iter(x)))
不转 list 也可以实现 |
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8
yucongo Aug 21, 2020 via Android
sorted(data_list, key=lambda x: [*x])
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duyuyouci OP @lithbitren 高级
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yucongo Aug 21, 2020
In [43]: data_list = [{1: "1"}, {3: "3"}, {4: "4"}, {10: "10"}, {7: "7"}, {6: "6"}]
In [44]: sorted(data_list, key=lambda x: [*x]) Out[44]: [{1: '1'}, {3: '3'}, {4: '4'}, {6: '6'}, {7: '7'}, {10: '10'}] 完全用你的数据，python3.6, 顺序怎么没有变呢 |
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12
yucongo Aug 21, 2020
或（ in-place ）:
data_list.sort(key=lambda x: [*x]) |
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14
lithbitren Aug 22, 2020
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duyuyouci OP @lithbitren 原来如此，受教了
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yucongo Aug 22, 2020
对啊，key=list 就行了……
那么可以来一个最短的:) sorted(data_list, key=set) 或 data_list.sort(key=set) |
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