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这是一个创建于 1558 天前的主题，其中的信息可能已经有所发展或是发生改变。
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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Akikiki 2020-08-19 16:48:37 +08:00
data_list.sort(key=lambda x: x.keys()[0])
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duyuyouci OP @Akikiki 会报错的，TypeError: 'dict_keys' object is not subscriptable
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h272377502 2020-08-19 16:54:00 +08:00
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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mahonejolla 2020-08-19 17:05:37 +08:00
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 2020-08-21 09:21:50 +08:00
data_list.sort(key=lambda x: next(iter(x)))
不转 list 也可以实现 |
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yucongo 2020-08-21 14:57:38 +08:00 via Android
sorted(data_list, key=lambda x: [*x])
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duyuyouci OP @lithbitren 高级
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yucongo 2020-08-21 19:31:49 +08:00
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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yucongo 2020-08-21 19:40:57 +08:00
或（ in-place ）:
data_list.sort(key=lambda x: [*x]) |
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duyuyouci OP @yucongo 哦，sorted 是创建了一个副本，我打印的原列表，哈哈，高级，但是这个语法不太懂，老哥能解释吗
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lithbitren 2020-08-22 10:53:02 +08:00
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duyuyouci OP @lithbitren 原来如此，受教了
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yucongo 2020-08-22 16:03:21 +08:00
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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