一个关于排序的问题，请各位大佬赐教

data_list.sort(key=lambda x: x.keys()[0])

@Akikiki 会报错的，TypeError: 'dict_keys' object is not subscriptable

就假设你的字典都是一个 key，sorted(data_list, key=lambda k: list(k.keys())[0])

@h272377502 对呀，转化一下类型就好了，厉害

@duyuyouci 哦 你是 python3 吧

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'}]

data_list.sort(key=lambda x: next(iter(x)))

不转 list 也可以实现

sorted(data_list, key=lambda x: [*x])

@yucongo 这个好像不行，顺序没有变

@lithbitren 高级

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

或（ in-place ）:
data_list.sort(key=lambda x: [*x])

@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

@duyuyouci

[*x]相当于[i for i in x]，也相当于 list(x)

他这个一行其实可以直接写成 data_list.sort(key=list)，本质还是转数组

@lithbitren 原来如此，受教了

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)