Python3使用Lambda和Sort排序元组列表

时间:2016-11-29 01:00:09

标签: python list sorting lambda

我有一个按第二个元素排序的元组列表,如下所示:

[('7', 10), ('2', 9), ('8', 9)]

我想按照第一个元素对它们进行排序:

[('2',9),('7', 10),('8', 9)]

我试过了:

test_val = [('7', 10), ('2', 9), ('8', 9)]
test_val.sort(key=lambda x: x[0])

我从here

找到的

但是,列表未排序。

真实数据示例:

尝试前:

[('7', 10), ('2', 9), ('8', 9), ('6', 8), ('24', 8), ('3', 7), ('5', 7), ('9', 6), ('35', 6), ('15', 5), ('16', 5), ('1', 4), ('14', 4), ('17', 3), ('19', 3), ('12', 2), ('39', 2), ('26', 1), ('25', 1), ('22', 0)]

尝试后:

[('1', 4), ('12', 2), ('14', 4), ('15', 5), ('16', 5), ('17', 3), ('19', 3), ('2', 9), ('22', 0), ('24', 8), ('25', 1), ('26', 1), ('3', 7), ('35', 6), ('39', 2), ('5', 7), ('6', 8), ('7', 10), ('8', 9), ('9', 6)]

1 个答案:

答案 0 :(得分:5)

列表也是如此排序。请注意,您的排序键是字符串,而不是整数。也许你想要的是

test_val.sort(key=lambda x: int(x[0]))