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Python实现字典依据value排序

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WBOYOriginal
2016-06-10 15:06:151284browse

具体内容如下:

使用sorted将字典按照其value大小排序

>>> record = {'a':89, 'b':86, 'c':99, 'd':100}
>>> sorted(record.items(), key=lambda x:x[1])
[('b', 86), ('a', 89), ('c', 99), ('d', 100)]

sorted第一个参数要可迭代,可以为tuple, list

>>> items = [(1, 'B'), (1, 'A'), (2, 'A'), (0, 'B'), (0, 'a')]
>>> sorted(items)
[(0, 'B'), (0, 'a'), (1, 'A'), (1, 'B'), (2, 'A')]

为什么(0, 'B')在(0, 'a')前面?

因为ASCII码中大写字母排在在小写字母前面,使用str.lower()方法改变其顺序

>>> sorted(items, key=lambda x:(x[0], x[1].lower()))
[(0, 'a'), (0, 'B'), (1, 'A'), (1, 'B'), (2, 'A')]

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