Python 如何根据子列表的长度对列表列表进行排序
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How to sort list of lists according to length of sublists
提问by rajeshv90
I have the following list
我有以下清单
a = [['a', 'b', 'c'], ['d', 'e'], ['f', 'g', 'h'], ['i', 'j', 'k', 'l'], ['m', 'n'], ['o']]
I would like to sort the list based on the length of their sub lists. The result should be like:
我想根据他们的子列表的长度对列表进行排序。结果应该是这样的：
a = [['o'],['d','e'],['m', 'n'],['a', 'b', 'c'],['f', 'g', 'h'], ['i','j','k','l']]
采纳答案by Alik
Use keyparameter available in sortand sorted. It specifies a function of one argument that is used to extract a comparison key from each list element
使用和 中key可用的参数。它指定了一个参数的函数，用于从每个列表元素中提取比较键sortsorted
In [6]: a = [['a', 'b', 'c'], ['d', 'e'], ['f', 'g', 'h'], ['i', 'j', 'k', 'l'], ['m', 'n'], ['o']]
In [7]: a.sort(key=len)
In [8]: print a
[['o'], ['d', 'e'], ['m', 'n'], ['a', 'b', 'c'], ['f', 'g', 'h'], ['i', 'j', 'k', 'l']]
回答by sachin saxena
can be done by
可以通过
sorted(a, key=len)
