如何在python中对元组列表进行排序
例如:请注意,元组的最后一个元素是相同的
a = [
[(1, 5, 6), (2, 2, 6), (3, 6, 6)],
[(6, 1, 3), (5, 7, 3), (4, 1, 3)],
[(5, 7, 2), (7, 5, 2), (6, 3, 2)],
[(9, 1, 7), (1, 5, 7), (2, 6, 7)]
]
我想使用元组的最后一个元素作为键进行排序。所以预期的结果是
[
[(5, 7, 2), (7, 5, 2), (6, 3, 2)],
[(6, 1, 3), (5, 7, 3), (4, 1, 3)],
[(1, 5, 6), (2, 2, 6), (3, 6, 6)],
[(9, 1, 7), (1, 5, 7), (2, 6, 7)]
]
答案 0 :(得分:4)
In [27]: a = [
....:
....: [(1, 5, 6), (2, 2, 6), (3, 6, 6)],
....: [(6, 1, 3), (5, 7, 3), (4, 1, 3)],
....: [(5, 7, 2), (7, 5, 2), (6, 3, 2)],
....: [(9, 1, 7), (1, 5, 7), (2, 6, 7)]
....:
....: ]
# any tuple (first will do), last element
In [28]: a.sort(key=lambda l: l[0][-1])
In [29]: a
Out[29]:
[[(5, 7, 2), (7, 5, 2), (6, 3, 2)],
[(6, 1, 3), (5, 7, 3), (4, 1, 3)],
[(1, 5, 6), (2, 2, 6), (3, 6, 6)],
[(9, 1, 7), (1, 5, 7), (2, 6, 7)]]
答案 1 :(得分:4)
使用key
参数进行排序,并使用lambda
函数
a.sort(key=lambda x: x[-1][-1]). # sorts in-place
sorted(a, key=lambda x: x[-1][-1]). # new sorted list
编辑:根据您想要比较的元组来更改第一个索引。即,使用x[0][-1]
基于第一个元组进行比较