本文介绍了如何比较python中的多个元组列表?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!
问题描述
如何比较多个这样的元组列表:
How can I compare Multiple lists of tuples like this:
[[(1,2), (3,6), (5,3)], [(1,5), (3,5)], [(2,1), (1,8), (3,9)]]
输出应该是:
[(1,2), (1,5), (1,8)],[(3,6), (3,5), (3,9)]
这意味着我只想要那些 x 轴 值与其他值匹配的值.
(5,3) 和 (2,1) 应该被丢弃!
It means that i want just those values whose x-axis value matches others.
(5,3) and (2,1) should be discarded!
推荐答案
一种可能的选择
>>> def group(seq):
for k, v in groupby(sorted(chain(*seq), key = itemgetter(0)), itemgetter(0)):
v = list(v)
if len(v) > 1:
yield v
>>> list(group(some_list))
[[(1, 2), (1, 5), (1, 8)], [(3, 6), (3, 5), (3, 9)]]
另一个受欢迎的选择
>>> from collections import defaultdict
>>> def group(seq):
some_dict = defaultdict(list)
for e in chain(*seq):
some_dict[e[0]].append(e)
return (v for v in some_dict.values() if len(v) > 1)
>>> list(group(some_list))
[[(1, 2), (1, 5), (1, 8)], [(3, 6), (3, 5), (3, 9)]]
那么它们中的哪一个更适合示例数据?
So which of them fairs better with the example data?
>>> def group_sort(seq):
for k, v in groupby(sorted(chain(*seq), key = itemgetter(0)), itemgetter(0)):
v = list(v)
if len(v) > 1:
yield v
>>> def group_hash(seq):
some_dict = defaultdict(list)
for e in chain(*seq):
some_dict[e[0]].append(e)
return (v for v in some_dict.values() if len(v) > 1)
>>> t1_sort = Timer(stmt="list(group_sort(some_list))", setup = "from __main__ import some_list, group_sort, chain, groupby")
>>> t1_hash = Timer(stmt="list(group_hash(some_list))", setup = "from __main__ import some_list, group_hash,chain, defaultdict")
>>> t1_hash.timeit(100000)
3.340240917954361
>>> t1_sort.timeit(100000)
0.14324535970808938
还有一个更大的随机列表
And with a much larger random list
>>> some_list = [[sample(range(1000), 2) for _ in range(100)] for _ in range(100)]
>>> t1_sort.timeit(100)
1.3816694363194983
>>> t1_hash.timeit(1000)
34.015403087978484
>>>
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