如何像这样比较多个元组列表:
[[(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!
也许你正在寻找链接这个:
l = [[(1,2), (3,6), (5,3)], [(1,5), (3,5)], [(2,1), (1,8), (3,9)]]
output = [l[0][0], l[1][0], l[2][1]], [l[0][1], l[1][1], l[2][2]]
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)]]
那么他们中的哪一个更适合示例数据呢?
>>> 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
并且有一个更大的随机列表
>>> 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
>>>
>>> L=[[(1,2), (3,6), (5,3)], [(1,5), (3,5)], [(2,1), (1,8), (3,9)]]
>>> from collections import defaultdict
>>> from itertools import chain
>>> p = defaultdict(list)
>>> for i in chain.from_iterable(L):
... p[i[0]].append(i)
...
>>> p = {k:v for k,v in p.items() if len(v)>1}
>>> p.values()
[[(1, 2), (1, 5), (1, 8)], [(3, 6), (3, 5), (3, 9)]]
这很好用:
>>> l=[[(1,2), (3,6), (5,3)], [(1,5), (3,5)], [(2,1), (1,8), (3,9)]]
>>> [[t for t in [i for sub in l for i in sub] if t[0]==1]]+[[t for t in [i for sub in l for i in sub] if t[0]==3]]
[[(1, 2), (1, 5), (1, 8)], [(3, 6), (3, 5), (3, 9)]]
或者,不重复:
>>> flat=[i for sub in l for i in sub]
>>> [[t for t in flat if t[0]==1]]+[[t for t in flat if t[0]==3]]
[[(1, 2), (1, 5), (1, 8)], [(3, 6), (3, 5), (3, 9)]]