Question

如何比较多个这样的元组列表：

[[(1,2), (3,6), (5,3)], [(1,5), (3,5)], [(2,1), (1,8), (3,9)]]

输出应为：

[(1,2), (1,5), (1,8)],[(3,6), (3,5), (3,9)]

这意味着我只想要那些 x轴值与其他值相匹配的值（5,3）和（2,1）应该被丢弃！

Answer 1

也许你正在寻找一些链接：

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]]

Answer 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)]]

另一种流行选择

>>> 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)]]

那么他们哪个会更好地使用示例数据？

>>> 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
>>>

Answer 3

>>> 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)]]

Answer 4

这很有效：

>>> 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)]]

如何在python中比较多个元组列表？

4 个答案: