Question

从列表中

l =[(3,4),(2,3),(4,3),(3,2)]

我想以相反的顺序排列具有相同成员的对的所有第二次出现。即，结果应为

[(3,4),(2,3)]

在Python中最简洁的方法是什么？

Answer 1

或者，人们可能会以更冗长的方式做到这一点：

l = [(3,4),(2,3),(4,3),(3,2)]
L = []
omega = set([])
for a,b in l:
    key = (min(a,b), max(a,b))
    if key in omega:
        continue
    omega.add(key)
    L.append((a,b))

print(L)

Answer 2

如果我们只想保留每对中的第一个元组：

l =[(3,4),(2,3),(4,3),(3,2), (3, 3), (5, 6)]

def first_tuples(l):
    # We could use a list to keep track of the already seen
    # tuples, but checking if they are in a set is faster
    already_seen = set()
    out = []
    for tup in l:
        if set(tup) not in already_seen:
            out.append(tup)
            # As sets can only contain hashables, we use a 
            # frozenset here
            already_seen.add(frozenset(tup))
    return out

print(first_tuples(l))
# [(3, 4), (2, 3), (3, 3), (5, 6)]

Answer 3

这应该是诀窍：

[x for i, x in enumerate(l) if any(y[::-1] == x for y in l[i:])]
Out[23]: [(3, 4), (2, 3)]

使用不同的顺序扩展初始列表：

l =[(3,4),(2,3),(4,3),(3,2), (1,3), (3,1)]

[x for i, x in enumerate(l) if any(y[::-1] == x for y in l[i:])]
Out[25]: [(3, 4), (2, 3), (1, 3)]

并且，取决于每个元组是否保证有伴随的姐妹＆＃34;反转元组，逻辑可能会改变，以保持＆＃34;单身＆＃34;元组：

l = [(3, 4), (2, 3), (4, 3), (3, 2), (1, 3), (3, 1), (10, 11), (10, 12)]

[x for i, x in enumerate(l) if any(y[::-1] == x for y in l[i:]) or not any(y[::-1] == x for y in l)]
Out[35]: [(3, 4), (2, 3), (1, 3), (10, 11), (10, 12)]

Answer 4

恕我直言，这应该比迄今发布的任何内容都更短更清晰：

my_tuple_list = [(3,4),(2,3),(4,3),(3,2)]

set((left, right) if left < right else (right, left) for left, right in my_tuple_list)

>>> {(2, 3), (3, 4)}

它只是创建一组所有元组，如果第一个成员是＆gt;，则预先交换其成员。第二个成员。

从成对列表中对具有相同成员但不同顺序的对进行排序

4 个答案: