我正在嘗試合并具有公共元素的列表列表中的所有列表。我有一些工作代碼。但是，它似乎打破了這個例子：def merge_subs(lst_of_lsts): res = [] for row in lst_of_lsts: for i, resrow in enumerate(res): if row[0]==resrow[0]: res[i] += row[1:] break else: res.append(sorted(row)) return sorted(res)輸入是：merge_subs([[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]])我的結(jié)果是：[[0, 2, 4, 6], [1, 3, 5, 7], [3, 5, 7]]但我應該得到：[[0, 2, 4, 6], [1, 3, 5, 7]]
2 回答
交互式愛情
TA貢獻1712條經(jīng)驗 獲得超3個贊
您需要使用遞歸：
def group(d, _start, _c = [], _seen = [], _used=[]):
r = [i for i in d if any(c in _start for c in i) and i not in _seen and i not in _used]
if not r:
yield set(_c)
for i in d:
if i != _start and i not in _used:
yield from group(d, i, _c=[], _seen=[], _used=_used+[i, *r])
else:
yield from group(d, _start, _c=_c+[i for b in r for i in b], _seen=_seen+r, _used=_used+r)
data = [[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]]
result = list(map(list, {tuple(i) for i in group(data, data[0], _seen=[data[0]]) if i}))
輸出：
[[0, 2, 4, 6], [1, 3, 5, 7]]
吃雞游戲
TA貢獻1829條經(jīng)驗 獲得超7個贊
這個問題可以使用遞歸解決，特別是尾遞歸：
def merge(lists, results=None):
if results is None:
results = []
if not lists:
return results
first = lists[0]
merged = []
output = []
for li in lists[1:]:
for i in first:
if i in li:
merged = merged + li
break
else:
output.append(li)
merged = merged + first
results.append(list(set(merged)))
return merge(output, results)
結(jié)果如下所示：
>>> lists = [[1, 7, 3], [1, 7, 5], [2, 0, 4], [2, 0, 6], [3, 7, 1], [3, 7, 5], [4, 0, 2], [4, 0, 6], [5, 7, 1], [5, 7, 3], [6, 0, 2], [6, 0, 4]]
>>> merge(lists)
[[1, 3, 5, 7], [0, 2, 4, 6]]
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