这是一个相对优化的朴素算法。首先，将每个序列转换为所有ngram的集合。然后，将所有集合相交，并在相交中找到最长的ngram。@H_403_1@
from functools import partial, reduce
from itertools import chain
from typing import Iterator
def ngram(seq: str, n: int) -> Iterator[str]:
return (seq[i: i+n] for i in range(0, len(seq)-n+1))
def allngram(seq: str) -> set:
lengths = range(len(seq))
ngrams = map(partial(ngram, seq), lengths)
return set(chain.from_iterable(ngrams))
sequences = ["brownasdfoersjumps",
"foxsxzxasis12sa[[#brown",
"thissasbrownxc-34a@s;"]
seqs_ngrams = map(allngram, sequences)
intersection = reduce(set.intersection, seqs_ngrams)
longest = max(intersection, key=len) # -> brown
虽然这可能使您了解短序列，但此算法在长序列上效率极低。如果序列很长，则可以添加启发式方法以限制最大可能的ngram长度（即，可能的最长公共子串）。这种启发式方法的一个显而易见的价值可能是最短序列的长度。@H_403_1@
def allngram(seq: str, minn=1, maxn=None) -> Iterator[str]:
lengths = range(minn, maxn) if maxn else range(minn, len(seq))
ngrams = map(partial(ngram, seq), lengths)
return set(chain.from_iterable(ngrams))
sequences = ["brownasdfoersjumps",
"foxsxzxasis12sa[[#brown",
"thissasbrownxc-34a@s;"]
maxn = min(map(len, sequences))
seqs_ngrams = map(partial(allngram, maxn=maxn), sequences)
intersection = reduce(set.intersection, seqs_ngrams)
longest = max(intersection, key=len) # -> brown
这可能仍会花费太长时间（或使您的计算机用完RAM），因此您可能需要阅读一些最佳算法（请参阅我在评论中留给您的问题的链接）。@H_403_1@
@H_403_1@
计算每个ngram出现的字符串数@H_403_1@
from collections import Counter
sequences = ["brownasdfoersjumps",
"foxsxzxasis12sa[[#brown",
"thissasbrownxc-34a@s;"]
seqs_ngrams = map(allngram, sequences)
counts = Counter(chain.from_iterable(seqs_ngrams))
Counter是的子类dict，因此其实例具有相似的接口：@H_403_1@
print(counts)
Counter({'#': 1,
'#b': 1,
'#br': 1,
'#bro': 1,
'#brow': 1,
'#brown': 1,
'-': 1,
'-3': 1,
'-34': 1,
'-34a': 1,
'-34a@': 1,
'-34a@s': 1,
'-34a@s;': 1,
...
您可以过滤计数以使子字符串至少出现在n字符串中：{string: count for string, count in counts.items() if count >= n}@H_403_1@
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