from itertools import islice
def is_sublist(source, target):
slen = len(source)
return any(all(item1 == item2 for (item1, item2) in zip(source, islice(target, i, i+slen))) for i in range(len(target) - slen + 1))
def long_substr_by_word(data):
subseq = []
data_seqs = [s.split(' ') for s in data]
if len(data_seqs) > 1 and len(data_seqs[0]) > 0:
for i in range(len(data_seqs[0])):
for j in range(len(data_seqs[0])-i+1):
if j > len(subseq) and all(is_sublist(data_seqs[0][i:i+j], x) for x in data_seqs):
subseq = data_seqs[0][i:i+j]
return ' '.join(subseq)
from itertools import combinations
mylist = ['commercial van for movers',
'partial van for movers',
'commercial van for moving' ]
s0 = mylist[0].split()
candidates = [' '.join(s0[slice(*c)]) for c in combinations(list(range(len(s0)+1)), 2)]
for s in mylist:
for i,c in reversed(list(enumerate(candidates.copy()))):
if not c in s:
candidates.pop(i)
max(candidates, key=lambda x: (x.count(' '), len(x)))
# returns:
'van for'
关键是修改为按整词子序列搜索。在
演示:
^{pr2}$您可以为第一个句子的所有子序列创建一个有序的powerset,然后在其他句子中搜索这些字符串中的每一个,删除找不到的子字符串。在
最后,选择具有最多空格的候选子字符串,如果出现平局,则选择最长的子字符串。在
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