有两个要匹配的列表,li_a
是给定的列表,由句子的字符序列组成,而li_b
是单词的集合。
li_a = ['T','h','e','s','e','45','a','r','e','c','a','r','s']
li_b = ['T','Th','The','Thes','These','a','ar','are','c','ca','car','cars']
该过程是迭代地将 li_a
项与 li_b
项进行匹配。如果li_a
的第一个字符与li_b
项相似,则li_a
的第一个字符与下一个字符连接,并重复该过程,直到达到到最长的匹配。然后,最长的期限应该被分割,这个过程将持续到最后。由于 li_a
中未出现在 li_b
中的未知字符和单词将按原样追加。
最终的作品应该是这样的:
new_li = ['These','45','are','cars']
尝试到目前为止,但这适用于两个字符串,不适用于列表,并且它不会检索未识别的单词。
def longest_matched_substring(s1, s2):
m = [[0] * (1 + len(s2)) for i in xrange(1 + len(s1))]
longest, x_longest = 0, 0
for x in xrange(1, 1 + len(s1)):
for y in xrange(1, 1 + len(s2)):
if s1[x - 1] == s2[y - 1]:
m[x][y] = m[x - 1][y - 1] + 1
if m[x][y] > longest:
longest = m[x][y]
x_longest = x
else:
m[x][y] = 0
return s1[x_longest - longest: x_longest]
最佳答案
您可以使用两个 for 循环
和一个临时变量
来完成此操作,如下所示:
def longest_matched_substring(li1, li2):
new_li = []
tmp = ''
for a in li1:
tmp += a
count = 0
for b in li2:
if tmp == b:
count += 1
if count == 0:
tmp1 = tmp.replace(a, '')
new_li.append(tmp1)
tmp = a
if li2.__contains__(tmp):
new_li.append(tmp)
return new_li
输入:
li_a = ['T','h','e','s','e','45','a','r','e','c','a','r','s']
li_b = ['T','Th','The','Thes','These','a','ar','are','c','ca','car','cars']
print longest_matched_substring(li_a, li_b)
输出:
['These', '45', 'are', 'cars']
对于新的场景,可以修改函数如下:
def longest_matched_substring(li1, li2):
new_li = []
tmp = ''
for a in li1:
tmp += a
count = 0
for b in li2:
if tmp == b:
count += 1
if count == 0:
tmp1 = tmp.replace(a, '')
new_li.append(tmp1)
tmp = a
if li_b.__contains__(tmp):
new_li.append(tmp)
for e1 in new_li:
tmp2 = e1
rm = []
for e2 in new_li:
if e1 != e2:
tmp2 += e2
rm.append(e2)
if tmp2 in li2:
new_li.insert(new_li.index(e1), tmp2) # if order matters
#new_li.append(tmp2) if order doesn't matter
for r in rm:
new_li.remove(r)
new_li.remove(e1)
rm = []
break
return new_li
输入:
li_a = ['T','h','e','s','e','45','a','r','e','c','a','r','s']
li_b = ['T','Th','The','These','a','ar','are','c','ca','car','cars']
print longest_matched_substring(li_a, li_b)
输出:
['These', '45', 'are', 'cars']
关于python - 通过与另一个列表匹配来检索列表中最长的匹配值 [Python 2.7],我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44048922/