比较字符串，找到每个字符串中存在的部分

Question

如何比较几行，并找到每行中存在的单词的单词/组合？使用纯Python，NLTK或其他任何东西。

few_strings = ('this is foo bar', 'this is not a foo bar', 'some other foo bar here')
# some magic
result = 'foo bar'

Answer 1

将每个字符串分开在白空间，然后将结果单词保存到集合中。然后，计算三组的交点：

few_strings = ('this is foo bar', 'this is not a foo bar', 'some other foo bar here')
sets = [set(s.split()) for s in few_strings]
common_words = sets[0].intersection(*sets[1:])
print(common_words)

输出：

{'bar', 'foo'}

Answer 2

您可能需要使用标准库 fifflib 进行序列比较，包括查找常见子字符串：

from difflib import SequenceMatcher

list_of_str = ['this is foo bar', 'this is not a foo bar', 'some other foo bar here']

result = list_of_str[0]
for next_string in list_of_str:
    match = SequenceMatcher(None, result, next_string).find_longest_match()
    result = result[match.a:match.a + match.size]

# result be 'foo bar'

• documentation
• ：

from difflib import SequenceMatcher

string1 = "apple pie available"
string2 = "come have some apple pies"

match = SequenceMatcher(None, string1, string2).find_longest_match()

print(match)  # -> Match(a=0, b=15, size=9)
print(string1[match.a:match.a + match.size])  # -> apple pie
print(string2[match.b:match.b + match.size])  # -> apple pie

Answer 3

few_strings = ('this is foo bar', 'this is not a foo bar', 'some other foo bar here')

1. 句子划分的每个句子（“” ）创建一组
2. 单词
3. 为每个 一个句子

# 1.
sets = [set(s.split(" ")) for s in few_strings]
# 2.
result = sets[0]
# 3.
for i in range(len(sets)):
    result = result.intersection(sets[i])

现在，您有一个python set 单词的单词>，这在所有句子中发生。
您可以将集合转换为列表：

result = list(result)

或与

result = " ".join(result)

Answer 4

您也可以在不使用库的情况下做到这一点

few_strings = ('this is foo bar', 'some other foo bar here', 'this is not a foo bar')
strings = [s.split() for s in few_strings]
strings.sort(key=len)
print(strings)
result = ''

for word in strings[0]:
    count = 0
    for string in strings:
        if word not in string:
            break
        else:
            count += 1
    if count == len(strings):
        result += word + ' '

print(result)