Question

所以我有一个字符串列表如下：

list = ["I love cat", "I love dog", "I love fish", "I hate banana", "I hate apple", "I hate orange"]

如何在没有给定关键字的情况下遍历列表并对部分匹配的字符串进行分组。结果如下：

list 1 = [["I love cat","I love dog","I love fish"],["I hate banana","I hate apple","I hate orange"]]

非常感谢你。

Answer 1

尝试构建反向索引，然后您可以选择自己喜欢的关键字。这种方法忽略了词序：

index = {}
for sentence in sentence_list:
    for word in set(sentence.split()):
        index.setdefault(word, set()).add(sentence)

或者这种方法，它通过所有可能的全字短语前缀来键入索引：

index = {}
for sentence in sentence_list:
    number_of_words = length(sentence.split())
    for i in xrange(1, number_of_words):
        key_phrase = sentence.rsplit(maxsplit=i)[0]
        index.setdefault(key_phrase, set()).add(sentence)

然后，如果你想找到包含关键字的所有句子（或者以短语开头，如果这是你的索引）：

match_sentences = index[key_term]

或一组给定的关键字：

matching_sentences = reduce(list_of_keywords[1:], lambda x, y: x & index[y], initializer = index[list_of_keywords[0]])

现在，您可以通过使用这些索引构建列表推导来生成句子，从而生成按几乎任何术语或短语组合分组的列表。例如，如果您构建了短语前缀索引并希望按前两个单词短语分组的所有内容：

return [list(index[k]) for k in index if len(k.split()) == 2]

Answer 2

序列匹配器将为您完成任务。调整得分比率以获得更好的结果。

试试这个：

from difflib import SequenceMatcher
sentence_list = ["I love cat", "I love dog", "I love fish", "I hate banana", "I hate apple", "I hate orange"]
result=[]
for sentence in sentence_list:
    if(len(result)==0):
        result.append([sentence])
    else:
        for i in range(0,len(result)):
            score=SequenceMatcher(None,sentence,result[i][0]).ratio()
            if(score<0.5):
                if(i==len(result)-1):
                    result.append([sentence])
            else:
                if(score != 1):
                    result[i].append(sentence)

输出：

[['I love cat', 'I love dog', 'I love fish'], ['I hate banana', 'I hate apple', 'I hate orange']]

Answer 3

在命名变量时避免像list这样的单词。 list 1也不是有效的python变量。

试试这个：

import sys
from itertools import groupby

#Assuming you group by the first two words in each string, e.g. 'I love', 'I hate'.

L = ["I love cat", "I love dog", "I love fish", "I hate banana", "I hate apple", "I hate orange"]

L = sorted(L)

result = []

for key,group in groupby(L, lambda x: x.split(' ')[0] + ' ' + x.split(' ')[1]):
    result.append(list(group))

print(result)

Answer 4

您可以尝试这种方法。虽然这不是最好的方法，但有助于以更有条理的方式理解问题。

from itertools import groupby

my_list = ["I love cat","I love dog","I love fish","I hate banana","I hate apple","I hate orange"];

each_word = sorted([x.split() for x in my_list])

# I assumed the keywords would be everything except the last word
grouped = [list(value) for key, value in groupby(each_word, lambda x: x[:-1])]

result = []
for group in grouped:
    temp = []
    for i in range(len(group)):
        temp.append(" ".join(group[i]))
    result.append(temp)

print(result)

输出：

[['I hate apple', 'I hate banana', 'I hate orange'], ['I love cat', 'I love dog', 'I love fish']]

Python - 遍历字符串列表并组合部分匹配字符串

4 个答案: