如何从列表中的一个元素检测字符串的某些部分，从列表中的另一个元素检测字符串的其他部分？

Question

upload = ['wish you happy birthday','take care','good night']

string = 'happy birthday take'

应为：：{'happy birthday':0,'take':1} ## 0 and 1 are index of list

Answer 1

如果我理解正确的话，这里的一个关键问题是为你的单词组合找到尽可能长的匹配。因此，我将使用递归来检查在上传中可以找到多少个单词：

upload = ['wish you happy birthday','take care','good night']
search_string = 'foo happy birthday take care bar good night good'

words = search_string.split()
len_word = len(words)
result = dict()

def findWordInUpload(search_str, upld):
    for j, u in enumerate(upld):
        if u.find(search_str) > -1:
            return j
    return -1

next_idx = 0
for i, w in enumerate(words):
    if i < next_idx:
        continue
    var_len = 1
    search_substr_new = w
    temp_res = -1
    while findWordInUpload(search_substr_new, upload) > -1:
        search_substr = search_substr_new      
        temp_res = findWordInUpload(search_substr, upload)
        if i + var_len < len_word:
            var_len += 1
            search_substr_new = " ".join(words[i:i + var_len])
        else:
            break
    
    if temp_res > -1:
        result[search_substr] = temp_res
    next_idx = i + var_len - 1

print(result)

通过前面提到的搜索字符串，它会输出{'happy birthday': 0, 'take care': 1, 'good night': 2, 'good': 2}这就是你要找的吗

编辑：简化了整个变量（第一版中的冗余）

Answer 2

您可以使用itertools：

import itertools

upload = ['wish you happy birthday','take care','good night']
string = 'happy birthday take'.split(' ')

combine = []    # combinations

for u in range(1, len(string)):
    for i in itertools.permutations(string, u):
        combine += [" ".join(list(i))]
print(combine)

result = {}
for count, words in enumerate(upload):
    for strings in combine:
        if strings in words:
            result[strings] = count

remove = []
# remove duplicate combinations
# e.g. if 'happy birthday' exist, remove 'happy' and 'birthday'
for items in result:
    if len(items.split(" ")) > 1:
        for u in items.split(" "):
            if u in result and u not in remove:
                remove += [u]
                
print(remove)
for items in remove:
    del result[items]

print(result)

输出：

['happy', 'birthday', 'take', 'happy birthday', 'happy take', 'birthday happy', 'birthday take', 'take happy', 'take birthday']
['happy', 'birthday']
{'happy birthday': 0, 'take': 1}

Answer 3

蛮力法：

1. 最初，从源代码生成n*（n-1）个元素组合 列出并存储在二维数组中

2. 然后搜索后续的子字符串匹配（或字符串匹配） 先前生成的二维数组中所需的字符串

3. 与所需字符串匹配的生成字符串单元格的索引 给出了预期的结果