list_a = ['one', 'two', 'three', 'four', 'cat', 'dog', 'house', 'Spain', 'India', 'Portugal', 'hello', 'world']
list_b = ['PortugalSpainIndiahousefourtwo', 'fivesixseveneight', 'nineteneleventwelve', 'JapanBrazilthirtyseventeen']
matching_list = []

for b in list_b:
    search = b
    for a in list_a:
        if a in b:
            b = b.replace(a, '')
        if b == "":
            matching_list.append(search)
            break
print(matching_list)

不匹配列表print(set(list_b).difference(set(matching_list)))

您可以通过将单词替换为相应数量的占位符字符（例如'*'）来创建匹配位置的模式。如果字符串中的所有位置都至少替换了一次，则整个字符串由list_a中的单词组成。

list_a = ['one', 'two', 'three', 'four', 'cat', 'dog', 'house',
          'Spain', 'India', 'Portugal']
list_b = ['PortugalSpainIndiahousefourtwo', 'fivesixseveneight',
          'nineteneleventwelve', 'JapanBrazilthirtyseventeen']

r = [b for b in list_b
     if all("*" in z for z in zip(*(b.replace(a,"*"*len(a)) for a in list_a)))]

print(r)
['PortugalSpainIndiahousefourtwo']

要在此示例中说明“PortugalSpainIndiahousefourtwo”：

one      PortugalSpainIndiahousefourtwo # 1
two      PortugalSpainIndiahousefour*** # 2
three    PortugalSpainIndiahousefourtwo # 3
four     PortugalSpainIndiahouse****two # 4
cat      PortugalSpainIndiahousefourtwo # 5
dog      PortugalSpainIndiahousefourtwo # 6
house    PortugalSpainIndia*****fourtwo # 7
Spain    Portugal*****Indiahousefourtwo # 8
India    PortugalSpain*****housefourtwo # 9
Portugal ********SpainIndiahousefourtwo # 0
         ||||||||||||||||||||||||||||||
         vvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
matched: 000000008888899999777774444222  <- all positions have at least one *

• 注意，当字符串由单词覆盖自身形成时，这将不会检测自重叠单词，例如“edited”：'编辑'*

关于str.replace()函数有一点：如果没有找到搜索字符串，它将返回原始字符串。记住这一点，代码非常简单：

final_list = []
for word in list_b:
    for sub_word in list_a:
        word = word.replace(sub_word, "")
    final_list.append(word)