这里是您的解决方案:
lst1=[['hey','jude' ,'fox'],['how','you','do']]
lst2=[['hey','paul'],['how','do']]
new_list=[]
for item in lst1:
for item_1 in lst2:
if isinstance(item_1,list):
for sub_item in item_1:
if sub_item in item:
if len(sub_item)>=1:
if len(item)>len(item_1):
if item not in new_list:
new_list.append(item)
elif len(item_1)>len(item):
if item_1 not in new_list:
new_list.append(item_1)
if len(sub_item)<2:
if sub_item not in new_list:
new_list.append(sub_item)
new_list.append(item)
print([j for i,j in enumerate(new_list) if j not in new_list[:i]])
P.S:你说有困惑的问题,你是什么 “字样大于或等于1”的意思你是指单词的长度还是单词的匹配?我接受了这个单词。
测试:
与你的列表:
输出:
[['hey', 'jude', 'fox'], ['how', 'you', 'do']]
当列表中的一个有comman小于2 LEN信然:
lst1=[['hey','jude' ,'fox'],['how','you','do'],["wow"],["wllla","wlahi","aasma"],["h","i","j"]]
lst2=[['hey','paul'],['how','do'],["wllla"],["h"]]
输出:
[['hey', 'jude', 'fox'], ['how', 'you', 'do'], ['wllla', 'wlahi', 'aasma'], ['h', 'i', 'j'], 'h']
当第二个列表具有最大公共列表的话它的匹配情况:
lst1=[['hey','jude' ,'fox'],['how','you','do'],["wow"],["wllla","wlahi","aasma"],["h","i","j"],["abc","def"]]
lst2=[['hey','paul'],['how','do'],["wllla"],["h"],["abc","def","ghi"]]
输出:
[['hey', 'jude', 'fox'], ['how', 'you', 'do'], ['wllla', 'wlahi', 'aasma'], ['h', 'i', 'j'], 'h', ['abc', 'def', 'ghi']]
如果你想避免嵌套打破逻辑
出功能 减少嵌套。
请显示所需的输出。 –
你有什么尝试?请参阅[如何创建最小,完整和可验证示例](https://stackoverflow.com/help/mcve) –