问题：给出abc的信息

a = ['abc-123', 'def-456', 'ghi-789', 'abc-456']


aa = [ string for string in a if  "abc" in string]
print(aa)


Output =>  ['abc-123', 'abc-456']

只是抛出这个：如果你恰好需要匹配多个字符串，例如abc和def，你可以按如下方式组合两个理解：

matchers = ['abc','def']
matching = [s for s in my_list if any(xs in s for xs in matchers)]

输出：

['abc-123', 'def-456', 'abc-456']

使用筛选器获取具有“abc”的所有元素：

>>> xs = ['abc-123', 'def-456', 'ghi-789', 'abc-456']
>>> list(filter(lambda x: 'abc' in x, xs))
['abc-123', 'abc-456']

还可以使用列表理解：

>>> [x for x in xs if 'abc' in x]

def find_dog(new_ls):
    splt = new_ls.split()
    if 'dog' in splt:
        print("True")
    else:
        print('False')


find_dog("Is there a dog here?")

我需要与以下匹配项对应的列表索引：

lst=['abc-123', 'def-456', 'ghi-789', 'abc-456']

[n for n, x in enumerate(lst) if 'abc' in x]

输出

[0, 3]

mylist=['abc','def','ghi','abc']

pattern=re.compile(r'abc') 

pattern.findall(mylist)