英文:
More pythonic way to filter objects from a list of dicts based on a dict value that must not contain a string from a list of strings
问题
作为来自Java的开发者，我想知道是否有一种更具Python风格的方法来实现以下逻辑：
movies = [{'original_title': 'Star Wars'}, {'original_title': 'Avengers'}, {'original_title': 'After Love'}]
blacklist = ['Star', 'Love']
filtered_movies = []
for movie in movies:
blacklisted = False
for item in blacklist:
if item in movie['original_title']:
blacklisted = True
break
if not blacklisted:
filtered_movies.append(movie)
return filtered_movies
然后，filtered_movies 仅包含项目 {'original_title': 'Avengers'}。
英文:
As I'm coming from Java, I wonder if there is a more pythonic approach to achieve the following logic:
movies = [{'original_title': 'Star Wars'}, {'original_title': 'Avengers'}, {'original_title': 'After Love'}]
blacklist = ['Star', 'Love']
filtered_movies = []
for movie in movies:
blacklisted = False
for item in blacklist:
if item in movie['original_title']:
blacklisted = True
break
if not blacklisted:
filtered_movies.append(movie)
return filtered_movies
filtered_movies then only contains the item {'original_title': 'Avengers'}.
答案1
得分: 3
你可以使用 列表推导式 来实现这个操作：
filtered_movies = [movie for movie in movies
if not any(
item in movie['original_title']
for item in blacklist
)]
例如：
>>> movies = [{'original_title': 'Star Wars'}, {'original_title': 'Avengers'}, {'original_title': 'After Love'}]
>>> blacklist = ['Star', 'Love']
>>> filtered_movies = [movie for movie in movies
... if not any(item in movie['original_title'] for item in blacklist)]
>>> filtered_movies
[{'original_title': 'Avengers'}]
这是执行此类过滤操作的一种常见方法；内置的 any 方法（在此处使用）和 all 方法在此非常有用。
英文:
You can do this with a list comprehension:
filtered_movies = [movie for movie in movies
if not any(
item in movie['original_title']
for item in blacklist
)]
For example:
>>> movies = [{'original_title': 'Star Wars'}, {'original_title': 'Avengers'}, {'original_title': 'After Love'}]
>>> blacklist = ['Star', 'Love']
>>> filtered_movies = [movie for movie in movies
... if not any(item in movie['original_title'] for item in blacklist)]
>>> filtered_movies
[{'original_title': 'Avengers'}]
This is a pretty canonical way of performing this sort of filtering operation; the builtin any (used here) and all methods are particularly useful for this.
答案2
得分: 1
我会这样做，但不确定是否更符合Python的风格。
movies = [
{'original_title': 'Star Wars'},
{'original_title': 'Avengers'},
{'original_title': 'After Love'},
]
blacklist = ['Star', 'Love']
filtered_movies = []
for movie in movies:
if any(word in movie['original_title'] for word in blacklist):
continue
filtered_movies.append(movie)
print(filtered_movies)
英文:
I would have done it like this, but no idea if this is a more pythonic approach.
movies = [
{'original_title': 'Star Wars'},
{'original_title': 'Avengers'},
{'original_title': 'After Love'},
]
blacklist = ['Star', 'Love']
filtered_movies = []
for movie in movies:
if any(word in movie['original_title'] for word in blacklist):
continue
filtered_movies.append(movie)
print(filtered_movies)
答案3
得分: -2
以下是翻译好的部分：
这里使用了filter和lambda函数以及set
>>> list(filter(lambda x: len(set(blacklist) & set(x["original_title"].split())) == 0, movies))
[{'original_title': 'Avengers'}]
如果不允许拆分，则：
>>> list(filter(lambda x: len([1 for y in blacklist if y in x["original_title"]])==0, movies))
[{'original_title': 'Avengers'}]
>>> list(filter(lambda x: not any(1 for y in blacklist if y in x["original_title"]), movies))
[{'original_title': 'Avengers'}]
英文:
Here you go using filter and lambda function and set
>>> list(filter(lambda x: len(set(blacklist) & set(x["original_title"].split())) == 0, movies))
[{'original_title': 'Avengers'}]
if splitting is not allowed then:
>>> list(filter(lambda x: len([1 for y in blacklist if y in x["original_title"]])==0, movies))
[{'original_title': 'Avengers'}]
>>> list(filter(lambda x: not any(1 for y in blacklist if y in x["original_title"]), movies))
[{'original_title': 'Avengers'}]
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