从另一个列表中存在的两个列表中查找2个项目的Python方法

Question

我有一些twitter数据，我将文本分为优雅和Python形式的带有快乐表情和悲伤表情的文本，如下所示：

happy_set = [":)",":-)","=)",":D",":-D","=D"]
sad_set = [":(",":-(","=("]

happy = [tweet.split() for tweet in data for face in happy_set if face in tweet]
sad = [tweet.split() for tweet in data for face in sad_set if face in tweet]

但是，这happy_set ，可能是在单个推文中同时找到了来自happy_set和sad_set的图释。 确保happy清单仅包含来自happy_set释，反之亦然的happy_set什么？

Answer 1

您可以尝试使用集合，特别是set.isdisjoint 。 检查快乐鸣叫中的令牌集是否与sad_set不相交。 如果是这样，它肯定属于happy ：

happy_set = set([":)",":-)","=)",":D",":-D","=D"])
sad_set = set([":(",":-(","=("])

# happy is your existing set of potentially happy tweets. To remove any tweets with sad tokens...
happy = [tweet for tweet in happy if sad_set.isdisjoint(set(tweet.split()))]

Answer 2

我会使用lambdas：

>>> is_happy = lambda tweet: any(map(lambda x: x in happy_set, tweet.split()))
>>> is_sad = lambda tweet: any(map(lambda x: x in sad_set, tweet.split()))

>>> data = ["Hi, I am sad :( but don't worry =D", "Happy day :-)", "Boooh :-("]
>>> filter(lambda tweet: is_happy(tweet) and not is_sad(tweet), data)
['Happy day :-)']
>>> filter(lambda tweet: is_sad(tweet) and not is_happy(tweet), data)
['Boooh :-(']

这样可以避免创建data中间副本。

而且，如果data确实很大，则可以从itertools包中的ifilter替换filter以获取迭代器而不是列表。

Answer 3

您在找吗？

happy_set = set([":)",":-)","=)",":D",":-D","=D"])
sad_set = set([":(",":-(","=("])

happy_maybe_sad = [tweet.split() for tweet in data for face in happy_set if face in tweet]
sad_maybe_happy = [tweet.split() for tweet in data for face in sad_set if face in tweet]

happy = [item for item in happy_maybe_sad if not in sad_maybe_happy]
sad = [item for item in sad_maybe_happy if not in happy_maybe_sad]

为了happy...和sad... ，我坚持使用列表解决方案，因为该项目的订单可能是相关的。 如果不是，最好使用set()来提高性能。 是加法，集合已提供基本集合操作 （联合，交集等）