如何计算熊猫列中单词实例的数量？
我有一个熊猫的数据框架，其中包含额定电影的流派。有些电影属于多种类型，每种类型都被“ |”隔开。您可以在下面的代码中看到此示例。
import pandas as pd
unames = ['user_id', 'gender', 'age', 'occupation', 'zip']
users = pd.read_table('ml-1m/users.dat', sep='::', header=None, names=unames, engine='python')
rnames = ['user_id', 'movie_id', 'rating', 'timestamp']
ratings = pd.read_table('ml-1m/ratings.dat', sep='::', header=None, names=rnames, engine='python')
mnames = ['movie_id', 'title', 'genres']
movies = pd.read_table('ml-1m/movies.dat', sep='::', header=None, names=mnames, engine='python')
df = pd.merge(pd.merge(ratings, users), movies)
df["genres"].value_counts()
如您所见，value_counts（）方法不是计算每个唯一类型的次数的有效方法。 是否有一种熊猫方法可以使我计算出每种唯一类型的“词”的次数，或者我需要使用一些循环将所有组合流派分开？
I have a Pandas dataframe that contains genres of rated movies. Some movies fall under multiple genres, each genre separated by a "|". You can see examples of this in the code below.
import pandas as pd
unames = ['user_id', 'gender', 'age', 'occupation', 'zip']
users = pd.read_table('ml-1m/users.dat', sep='::', header=None, names=unames, engine='python')
rnames = ['user_id', 'movie_id', 'rating', 'timestamp']
ratings = pd.read_table('ml-1m/ratings.dat', sep='::', header=None, names=rnames, engine='python')
mnames = ['movie_id', 'title', 'genres']
movies = pd.read_table('ml-1m/movies.dat', sep='::', header=None, names=mnames, engine='python')
df = pd.merge(pd.merge(ratings, users), movies)
df["genres"].value_counts()
As you can see, the value_counts() method isn't an effective way of counting the number of times each unique genre is rated. Is there a pandas method that would allow me to count the number of times each unique genre "word" appears or do I need to use some loops to separate all the combined genres out?
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您可以使用REGEX
r'\ s *\ | \ s *'，甚至*[|] *将类型列分开，然后爆炸并进行计数。请注意，\ s代表空间。而且由于|是一个metacharacter，因此您需要通过backspace或将其放置在字符类中，IE[]You could use the regex
r'\s*\|\s*'or even*[|] *to split your genre column then explode the column and do the count. Note that\sstands for space. and since|is a metacharacter, you need to escape it by a backspace or by placing it in a character class ie[]