如何解决熊猫：如何通过考虑索引而不是索引来过滤列
我有一个数据框，代表顾客对餐厅的评价。 star_rating是此数据框中客户的评价。
- 我想做的是在同一数据框中添加一列
nb_fave_rating，该列表示对餐厅的好评数量。如果它的星星数是> = 3。
data = {'rating_id': ['1','2','3','4','5','6','7','8','9'],'user_id': ['56','13','56','99','12','88','45'],'restaurant_id': ['xxx','xxx','yyy','zzz','eee','eee'],'star_rating': ['2.3','3.7','1.2','5.0','1.0','3.2','2.2','0.2'],'rating_year': ['2012','2012','2020','2001','2015','2000','2003','2004'],'first_year': ['2012','2001'],'last_year': ['2020','2020'],}
df = pd.DataFrame (data,columns = ['rating_id','user_id','restaurant_id','star_rating','rating_year','first_year','last_year'])
df['star_rating'] = df['star_rating'].astype(float)
positive_reviews = df[df.star_rating >= 3.0 ].groupby('restaurant_id')
positive_reviews.head()
从这里开始，我不知道要计算一家餐馆的正面评论数量，并将其添加到我的初始数据框df的新列中。
期望的输出将是这样的。
data = {'rating_id': ['1','nb_fave_rating': ['1','1','0','0'],}
所以我尝试了这个，并得到了一堆NaN
df['nb_fave_rating']=df[df.star_rating >= 3.0 ].groupby('restaurant_id').agg({'star_rating': 'count'})
df.head()
解决方法
#filtering the data with >=3 ratings
filtered_data = df[df['star_rating'] >= 3]
#creating a dict containing the counts of the all the favorable reviews
d = filtered_data.groupby('restaurant_id')['star_rating'].count().to_dict()
#mapping the dictionary to the restaurant_id to generate 'nb_fave_rating'
df['nb_fave_rating'] = df['restaurant_id'].map(d)
#taking care of `NaN` values
df.fillna(0,inplace=True)
#making the column integer (just to match the requirements)
df['nb_fave_rating'] = df['nb_fave_rating'].astype(int)
print(df)
输出：
rating_id user_id restaurant_id star_rating rating_year first_year last_year nb_fave_rating
0 1 56 xxx 2.3 2012 2012 2020 1
1 2 13 xxx 3.7 2012 2012 2020 1
2 3 56 yyy 1.2 2020 2001 2020 1
3 4 99 yyy 5.0 2001 2001 2020 1
4 5 99 xxx 1.0 2020 2012 2020 1
5 6 13 zzz 3.2 2015 2000 2015 1
6 7 12 zzz 1.0 2000 2000 2015 1
7 8 88 eee 2.2 2003 2001 2020 0
8 9 45 eee 0.2 2004 2001 2020 0
一行完成。
groupby()，transform布尔选择并将结果转换为integer。
df['nb_fave_rating']=df.groupby('restaurant_id')['star_rating'].transform(lambda x: int((x>=3).sum()))
rating_id user_id restaurant_id star_rating rating_year first_year \
0 1 56 xxx 2.3 2012 2012
1 2 13 xxx 3.7 2012 2012
2 3 56 yyy 1.2 2020 2001
3 4 99 yyy 5.0 2001 2001
4 5 99 xxx 1.0 2020 2012
5 6 13 zzz 3.2 2015 2000
6 7 12 zzz 1.0 2000 2000
7 8 88 eee 2.2 2003 2001
8 9 45 eee 0.2 2004 2001
last_year nb_fave_rating
0 2020 1.0
1 2020 1.0
2 2020 1.0
3 2020 1.0
4 2020 1.0
5 2015 1.0
6 2015 1.0
7 2020 0.0
8 2020 0.0
positive_reviews = df[df.star_rating >= 3.0 ].groupby('restaurant_id',as_index=False).agg({'star_rating': 'count'}).rename(columns={'star_rating': 'nb_fave_rating'})
# join back to df
df = df.merge(positive_reviews,how='left',on='restaurant_id').fillna(0)
# display(df)
rating_id user_id restaurant_id star_rating rating_year first_year last_year nb_fave_rating
0 1 56 xxx 2.3 2012 2012 2020 1.0
1 2 13 xxx 3.7 2012 2012 2020 1.0
2 3 56 yyy 1.2 2020 2001 2020 1.0
3 4 99 yyy 5.0 2001 2001 2020 1.0
4 5 99 xxx 1.0 2020 2012 2020 1.0
5 6 13 zzz 3.2 2015 2000 2015 1.0
6 7 12 zzz 1.0 2000 2000 2015 1.0
7 8 88 eee 2.2 2003 2001 2020 0.0
8 9 45 eee 0.2 2004 2001 2020 0.0
%timeit比较
- 给出问题中的9行数据框
df
# create a test dataframe of 1,125,000 rows
dfl = pd.concat([df] * 125000).reset_index(drop=True)
# test with transform
def add_rating_transform(df):
return df.groupby('restaurant_id')['star_rating'].transform(lambda x: int((x>=3).sum()))
%timeit add_rating_transform(dfl)
[out]:
222 ms ± 9.01 ms per loop (mean ± std. dev. of 7 runs,1 loop each)
# test with map
def add_rating_map(df):
filtered_data = df[df['star_rating'] >= 3]
d = filtered_data.groupby('restaurant_id')['star_rating'].count().to_dict()
return df['restaurant_id'].map(d).fillna(0).astype(int)
%timeit add_rating_map(dfl)
[out]:
105 ms ± 1.56 ms per loop (mean ± std. dev. of 7 runs,10 loops each)
# test with merge
def add_rating_merge(df):
positive_reviews = df[df.star_rating >= 3.0 ].groupby('restaurant_id',as_index=False).agg({'star_rating': 'count'}).rename(columns={'star_rating': 'nb_fave_rating'})
return df.merge(positive_reviews,on='restaurant_id').fillna(0)
%timeit add_rating_merge(dfl)
[out]:
639 ms ± 26.5 ms per loop (mean ± std. dev. of 7 runs,1 loop each)
计算评分大于等于3.0的情况
df['nb_fave_rating'] = df.groupby('restaurant_id')['star_rating'].transform(lambda x: x.ge(3.0).sum()).astype(np.int)
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