Pandas：按满足条件的列分组
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Pandas: Group by a column that meets a condition
提问by seisgradox
I have a data set with three colums: rating , breed, and dog.
我有一个包含三列的数据集： rating 、breed 和 dog。
import pandas as pd
dogs = {'breed': ['Chihuahua', 'Chihuahua', 'Dalmatian', 'Sphynx'],
'dog': [True, True, True, False],
'rating': [8.0, 9.0, 10.0, 7.0]}
df = pd.DataFrame(data=dogs)
I would like to calculate the meanrating per breed where dog is True. This would be the expected:
我想计算每个品种的平均评分，其中狗是真的。这将是预期的：
breed rating
0 Chihuahua 8.5
1 Dalmatian 10.0
This has been my attempt:
这是我的尝试：
df.groupby('breed')['rating'].mean().where(dog == True)
And this is the error that I get:
这是我得到的错误：
NameError: name 'dog' is not defined
But when I try add the wherecondition I only get errors. Can anyone advise a solution? TIA
但是当我尝试添加where条件时，我只会得到错误。任何人都可以建议解决方案吗？TIA
采纳答案by user3483203
Once you groupby and select a column, your dogcolumn doesn't exist anymore in the context you have selected (and even if it did you are not accessing it correctly).
一旦您分组并选择一列，您的dog列将不再存在于您选择的上下文中（即使存在，您也没有正确访问它）。
Filter your dataframe first, thenuse groupbywith mean
第一过滤您的数据帧，然后用groupby用mean
df[df.dog].groupby('breed')['rating'].mean().reset_index()
breed rating
0 Chihuahua 8.5
1 Dalmatian 10.0
回答by jpp
An alternative solution is to make dogone of your grouper keys. Then filter by dogin a separate step. This is more efficient if you do not want to lose aggregated data for non-dogs.
另一种解决方案是制作dog您的石斑鱼钥匙之一。然后dog在单独的步骤中过滤。如果您不想丢失非狗的聚合数据，这会更有效。
res = df.groupby(['dog', 'breed'])['rating'].mean().reset_index()
print(res)
dog breed rating
0 False Sphynx 7.0
1 True Chihuahua 8.5
2 True Dalmatian 10.0
print(res[res['dog']])
dog breed rating
1 True Chihuahua 8.5
2 True Dalmatian 10.0
