使用Pandas链式方法筛选组内行，排除两类创建量均为0的用户

Got it, let's solve this problem using pandas' chained methods as you requested. The goal is to exclude users where all their rows (both is_manually=True and is_manually=False) have created_per_week=0—like user_id 50 in your example.

Step 1: Recreate your sample DataFrame #

First, let's set up the data to work with:

import pandas as pd

df = pd.DataFrame({
    'user_id': [10, 10, 33, 33, 50, 50],
    'is_manually': [True, False, True, False, True, False],
    'created_per_week': [59, 90, 0, 64, 0, 0]
})

Step 2: Chained Groupby Filter Method #

The cleanest way to do this is using groupby().filter() in a chain. This method lets you apply a condition to each user group and keep only the groups that meet your criteria:

filtered_df = df.groupby('user_id').filter(lambda group: not (group['created_per_week'] == 0).all())

How this works:

• groupby('user_id'): Groups the DataFrame by each unique user.
• lambda group: not (group['created_per_week'] == 0).all(): For each group, we check if all values in created_per_week are 0. We negate this condition (not) so we keep groups where at least one row has a non-zero value.

Step 3: Alternative Chained Method with Transform #

If you prefer using a mask to filter directly on the original DataFrame, you can use transform() in a chain:

filtered_df = df[df.groupby('user_id')['created_per_week'].transform(lambda x: not (x == 0).all())]

How this works:

• transform() returns a boolean array with the same length as the original DataFrame, marking which rows belong to groups that aren't all zeros.
• We use this array to index the original DataFrame and keep only the relevant rows.

Result #

Both methods will give you this filtered DataFrame:

user_id  is_manually  created_per_week
0       10         True                59
1       10        False                90
2       33         True                 0
3       33        False                64

Notice user_id 50 is excluded, as both their rows had created_per_week=0.

内容的提问来源于stack exchange，提问作者ytu