Question

我已经删除了公司未向给定period收取任何费用的行（例如，revenue == 0的行）。

以下是结算数据的示例：

import numpy as np
import pandas as pd

data = {
    'account_id': ['111','111','222','333','666','666','111','222','333','666','666'],
    'company': ['initech','initech','jackson steinem & co','ingen','enron','enron','initech','jackson steinem & co','ingen','enron','enron'],
    'billing_type': ['subscription','discount','subscription','subscription','subscription','discount','subscription','subscription','subscription','subscription','discount'],
    'period': ['2012-10-31','2012-10-31','2012-10-31','2012-10-31','2012-10-31','2012-10-31','2012-11-30','2012-11-30','2012-11-30','2012-11-30','2012-11-30'],
    'revenue':[39.95,-39.95,199.95,299.95,499.95,-499.95,39.95,199.95,299.95,499.95,-499.95]
}
df = pd.DataFrame(data)
df['period'] = pd.to_datetime(df['period'],format='%Y-%m-%d')

这会生成如下数据框：

In [16]: df
Out[16]:
   account_id  billing_type               company     period  revenue
0         111  subscription               initech 2012-10-31    39.95
1         111      discount               initech 2012-10-31   -39.95
2         222  subscription  jackson steinem & co 2012-10-31   199.95
3         333  subscription                 ingen 2012-10-31   299.95
4         666  subscription                 enron 2012-10-31   499.95
5         666      discount                 enron 2012-10-31  -499.95
6         111  subscription               initech 2012-11-30    39.95
7         222  subscription  jackson steinem & co 2012-11-30   199.95
8         333  subscription                 ingen 2012-11-30   299.95
9         666  subscription                 enron 2012-11-30   499.95
10        666      discount                 enron 2012-11-30  -499.95

我需要做的是删除给定revenue / company period加0的行。因此，举例来说，我需要删除所有安然的行，但只删除2012年10月的Initech时段：

In [17]: df.groupby(['company','period'])['revenue'].sum()
Out[17]:
company               period
enron                 2012-10-31      0.00
                      2012-11-30      0.00
ingen                 2012-10-31    299.95
                      2012-11-30    299.95
initech               2012-10-31      0.00
                      2012-11-30     39.95
jackson steinem & co  2012-10-31    199.95
                      2012-11-30    199.95

number other posts解决了类似情况，但我无法找到任何可以帮助/解释如何完成此请求的内容。

Answer 1

你可以使用transform制作一个帧大小的掩码，然后用它来选择：

>>> keep = df.groupby(["company", "period"])["revenue"].transform(sum) != 0
>>> df.loc[keep]
  account_id  billing_type               company     period  revenue
2        222  subscription  jackson steinem & co 2012-10-31   199.95
3        333  subscription                 ingen 2012-10-31   299.95
6        111  subscription               initech 2012-11-30    39.95
7        222  subscription  jackson steinem & co 2012-11-30   199.95
8        333  subscription                 ingen 2012-11-30   299.95

这是有效的，因为transform获取groupby结果并将其“广播”回主索引：

>>> df.groupby(["company", "period"])["revenue"].transform(sum)
0       0.00
1       0.00
2     199.95
3     299.95
4       0.00
5       0.00
6      39.95
7     199.95
8     299.95
9       0.00
10      0.00
dtype: float64
>>> df.groupby(["company", "period"])["revenue"].transform(sum) != 0
0     False
1     False
2      True
3      True
4     False
5     False
6      True
7      True
8      True
9     False
10    False
dtype: bool

从数据框中删除总和为零的行

1 个答案: