英文:
Groupby and transform across a group, not within it
问题
df['CumulativeTotal'] = df.groupby('group')['weeklyTotal'].cumsum()
英文:
I'm trying to get the third column of this dataframe, given the first two columns.
I can't work out what to search for, it's like a within version of groupby('group')['weeklyTotal'].cumsum()??
I know I could pull out those two columns, make them distinct, then do the groupby cumsum, but would much prefer to have this within the same dataframe.
To save a tiny bit of pain, here's an example dataframe:
df = pd.DataFrame({'group':['A','A','A','B','B','B','C','C','C'], 'weeklyTotal':[1,1,1,3,3,3,2,2,2]})
| Group | WeeklyTotal | CumulativeTotal |
|---|---|---|
| A | 1 | 1 |
| A | 1 | 1 |
| A | 1 | 1 |
| B | 3 | 4 |
| B | 3 | 4 |
| B | 3 | 4 |
| C | 2 | 6 |
| C | 2 | 6 |
| C | 2 | 6 |
答案1
得分: 1
使用drop_duplicates函数，每个组保留一行，然后计算cumsum和map这些值：
df['CumulativeTotal'] = df['group'].map(df.drop_duplicates(subset='group').set_index('group')['weeklyTotal'].cumsum())
或者，使用mask和duplicated：
df['CumulativeTotal'] = (df['weeklyTotal'].mask(df['group'].duplicated(), 0).cumsum())
输出：
group weeklyTotal CumulativeTotal0 A 1 11 A 1 12 A 1 13 B 3 44 B 3 45 B 3 46 C 2 67 C 2 68 C 2 6
请注意，这是用Python Pandas进行数据处理的代码示例。
英文:
Keep only one row per group with drop_duplicates, compute the cumsum and map the values:
df['CumulativeTotal'] = df['group'].map(df.drop_duplicates(subset='group').set_index('group')['weeklyTotal'].cumsum())
Or, using a mask and duplicated:
df['CumulativeTotal'] = (df['weeklyTotal'].mask(df['group'].duplicated(), 0).cumsum())
Output:
group weeklyTotal CumulativeTotal0 A 1 11 A 1 12 A 1 13 B 3 44 B 3 45 B 3 46 C 2 67 C 2 68 C 2 6
答案2
得分: 0
这是另一种方式：
m = df['group'].ne(df['group'].shift())m.mul(df['weeklyTotal']).cumsum()
输出：
0 11 12 13 44 45 46 67 68 6
英文:
Here is another way:
m = df['group'].ne(df['group'].shift())m.mul(df['weeklyTotal']).cumsum()
Output:
0 11 12 13 44 45 46 67 68 6
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论