英文:
Calculation problem after pandas grouping
问题
在pandas中，要实现按日期分组，在每个日期组中，将列A和B的每一行相乘，然后相加，再除以日期组中所有B列的总和，可以尝试以下方式：
(df.groupby('date').apply(lambda group: (group['A'] * group['B']).sum() / group['B'].sum()).reset_index(name='result'))
这将返回一个包含日期、结果的DataFrame。
英文:
In pandas, how to achieve, grouped by date, in each date group, each row of column A and B multiplied and then summed, and then divided by the sum of all B columns in the date group.
I have tried:
(df.groupby('date')['A','B'].transform(lambda x: (x['A'] * x['B']).sum()).div(df.groupby('date')['B'].agg('sum')))
and:
(df.groupby('date').transform(lambda x: (x['A'] * x['B']).sum()).div(df.groupby('date')['B'].agg('sum')))
both showed:
> KeyError: 'A'
答案1
得分: 1
你应该使用 .apply() 而不是 .transform()，如下所示：
df.groupby('date').apply(lambda x: (x['A'] * x['B']).sum() / x['B'].sum())
通过使用 .apply()，lambda 函数中的 x 变量将被识别为代表每个分组的 DataFrame 对象，可以分别索引各个列。
.transform() 方法只将 x 视为代表一次仅一个列的 Series 对象。
来自这个答案：
`apply` 和 `transform` 之间的两个主要区别
transform和applygroupby 方法之间有两个主要区别。
- 输入:
apply隐式地将每个组的所有列作为DataFrame传递给自定义函数。- 而
transform将每个组的每列分别作为Series传递给自定义函数。- 输出:
- 传递给**
apply的自定义函数可以返回标量，或Series或DataFrame（甚至是numpy数组或列表）。- 传递给**
transform的自定义函数必须返回一个序列**（长度与组相同的一维Series、数组或列表）。因此，
transform仅逐一处理一列Series，而apply一次处理整个DataFrame。
示例
示例数据：
import pandas as pddata = {'date': ['01/01/1999', '02/01/1999', '03/01/1999', '03/01/1999'],'A': [2, 4, 7, 4],'B': [5, 7, 9, 6]}df = pd.DataFrame(data)print(df)
date A B0 01/01/1999 2 51 02/01/1999 4 72 03/01/1999 7 93 03/01/1999 4 6
对于 03/01/1999 的计算将是：
((7 * 9) + (4 * 6)) / (9 + 6) # = 5.8
使用 .apply() 计算每个日期组的结果：
df_ab_calc = df.groupby('date').apply(lambda x: (x['A'] * x['B']).sum() / x['B'].sum())print(df_ab_calc)
date01/01/1999 2.002/01/1999 4.003/01/1999 5.8
英文:
You should use .apply() instead of .transform(), as follows:
df.groupby('date').apply(lambda x: (x['A'] * x['B']).sum() / x['B'].sum())
By using .apply(), the x variables in the lambda function will be recognised as DataFrame objects representing each group, which can each be indexed for individual columns.
The .transform() method only treats x as a Series object representing only one column at a time.
From this answer:
<blockquote>
<h3>Two major differences between apply and transform</h3>
There are two major differences between the transform and apply groupby methods.
- Input:
applyimplicitly passes all the columns for each group as a DataFrame to the custom function.- while
transformpasses each column for each group individually as a Series to the custom function.
- Output:
- The custom function passed to
applycan return a scalar, or a Series or DataFrame (or numpy array or even list). - The custom function passed to
transformmust return a sequence (a one dimensional Series, array or list) the same length as the group.
- The custom function passed to
So, transform works on just one Series at a time and apply works on the entire DataFrame at once.
</blockquote>
Example
Sample data:
import pandas as pddata = {'date': ['01/01/1999', '02/01/1999', '03/01/1999', '03/01/1999'],'A': [2, 4, 7, 4],'B': [5, 7, 9, 6]}df = pd.DataFrame(data)print(df)
date A B0 01/01/1999 2 51 02/01/1999 4 72 03/01/1999 7 93 03/01/1999 4 6
The calculation for '03/01/1999' would be:
((7 * 9) + (4 * 6)) / (9 + 6) # = 5.8
Calculation for each date group using .apply():
df_ab_calc = df.groupby('date').apply(lambda x: (x['A'] * x['B']).sum() / x['B'].sum())print(df_ab_calc)
date01/01/1999 2.002/01/1999 4.003/01/1999 5.8
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