[英]Pandas groupby multiple fields then diff
So my dataframe looks like this:所以我的 dataframe 看起来像这样:
date site country score
0 2018-01-01 google us 100
1 2018-01-01 google ch 50
2 2018-01-02 google us 70
3 2018-01-03 google us 60
4 2018-01-02 google ch 10
5 2018-01-01 fb us 50
6 2018-01-02 fb us 55
7 2018-01-03 fb us 100
8 2018-01-01 fb es 100
9 2018-01-02 fb gb 100
Each site
has a different score depending on the country
.每个
site
都有不同的分数,具体取决于country
。 I'm trying to find the 1/3/5-day difference of score
s for each site
/ country
combination.我正在尝试为每个
site
/ country
/地区组合找到score
的 1/3/5 天差异。
Output should be: Output 应该是:
date site country score diff
8 2018-01-01 fb es 100 0.0
9 2018-01-02 fb gb 100 0.0
5 2018-01-01 fb us 50 0.0
6 2018-01-02 fb us 55 5.0
7 2018-01-03 fb us 100 45.0
1 2018-01-01 google ch 50 0.0
4 2018-01-02 google ch 10 -40.0
0 2018-01-01 google us 100 0.0
2 2018-01-02 google us 70 -30.0
3 2018-01-03 google us 60 -10.0
I first tried sorting by site
/ country
/ date
, then grouping by site
and country
but I'm not able to wrap my head around getting a difference from a grouped object.我首先尝试按
site
/ country
/地区/ date
进行排序,然后按site
和country
/地区分组,但我无法理解与分组的 object 的区别。
First, sort the DataFrame and then all you need is groupby.diff()
:首先,对 DataFrame 进行排序,然后您只需要
groupby.diff()
:
df = df.sort_values(by=['site', 'country', 'date'])
df['diff'] = df.groupby(['site', 'country'])['score'].diff().fillna(0)
df
Out:
date site country score diff
8 2018-01-01 fb es 100 0.0
9 2018-01-02 fb gb 100 0.0
5 2018-01-01 fb us 50 0.0
6 2018-01-02 fb us 55 5.0
7 2018-01-03 fb us 100 45.0
1 2018-01-01 google ch 50 0.0
4 2018-01-02 google ch 10 -40.0
0 2018-01-01 google us 100 0.0
2 2018-01-02 google us 70 -30.0
3 2018-01-03 google us 60 -10.0
sort_values
doesn't support arbitrary orderings. sort_values
不支持任意排序。 If you need to sort arbitrarily (google before fb for example) you need to store them in a collection and set your column as categorical.如果您需要任意排序(例如在 fb 之前使用 google),您需要将它们存储在一个集合中并将您的列设置为分类。 Then sort_values will respect the ordering you provided there.
然后 sort_values 将尊重您在那里提供的排序。
You can shift and substract grouped values:您可以移动和减去分组值:
df.sort_values(['site', 'country', 'date'], inplace=True)
df['diff'] = df['score'] - df.groupby(['site', 'country'])['score'].shift()
Result:结果:
date site country score diff
8 2018-01-01 fb es 100 NaN
9 2018-01-02 fb gb 100 NaN
5 2018-01-01 fb us 50 NaN
6 2018-01-02 fb us 55 5.0
7 2018-01-03 fb us 100 45.0
1 2018-01-01 google ch 50 NaN
4 2018-01-02 google ch 10 -40.0
0 2018-01-01 google us 100 NaN
2 2018-01-02 google us 70 -30.0
3 2018-01-03 google us 60 -10.0
To fill NaN
with 0
use df['diff'].fillna(0, inplace=True)
.要用
0
填充NaN
使用df['diff'].fillna(0, inplace=True)
。
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