Groupby year并将函数应用于另一列—Python、Pandas
我有一个多索引、多列数据帧:Groupby year并将函数应用于另一列—Python、Pandas,python,pandas,group-by,apply,Python,Pandas,Group By,Apply,我有一个多索引、多列数据帧: start A B C D 0 2019 35.156667 51.621111 18.858684 1 1 2019 NaN 50.211905 18.991290 -1 2 2019 42.836250 58.778235 18.788889 1 3 2020 NaN 8.188000 17.805833
start A B C D
0 2019 35.156667 51.621111 18.858684 1
1 2019 NaN 50.211905 18.991290 -1
2 2019 42.836250 58.778235 18.788889 1
3 2020 NaN 8.188000 17.805833 1
4 2020 42.568000 55.907143 17.300000 -1
5 2021 46.458333 42.293750 26.322500 1
6 2021 43.675000 60.475000 29.520000 1
每年(列“开始”),如果D>0,我想用正向值填充列a中的NaN,如果D则用反向值填充列a中的NaN,并使用每组的正向和反向填充,然后通过以下方式设置值:
谢谢你的回答,你确定这一点:df1['ffill',df1['bfill']?我得到键错误,因为它将其视为列标签。顺便说一句,数据框更大,每年都有许多行。@Luca91-是的,它位于另一个新的数据框
df1
列中。@Luca91-答案已编辑,如何为您print(df1)
?正确!我仍然在挣扎,因为这是一个多列数据帧,其中a是上一列的子列,通常我通过df[('UpColumn','a')]调用,所以现在它变成了一个3级列,你知道如何调用它吗?它与df[('UpColumn','a','ffill')]一起工作谢谢!
start A B C D
0 2019 35.156667 51.621111 18.858684 1
1 2019 35.156667 50.211905 18.991290 -1
2 2019 42.836250 58.778235 18.788889 1
3 2020 42.568000 8.188000 17.805833 1
4 2020 42.568000 55.907143 17.300000 -1
5 2021 46.458333 42.293750 26.322500 1
6 2021 43.675000 60.475000 29.520000 1
df[['A','D']] = df.groupby('start').apply(lambda x: x['A'].fillna(method='ffill') if x['D']>0 else x['A'].fillna(method='bfill'))
The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
df1 = df.groupby('start')['A'].agg(['ffill','bfill'])
print (df1)
ffill bfill
0 35.156667 35.156667
1 35.156667 42.836250
2 42.836250 42.836250
3 NaN 42.568000
4 42.568000 42.568000
5 46.458333 46.458333
6 43.675000 43.675000
print (df1.columns)
Index(['ffill', 'bfill'], dtype='object')
df['A'] = np.where(df['D'] < 0, df1['ffill'], df1['bfill'])
print (df)
start A B C D
0 2019 35.156667 51.621111 18.858684 1
1 2019 35.156667 50.211905 18.991290 -1
2 2019 42.836250 58.778235 18.788889 1
3 2020 42.568000 8.188000 17.805833 1
4 2020 42.568000 55.907143 17.300000 -1
5 2021 46.458333 42.293750 26.322500 1
6 2021 43.675000 60.475000 29.520000 1
df['A'] = df.groupby('start')['A'].apply(lambda x: x.ffill().bfill())
print (df)
start A B C D
0 2019 35.156667 51.621111 18.858684 1
1 2019 35.156667 50.211905 18.991290 -1
2 2019 42.836250 58.778235 18.788889 1
3 2020 42.568000 8.188000 17.805833 1
4 2020 42.568000 55.907143 17.300000 -1
5 2021 46.458333 42.293750 26.322500 1
6 2021 43.675000 60.475000 29.520000 1