pandas 数据框groupby并加入 [英] pandas dataframe groupby and join
本文介绍了 pandas 数据框groupby并加入的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
让我们假设有这个:
np.random.seed(123)
df = pd.DataFrame({'A' : ['foo', 'bar', 'foo', 'bar',
'foo', 'bar', 'foo', 'foo'],
'B' : ['one', 'one', 'two', 'three',
'two', 'two', 'one', 'three'],
'C' : np.random.randn(8),
'D' : np.random.randn(8)})
因此数据框如下所示:
A B C D
0 foo one -1.085631 1.265936
1 bar one 0.997345 -0.866740
2 foo two 0.282978 -0.678886
3 bar three -1.506295 -0.094709
4 foo two -0.578600 1.491390
5 bar two 1.651437 -0.638902
6 foo one -2.426679 -0.443982
7 foo three -0.428913 -0.434351
我想将df
按B
分组,计算C
列的总和乘以D
列的总和,最后将每个分组结果与原始df
合并.
在Python中:
I want to group the df
by B
, calculate the sum of C
column multiplied by the sum of D
column for each group and finally joining this grouped-by result with the original df
.
In Python:
grouped = df.groupby('B').apply(lambda group: sum(group['C'])*sum(group['D'])).reset_index()
grouped.columns = ['B', 'new_value']
df.join(grouped.set_index('B'), on='B')
还有一种更有效的 pythonic 方法来解决此类问题吗?
There exists a more pythonic and efficient way to solve this kind of problem?
推荐答案
解决方案1:
In [25]: df.groupby('B')['C','D'].transform('sum').prod(1)
Out[25]:
0 0.112635
1 0.112635
2 0.235371
3 1.023841
4 0.235371
5 0.235371
6 0.112635
7 1.023841
dtype: float64
解决方案2:
In [18]: grp = df.groupby('B')
In [19]: grp['C'].transform('sum') * grp['D'].transform('sum')
Out[19]:
0 0.112635
1 0.112635
2 0.235371
3 1.023841
4 0.235371
5 0.235371
6 0.112635
7 1.023841
dtype: float64
演示:
In [20]: df
Out[20]:
A B C D
0 foo one -1.085631 1.265936
1 bar one 0.997345 -0.866740
2 foo two 0.282978 -0.678886
3 bar three -1.506295 -0.094709
4 foo two -0.578600 1.491390
5 bar two 1.651437 -0.638902
6 foo one -2.426679 -0.443982
7 foo three -0.428913 -0.434351
In [21]: grp = df.groupby('B')
In [22]: df['new'] = grp['C'].transform('sum') * grp['D'].transform('sum')
In [23]: df
Out[23]:
A B C D new
0 foo one -1.085631 1.265936 0.112635
1 bar one 0.997345 -0.866740 0.112635
2 foo two 0.282978 -0.678886 0.235371
3 bar three -1.506295 -0.094709 1.023841
4 foo two -0.578600 1.491390 0.235371
5 bar two 1.651437 -0.638902 0.235371
6 foo one -2.426679 -0.443982 0.112635
7 foo three -0.428913 -0.434351 1.023841
In [26]: df['new2'] = df.groupby('B')['C','D'].transform('sum').prod(1)
In [27]: df
Out[27]:
A B C D new new2
0 foo one -1.085631 1.265936 0.112635 0.112635
1 bar one 0.997345 -0.866740 0.112635 0.112635
2 foo two 0.282978 -0.678886 0.235371 0.235371
3 bar three -1.506295 -0.094709 1.023841 1.023841
4 foo two -0.578600 1.491390 0.235371 0.235371
5 bar two 1.651437 -0.638902 0.235371 0.235371
6 foo one -2.426679 -0.443982 0.112635 0.112635
7 foo three -0.428913 -0.434351 1.023841 1.023841
检查:
In [28]: df.new.eq(df.new2).all()
Out[28]: True
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