跨行分组 Pandas 数据框

Question

我有一个这样的 csv：

client1,client2,client3,client4,client5,client6,amount
,,,Comp1,,,4.475000
,,,Comp2,,,16.305584
,,,Comp3,,,4.050000
Comp2,Comp1,,Comp4,,,21.000000
,,,Comp4,,,30.000000
,Comp1,,Comp2,,,5.137500
,,,Comp3,,,52.650000
,,,Comp1,,,2.650000
Comp3,,,Comp3,,,29.000000
Comp5,,,Comp2,,,20.809000
Comp5,,,Comp2,,,15.100000
Comp5,,,Comp2,,,52.404000

将其读入pandas数据框df后，我想分两步进行聚合：

第一步：

首先，我总结了金额：

client1 client2 client3 client4 client5 client6  amount
                        Comp1                    7.125000
                        Comp2                    16.305584
                        Comp3                    56.700000
                        Comp4                    30.000000
         Comp1          Comp2                    5.137500
Comp2    Comp1          Comp4                    21.000000
Comp3                   Comp3                    29.000000
Comp5                   Comp2                    88.313000  

然后，我想按每个客户端名称进行聚合，这样如果涉及多个客户端，如第 5 组，则 5.1375 必须在 Comp1 和 Comp2 之间平均分配。试过这样：

df.groupby(['client1','client2','client3','client4','client5','client6']).apply(lambda x: x['amount'].sum()/len(x) if x.any().nunique()>=1 else x['amount'].sum())



client1 client2 client3 client4 client5 client6 0
0                           Comp1                   3.562500
1                           Comp2                   16.305584
2                           Comp3                   28.350000
3                           Comp4                   30.000000
4           Comp1           Comp2                   5.137500
5   Comp2   Comp1           Comp4                   21.000000
6   Comp3                   Comp3                   29.000000
7   Comp5                   Comp2                   29.437667

预期输出是：

Client Amount 
Comp1  4.475+21/3+5.1375/2+2.65 = 16.69375
Comp2  16.305584+21/3+20.809/2+15.10/2+52.404/2 = 67.462084
Comp3  4.05+52.65+29 = 85.7
Comp4  21/3+30 = 37
Comp5  20.809/2+15.10/2+52.404/2 = 44.1565

我尝试使用sum(axis=0)，但没有用。

Answer 1

我们可以在这里使用一点数学

cols = ['amount'] 
# Divide the amount by non null fields 
df['new'] = df['amount']/df.drop(cols,1).notnull().sum(1)

#Set the index as new by droping amount column, unstack and drop the nans.
x = df.drop(cols,1).set_index('new').unstack().dropna()

#Create dataframe just from amount and the clients
ndf = pd.DataFrame({'amount':x.index.droplevel(0).values,'clients':x.values})

#Groupby client and get the sum 
ndf.groupby('clients').sum()

输出：

数量 客户 Comp1 16.360417 比较 2 69.697501 Comp3 85.700000 比较 4 36.666667 Comp5 44.156500

Answer 2

我会这样组织它：

d = df.drop('amount', 1)  # new df without `amount`
a = df.amount             # separate series of `amount`
c = d.count(1)            # count of non-null values

a.div(c).repeat(c).groupby(d.stack().values).sum()

Comp1    16.693750
Comp2    70.030834
Comp3    85.700000
Comp4    37.000000
Comp5    44.156500
dtype: float64