我正在使用python3.7在pandas中使用以下数据框
data = {'s':['a','a','a','a','b','b'],
'cp':['C','P','C','C','C','P'],
'st':[300,300,300,300,310,310],
'qty':[3000,3000,3000,6000,9000,3000],
'p':[16,15,14,10,8,12]}
df=pd.DataFrame(data)
df['t']=df['p']*df['qty']
df['ct']=df['t'].cumsum()
df
s cp st qty p t ct
0 a C 300 3000 16 48000 48000
1 a P 300 3000 15 45000 93000
2 a C 300 3000 14 42000 135000
3 a C 300 6000 10 60000 195000
4 b C 310 9000 8 72000 267000
5 b P 310 3000 12 36000 303000
我想基于S创建两个单独的列,分别为x和y,CP值具有累计数量qty
col x = cumm qty where cp="c" group by col s
col y = cumm qty where cp=P group by col s
s cp st qty p t ct x y
0 a C 300 3000 16 48000 48000 3000 0
1 a P 300 3000 15 45000 93000 3000 3000
2 a C 300 3000 14 42000 135000 6000 3000
3 a C 300 6000 10 60000 195000 12000 3000
4 b C 310 9000 8 72000 267000 9000 0
5 b P 310 3000 12 36000 303000 9000 3000
I tried something like this
df['x']=df.loc[df['p']>0].groupby(['s'])['s','cp','qty','ct'].apply(lambda x:x['qty'].cumsum() if x['cp']=="C" else 0)
出现以下错误 系列的真实值是不明确的。使用a.empty,a.bool(),a.item(),a.any()或a.all()。
我也不知道它将在哪里给我预期的输出。你能帮我吗?
答案 0 :(得分:4)
这是我的解决方法
@Nullable
@Override
public View onCreateView(LayoutInflater inflater, @Nullable ViewGroup
container, @Nullable Bundle savedInstanceState) {
View v = inflater.inflate(R.layout.fragment_home, container, false);
btnFertilizers = (Button)v.findViewById(R.id.btnFertilizers);
return v;
}
输出:
df['x'] = df['qty'].mul(df['cp'].eq('C')).groupby(df['s']).cumsum()
df['y'] = df['qty'].mul(df['cp'].eq('P')).groupby(df['s']).cumsum()
答案 1 :(得分:2)
您可以使用:
df['X']=df.where(df['cp'].eq('C')).groupby('s')['qty'].cumsum().fillna(df['qty'])
df['Y']=0
df.loc[~df['cp'].shift(-1).eq('P'),'Y']=df.loc[df['cp'].eq('P'),'qty']
df=df.ffill()
s cp st qty p t ct x y
0 a C 300 3000 16 48000 48000 3000 0
1 a P 300 3000 15 45000 93000 3000 3000
2 a C 300 3000 14 42000 135000 6000 3000
3 a C 300 6000 10 60000 195000 12000 3000
4 b C 310 9000 8 72000 267000 9000 0
5 b P 310 3000 12 36000 303000 9000 3000