我有一个函数minmax,它基本上迭代事务的数据帧。我想计算一组包括id的计算,所以accountstart和accountend是计算的两个字段。其目的是将此计算作为我的月份和账目
因此,当我这样做时:
df1 = df.loc[df['accountNo']==10]
minmax(df1) it works.
我不能做的是:
df.groupby('accountNo').apply(minmax)
当我这样做时:
grouped = df.groupby('accountNo')
for i,j in grouped:
print(minmax(j))
它进行计算,打印结果,但不打印它会抱怨KeyError:-1,即itertools。那么阿克沃德
如何在熊猫身上解决这个问题
def minmax(x):
dfminmax = {}
accno = set(x['accountNo'])
accno = repr(accno)
kgroup = x.groupby('monthStart')['cumsum'].sum()
maxt = x['startbalance'].max()
kgroup = pd.DataFrame(kgroup)
kgroup['startbalance'] = 0
kgroup['startbalance'][0] = maxt
kgroup['endbalance'] = 0
kgroup['accountNo'] = accno
kgroup['accountNo'] = kgroup['accountNo'].str.strip('{}.0')
kgroup.reset_index(inplace=True)
for idx, row in kgroup.iterrows():
if kgroup.loc[idx,'startbalance']==0:
kgroup.loc[idx,'startbalance']=kgroup.loc[idx-1,'endbalance'],
if kgroup.loc[idx,'endbalance']==0:
kgroup.loc[idx,'endbalance'] =
kgroup.loc[idx,'cumsum']+kgroup.loc[idx,'startbalance']
dfminmax['monthStart'].append(kgroup['monthStart'])
dfminmax['startbalance'].append(kgroup['startbalance'])
dfminmax['endbalance'].append(kgroup['endbalance'])
dfminmax['accountNo'].append(kgroup['accountNo'])
return dfminmax
.apply()
将pandas
{DataFrames
。使用.agg
,如df.groupby('accountNo').agg(yourfunction)
应该会产生更好的结果。请务必查看documentation以了解有关实现的详细信息相关问题 更多 >
编程相关推荐