python - 用于 cumsum 的 Groupby 函数并重置其索引

Question

我的条件很简单:

如果该月的总和//100:

print sum and reset cumsum index

其他:

keep cumsumming

这是我的数据:

data = dict(
    Year=['2018', '2018', '2018', '2018', '2018', '2017', '2017', '2017'],
    Month=['08', '08', '04', '05', '05', '06', '02', '01'],
    Money=[26, 50, 25, 45, 20, 36, 84, 24]
)

这是我的尝试:

df = pd.DataFrame(data)
df = df.groupby(['Year', 'Month']).sum()
df['cum_y'] = df.groupby(['Year']).Money.cumsum() 


df['cum_m'] = df.groupby([lambda x: x // 100], level=0).Money.cumsum()

df['cum_m'] = df.groupby(lambda x: [x if x // 100 else None]).Money.cumsum()

df['cum_m'] = df.groupby(['Money']).agg(lambda x: x // 100).cumsum()

我想要类似的东西:

            Money  cum_y  cum_m (Payout actually)
Year Month
2017 01        24    24     x    (means None)
     02        84   108    108 - reset cumsum counter()
     06        36   144     x    (36)
2018 04        25    25     x    (61)
     05        65    90    126 - reset cumsum counter()
     08        76   166     x    (76)

Answer 1

我知道应该尽可能避免迭代，但这里有一个使用迭代的解决方案:

total = 0
Cumsum = []
for item in df.Money:
    total += item
    if total < 100:
        Cumsum.append(np.nan)
    else:
        Cumsum.append(total)
        total = 0

df['Cumsum'] = Cumsum

输出:

               Money    Cumsum
Year    Month       
2017    01     24       NaN
        02     84       108.0
        06     36       NaN
2018    04     25       NaN
        05     65       126.0
        08     76       NaN