我有一个熊猫df，其中有2列 Day （日期时间格式的日期）和故障数。
我想创建两个新列，第一列是 Previous 故障事件的天数（已经完成），第二列是 Next 的天数崩溃事件（我很难做到）。
Day Number of breakdowns Days from Previous Breakdown Event
2017-01-09 0.0 0
2017-01-12 0.0 0
2017-01-13 0.0 0
2017-01-14 0.0 0
2017-01-16 1.0 0
2017-01-17 0.0 1
2017-01-18 0.0 2
2017-01-19 1.0 0
2017-01-20 0.0 1
2017-01-21 0.0 2
2017-01-23 1.0 0
上次发生故障的天数计算发生故障以来经过的天数。
代码：
s = df.groupby(df['Number of breakdowns'].ne(0).cumsum())['Day'].transform('first')
df['Days from Previous Breakdown Event'] = (df['Day'] - s).dt.days
zeros_index = df['Number of breakdowns'].ne(0).idxmax()
df.loc[:zeros_index,'Days from Previous Breakdown Event'] = 0
我需要“下一次的天数”细分列，我希望这样：
Day Number of breakdowns Days from Next Breakdown Event
2017-01-09 0.0 7
2017-01-12 0.0 4
2017-01-13 0.0 3
2017-01-14 0.0 2
2017-01-16 1.0 0
2017-01-17 0.0 2
2017-01-18 0.0 1
2017-01-19 1.0 0
2017-01-20 0.0 3
2017-01-21 0.0 2
2017-01-23 1.0 0
答案 0 :(得分：1)
通过iloc[::-1]与transform和last交换订单，然后也交换s - df['Day']：
s = df.groupby(df['Number of breakdowns'].iloc[::-1].ne(0).cumsum())['Day'].transform('last')
df['Days from Next Breakdown Event'] = (s - df['Day']).dt.days
print (df)
Day Number of breakdowns Days from Previous Breakdown Event \
0 2017-01-09 0.0 0
1 2017-01-12 0.0 0
2 2017-01-13 0.0 0
3 2017-01-14 0.0 0
4 2017-01-16 1.0 0
5 2017-01-17 0.0 1
6 2017-01-18 0.0 2
7 2017-01-19 1.0 0
8 2017-01-20 0.0 1
9 2017-01-21 0.0 2
10 2017-01-23 1.0 0
Days from Next Breakdown Event
0 7
1 4
2 3
3 2
4 0
5 2
6 1
7 0
8 3
9 2
10 0
详细信息：
print (s)
0 2017-01-16
1 2017-01-16
2 2017-01-16
3 2017-01-16
4 2017-01-16
5 2017-01-19
6 2017-01-19
7 2017-01-19
8 2017-01-23
9 2017-01-23
10 2017-01-23
Name: Day, dtype: datetime64[ns]