我有一个如下数据框：
row_no stock_name last_price Var1
1 SAIL 501.00 0
2 SAIL 501.60 23
3 SAIL 500.00 0
4 SAIL 499.10 0
5 SAIL 499.40 0
6 SAIL 499.40 0
7 SAIL 502.00 0
8 SAIL 497.95 0
9 SAIL 495.55 20
10 SAIL 496.75 0
11 SAIL 496.75 0
12 SAIL 513.00 0
13 SAIL 497.00 0
14 SAIL 497.20 0
15 SAIL 497.00 0
16 SAIL 494.00 0
17 SAIL 497.00 0
18 SAIL 497.00 0
19 SAIL 497.00 0
20 SAIL 496.60 -9
21 SAIL 497.25 0
当Var1不为零时，需要计算最大和最小last_price。
当当前Var1不等于零时，将在Var1的先前非零值之间计算最大和最小last_price。
示例： 对于第9行：计算的最大last_price为Max（从第9行到第2行的last_price）= 502，最小的last_price作为Min（从第9行到第2行的last_price）= 495.55
我想要如下所示的最终数据框：
row_no stock_name last_price Var1 Max_LTP Min_LTP
1 SAIL 501 0 0 0 0
2 SAIL 501.6 23 501.6 501
3 SAIL 500 0 0 0 0
4 SAIL 499.1 0 0 0
5 SAIL 499.4 0 0 0
6 SAIL 499.4 0 0 0
7 SAIL 502 0 0 0 0
8 SAIL 497.95 0 0 0
9 SAIL 495.55 20 502 495.55
10 SAIL 496.75 0 0 0
11 SAIL 496.75 0 0 0
12 SAIL 513 0 0 0 0
13 SAIL 497 0 0 0 0
14 SAIL 497.2 0 0 0
15 SAIL 497 0 0 0 0
16 SAIL 494 0 0 0 0
17 SAIL 497 0 0 0 0
18 SAIL 497 0 0 0 0
19 SAIL 497 0 0 0 0
20 SAIL 496.6 -9 513 494
21 SAIL 497.25 0 0 0 0
答案 0 :(得分：3)
使用带有shift和cumsum的Var1创建组密钥，然后创建groupby，并将min和max的值分配回原始数据帧< / p>
mask=df.Var1.ne(0).shift().fillna(0).cumsum()
s=df.groupby(mask).last_price.agg(['min','max'])
s=s[df['Var1'].ne(0).groupby(mask).agg('any')]
s.index=df.index[df.Var1.ne(0)]
df[['Min_LTP','Max_LTP']]=s
df
Out[274]:
row_no stock_name last_price Var1 Max_LTP Min_LTP
0 1 SAIL 501.00 0 NaN NaN
1 2 SAIL 501.60 23 501.6 501.00
2 3 SAIL 500.00 0 NaN NaN
3 4 SAIL 499.10 0 NaN NaN
4 5 SAIL 499.40 0 NaN NaN
5 6 SAIL 499.40 0 NaN NaN
6 7 SAIL 502.00 0 NaN NaN
7 8 SAIL 497.95 0 NaN NaN
8 9 SAIL 495.55 20 502.0 495.55
9 10 SAIL 496.75 0 NaN NaN
10 11 SAIL 496.75 0 NaN NaN
11 12 SAIL 513.00 0 NaN NaN
12 13 SAIL 497.00 0 NaN NaN
13 14 SAIL 497.20 0 NaN NaN
14 15 SAIL 497.00 0 NaN NaN
15 16 SAIL 494.00 0 NaN NaN
16 17 SAIL 497.00 0 NaN NaN
17 18 SAIL 497.00 0 NaN NaN
18 19 SAIL 497.00 0 NaN NaN
19 20 SAIL 496.60 -9 513.0 494.00
20 21 SAIL 497.25 0 NaN NaN