Get first and last Index of a Pandas DataFrame subset - python-3.x

Question

I do got some data within a pandas DataFrame looking like this.
df =
A B
time
0.1 10.0 1
0.15 12.1 2
0.19 4.0 2
0.21 5.0 2
0.22 6.0 2
0.25 7.0 1
0.3 8.1 1
0.4 9.45 2
0.5 3.0 1
Based on the following condition I look for a generic solution to find the first and last index of every subset.
cond = df.B == 2
So far I tried using the groupby concept but without the expected result.
df_1 = cond.reset_index()
df_2 = df_1.groupby(df_1['B']).agg(['first','last']).reset_index()
This is the output I got.
B time
first last
0 False 0.1 0.5
1 True 0.15 0.4
This is the output I like to get.
B time
first last
0 False 0.1 0.1
1 True 0.15 0.22
2 False 0.25 0.3
3 True 0.4 0.4
3 False 0.5 0.5
How can I accomplish this by a more or less generic approach?

Answer 1

Create helper Series by Series.shift with Series.ne and cumulative sum by Series.cumsum for groups by consecutive values, then for aggregation is used dictionary:
df_1 = df_1.reset_index()
df_1.B = df_1.B == 2
g = df_1.B.ne(df_1.B.shift()).cumsum()
df_2 = df_1.groupby(g).agg({'B':'first','time': ['first','last']}).reset_index(drop=True)
print (df_2)
B time
first first last
0 False 0.10 0.10
1 True 0.15 0.22
2 False 0.25 0.30
3 True 0.40 0.40
4 False 0.50 0.50
If want avoid MultiIndex use named aggregations:
df_1 = df_1.reset_index()
df_1.B = df_1.B == 2
g = df_1.B.ne(df_1.B.shift()).cumsum()
df_2 = df_1.groupby(g).agg(B=('B','first'),
first=('time','first'),
last=('time','last')).reset_index(drop=True)
print (df_2)
B first last
0 False 0.10 0.10
1 True 0.15 0.22
2 False 0.25 0.30
3 True 0.40 0.40
4 False 0.50 0.50