I have following df
A B
0 1 10
1 2 20
2 NaN 5
3 3 1
4 NaN 2
5 NaN 3
6 1 10
7 2 50
8 Nan 80
9 3 5
Consisting of repeating sequences from 1-3 seperated by a variable number of NaN's.I want to groupby each this sequences from 1-3 and get the minimum value of column B within these sequences.
Desired Output something like:
B_min
0 1
6 5
Many thanks beforehand
draj
Idea is first remove rows by missing values by DataFrame.dropna, then use GroupBy.cummin by helper Series created by compare A for equal by Series.eq and Series.cumsum, last data cleaning to one column DataFrame:
df = (df.dropna(subset=['A'])
.groupby(df['A'].eq(1).cumsum())['B']
.min()
.reset_index(drop=True)
.to_frame(name='B_min'))
print (df)
B_min
0 1
1 5
All you need to df.groupby() and apply min(). Is this what you are expecting?
df.groupby('A')['B'].min()
Output:
A
1 10
2 20
3 1
Nan 80
If you don't want the NaNs in your group you can drop them using df.dropna()
df.dropna().groupby('A')['B'].min()
