Pandas Groupby Min and Max of last N rows of a column

Question

A time series data has 3 columns apart from index, which is time

indexTime,A,B,C

I want to list all As that have last 10 B and C +ve

This means I have to do a

groupby('A')

and then have an AND condition for

last N rows of B.min() > 0 AND last N rows of C.min() > 0

How do I do it ?

09:00,ABC,1,1
09:00,XYZ,15,2
09:01,ABC,2,4
09:01,XYZ,1,2
09:02,ABC,-1,2
09:02,XYZ,1,7
09:03,ABC,3,5
09:03,XYZ,5,2

let us say last 3 in this case XYZ would satisfy the condition as it has both B and C column last 3 rows positive where as ABC does not have all last 3 rows positive

Column B of ABC 09:02 is -1 so it would fail the test even though column C of ABC is all positive. But because of AND condition it would fail

Thus for the condition output would be XYZ as only that satisfies the condition

Answer 1

Use groupby with tail and all for check all True s:

a = df.groupby('A').apply(lambda x: (x.tail(3) > 0).all(1))
print (a)
     09:01  09:02  09:03
A                       
ABC   True  False   True
XYZ   True   True   True

b = a.index[a.all(1)]
print (b)
Index(['XYZ'], dtype='object', name='A')

---

print (a)
A        ABC   XYZ
09:01   True  True
09:02  False  True
09:03   True  True

b = a.columns[a.all()].tolist()
print (b)
['XYZ']