Sum of Maximum Postive and Negative Consecutive rows in pandas

Question:

I have a dataframe df as below:

# Import pandas library
import pandas as pd
 
# initialize list elements
data = [10,-20,30,40,-50,60,12,-12,11,1,90,-20,-10,-5,-4]
 
# Create the pandas DataFrame with column name is provided explicitly
df = pd.DataFrame(data, columns=['Numbers'])
 
# print dataframe.
df

I want the sum of count of max consecutive positive and negative numbers.

I am able to get count of max consucutive positive and negative numbers, but unable to sum using below code.

my code:

streak = df['Numbers'].to_list()
from collections import defaultdict
from itertools import groupby
counter = defaultdict(list)
for key, val in groupby(streak, lambda ele: "plus" if ele >= 0 else "minus"):
    counter[key].append(len(list(val)))
lst = []
for key in ('plus', 'minus'):
    lst.append(counter[key])
print("Max Pos Count " + str(max(lst[0])))
print("Max Neg Count : " + str(max(lst[1])))

Current Output:

Max Pos Count 3
Max Neg Count : 4

I am struggling to get sum of max consuctive positive and negative.

Expected Output:

Sum Pos Max Consecutive: 102
Sum Neg Max Consecutive: -39
Asked By: Divyank

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Answers:

The logic is unclear, the way I understand it is:

  • group by successive negative/positive values
  • get the longest stretch per group
  • compute the sum

You can use:

m = df['Numbers'].gt(0).map({True: 'positive', False: 'negative'})
df2 = df.groupby([m, m.ne(m.shift()).cumsum()])['Numbers'].agg(['count', 'sum'])
out = df2.loc[df2.groupby(level=0)['count'].idxmax(), 'sum'].droplevel(1)

Output:

Numbers
negative    -39
positive    102
Name: sum, dtype: int64

Intermediate df2:

                  count  sum
Numbers  Numbers            
negative 2            1  -20
         4            1  -50
         6            1  -12
         8            4  -39 # longest negative stretch 
positive 1            1   10
         3            2   70
         5            2   72
         7            3  102 # longest positive stretch
Answered By: mozway
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