How to use Pandas vector methods based on rolling custom function that involves entire row and prior data - pandas

Question

While its easy to use pandas rolling method to apply standard formulas, but i find it hard if it involves multiple column with limited past rows. Using the following code to better elaborate: -
import numpy as np
import pandas as pd
#create dummy pandas
df=pd.DataFrame({'col1':np.arange(0,25),'col2':np.arange(100,125),'col3':np.nan})
def func1(shortdf):
#dummy formula
#use last row of col1 multiply by sum of col2
return (shortdf.col1.tail(1).values[0]+shortdf.col2.sum())*3.14
for idx, i in df.iterrows():
if idx>3:
#only interested in the last 3 rows from position of dataframe
df.loc[idx,'col3']=func1(df.iloc[idx-3:idx])
I currently use this iterrow method which needless to say is extremely slow. can anyone has better suggestion?

Answer 1

Option 1
So shift is the solution here. You do have to use rolling for the summation, and then shift that series after the addition and multiplication.
df = pd.DataFrame({'col1':np.arange(0,25),'col2':np.arange(100,125),'col3':np.nan})
ans = ((df['col1'] + df['col2'].rolling(3).sum()) * 3.14).shift(1)
You can check to see that ans is the same as df['col3'] by using ans.eq(df['col3']). Once you see that all but the first few are the same, just change ans to df['col3'] and you should be all set.
Option 2
Without additional information about the customized weight function, it is hard to help. However, this option may be a solution as it separates the rolling calculation at the cost of using more memory.
# df['col3'] = ((df['col1'] + df['col2'].rolling(3).sum()) * 3.14).shift(1)
s = df['col2']
stride = pd.DataFrame([s.shift(x).values[::-1][:3] for x in range(len(s))[::-1]])
res = pd.concat([df, stride], axis=1)
# here you can perform your custom weight function
res['final'] = ((res[0] + res[1] + res[2] + res['col1']) * 3.14).shift(1)
stride is adapted from this question and the calculation is concatenated row-wise to the original dataframe. In this way each column has the value needed to compute whatever it is you may need.
res['final'] is identical to option 1's ans