Question: I have dataframe which should be filled by understanding rows understanding like we do in excel. If its continious integer it fill by next number itself. Is there any function in python like this? import pandas as pd d = { 'year': [2019,2020,2019,2020,np.nan,np.nan], 'cat1': [1,2,3,4,np.nan,np.nan], 'cat2': ['c1','c1','c1','c2',np.nan,np.nan]} df = pd.DataFrame(data=d) df year cat1 cat2 0 2019.0 1.0 c1 1 2020.0 2.0 c1 2 2019.0 3.0 c1 3 2020.0 4.0 c2 4 NaN NaN NaN 5 NaN NaN NaN  output required:  year cat1 cat2 0 2019.0 1.0 c1 1 2020.0 2.0 c1 2 2019.0 3.0 c1 3 2020.0 4.0 c2 4 2019.0 5.0 c2 #here can be ignored if it can't understand the earlier pattern 5 2020.0 6.0 c2 #here can be ignored if it can't understand the earlier pattern  I tried df.interpolate(method='krogh') #it fill 1,2,3,4,5,6 but incorrect others.

Answer: Here is my solution for the specific use case you mention -  The code for these helper functions for categorical_repeat, continous_interpolate and other is provided below in EXPLANATION > Approach section.  config = {'year':categorical_repeat, #shortest repeating sequence 'cat1':continous_interpolate, #curve fitting (linear) 'cat2':other} #forward fill print(df.agg(config))   year cat1 cat2 0 2019.0 1 c1 1 2020.0 2 c1 2 2019.0 3 c1 3 2020.0 4 c2 4 2019.0 5 c2 5 2020.0 6 c2   EXPLANATION: As I understand, there is no direct way of handling all types of patterns in pandas as excel does. Excel involves linear interpolation for continuous sequences, but it involves other methods for other column patterns.  Continous integer array -> linear interpolation Repeated cycles -> Smallest repeating sequence Alphabet (and similar) -> Tiling fixed sequence until the length of df Unrecognizable pattern -> Forward fill  Here is the dummy dataset that I attempt my approach on - data = {'A': [2019, 2020, 2019, 2020, 2019, 2020], 'B': [1, 2, 3, 4, 5, 6], 'C': [6, 5, 4, 3, 2, 1], 'D': ['C', 'D', 'E', 'F', 'G', 'H'], 'E': ['A', 'B', 'C', 'A', 'B', 'C'], 'F': [1,2,3,3,4,2] } df = pd.DataFrame(data) empty = pd.DataFrame(columns=df.columns, index=df.index)[:4] df_new = df.append(empty).reset_index(drop=True) print(df_new)   A B C D E F 0 2019 1 6 C A 1 1 2020 2 5 D B 2 2 2019 3 4 E C 3 3 2020 4 3 F A 3 4 2019 5 2 G B 4 5 2020 6 1 H C 2 6 NaN NaN NaN NaN NaN NaN 7 NaN NaN NaN NaN NaN NaN 8 NaN NaN NaN NaN NaN NaN 9 NaN NaN NaN NaN NaN NaN  Approach: Let's start with some helper functions - import numpy as np import scipy as sp import pandas as pd #Curve fitting (linear) def f(x, m, c): return m*x+c #Modify to extrapolate for exponential sequences etc. #Interpolate continous linear def continous_interpolate(s): clean = s.dropna() popt, pcov = sp.optimize.curve_fit(f, clean.index, clean) output = [round(i) for i in f(s.index, *popt)] #Remove the round() for float values return pd.Series(output) #Smallest Repeating sub-sequence def pattern(inputv): ''' https://stackoverflow.com/questions/6021274/finding-shortest-repeating-cycle-in-word ''' pattern_end =0 for j in range(pattern_end+1,len(inputv)): pattern_dex = j%(pattern_end+1) if(inputv[pattern_dex] != inputv[j]): pattern_end = j; continue if(j == len(inputv)-1): return inputv[0:pattern_end+1]; return inputv; #Categorical repeat imputation def categorical_repeat(s): clean = s.dropna() cycle = pattern(clean) repetitions = (len(s)//len(cycle))+1 output = np.tile(cycle, repetitions)[:len(s)] return pd.Series(output) #continous sequence of alphabets def alphabet(s): alp = 'abcdefghijklmnopqrstuvwxyz' alp2 = alp*((len(s)//len(alp))+1) start = s[0] idx = alp2.find(start.lower()) output = alp2[idx:idx+len(s)] if start.isupper(): output = output.upper() return pd.Series(list(output)) #If no pattern then just ffill def other(s): return s.ffill()  Next, lets create a configuration based on what we want to solve and apply the methods required - config = {'A':categorical_repeat, 'B':continous_interpolate, 'C':continous_interpolate, 'D':alphabet, 'E':categorical_repeat, 'F':other} output_df = df_new.agg(config) print(output_df)   A B C D E F 0 2019 1 6 C A 1 1 2020 2 5 D B 2 2 2019 3 4 E C 3 3 2020 4 3 F A 3 4 2019 5 2 G B 4 5 2020 6 1 H C 2 6 2019 7 0 I A 2 7 2020 8 -1 J B 2 8 2019 9 -2 K C 2 9 2020 10 -3 L A 2