conditional fill in pandas dataframe

This can be done fairly efficiently with Numba. If you are not able to use Numba, just omit @njit and your logic will run as a Python-level loop.

import numpy as npimport pandas as pdfrom numba import njitnp.random.seed(0)df = pd.DataFrame(1000*(2+np.random.randn(500, 1)), columns=['A'])df.loc[1, 'A'] = np.nandf.loc[15, 'A'] = np.nandf.loc[240, 'A'] = np.nan@njitdef recurse_nb(x):    out = x.copy()    for i in range(1, x.shape[0]):        if not np.isnan(x[i]) and (abs(1 - x[i] / out[i-1]) < 0.3):            out[i] = out[i-1]    return outdf['B'] = recurse_nb(df['A'].values)print(df.head(10))             A            B0  3764.052346  3764.0523461          NaN          NaN2  2978.737984  2978.7379843  4240.893199  4240.8931994  3867.557990  4240.8931995  1022.722120  1022.7221206  2950.088418  2950.0884187  1848.642792  1848.6427928  1896.781148  1848.6427929  2410.598502  2410.598502

Not sure what you want to do with the first B-1 and the dividing by NaN situation:

df = pd.DataFrame([1,2,3,4,5,None,6,7,8,9,10], columns=['A'])b1 = df.A.shift(1)b1[0] = 1b = list(map(lambda a,b1: a if np.isnan(a) else (b1 if abs(b1-a)/b1 < 0.3 else a), df.A, b1 ))df['B'] = bdf       A    B0    1.0  1.01    2.0  2.02    3.0  3.03    4.0  4.04    5.0  4.05    NaN  NaN6    6.0  6.07    7.0  6.08    8.0  7.09    9.0  8.010  10.0  9.0

as per @jpp, you could also do a list comprehension version for list b:

b = [a if np.isnan(a) or abs(b-a)/b >= 0.3 else b for a,b in zip(df.A,b1)]

A simple solution that I could come up with is following. I was wondering if there is more pythonic way of doing things:

 a = df['A'].values b = [] b.append(t[0]) for i in range(1, len(a)):     if np.isnan(a[i]):         b.append(a[i])     else:         b.append(b[i-1] if abs(1 - a[i]/b[i-1]) < 0.3 else a[i]) df['B'] = b