How to replace NaNs by preceding or next values in pandas DataFrame?

You could use the fillna method on the DataFrame and specify the method as ffill (forward fill):

>>> df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])>>> df.fillna(method='ffill')   0  1  20  1  2  31  4  2  32  4  2  9

This method...

propagate[s] last valid observation forward to next valid

To go the opposite way, there's also a bfill method.

This method doesn't modify the DataFrame inplace - you'll need to rebind the returned DataFrame to a variable or else specify inplace=True:

df.fillna(method='ffill', inplace=True)

The accepted answer is perfect. I had a related but slightly different situation where I had to fill in forward but only within groups. In case someone has the same need, know that fillna works on a DataFrameGroupBy object.

>>> example = pd.DataFrame({'number':[0,1,2,nan,4,nan,6,7,8,9],'name':list('aaabbbcccc')})>>> example  name  number0    a     0.01    a     1.02    a     2.03    b     NaN4    b     4.05    b     NaN6    c     6.07    c     7.08    c     8.09    c     9.0>>> example.groupby('name')['number'].fillna(method='ffill') # fill in row 5 but not row 30    0.01    1.02    2.03    NaN4    4.05    4.06    6.07    7.08    8.09    9.0Name: number, dtype: float64

You can use pandas.DataFrame.fillna with the method='ffill' option. 'ffill' stands for 'forward fill' and will propagate last valid observation forward. The alternative is 'bfill' which works the same way, but backwards.

import pandas as pddf = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])df = df.fillna(method='ffill')print(df)#   0  1  2#0  1  2  3#1  4  2  3#2  4  2  9

There is also a direct synonym function for this, pandas.DataFrame.ffill, to make things simpler.