forward fill specific columns in pandas dataframe

Question

If I have a dataframe with multiple columns ['x', 'y', 'z'], how do I forward fill only one column 'x'? Or a group of columns ['x','y']?

I only know how to do it by axis.

Asked By: azuric

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Source

Answer 1

for col in ['X', 'Y']:
    df[col] = df[col].ffill()

Answered By: Woody Pride

Answer 2

tl;dr:

cols = ['X', 'Y']
df.loc[:,cols] = df.loc[:,cols].ffill()

And I have also added a self containing example:

>>> import pandas as pd
>>> import numpy as np
>>> 
>>> ## create dataframe
... ts1 = [0, 1, np.nan, np.nan, np.nan, np.nan]
>>> ts2 = [0, 2, np.nan, 3, np.nan, np.nan]
>>> d =  {'X': ts1, 'Y': ts2, 'Z': ts2}
>>> df = pd.DataFrame(data=d)
>>> print(df.head())
    X   Y   Z
0   0   0   0
1   1   2   2
2 NaN NaN NaN
3 NaN   3   3
4 NaN NaN NaN
>>> 
>>> ## apply forward fill
... cols = ['X', 'Y']
>>> df.loc[:,cols] = df.loc[:,cols].ffill()
>>> print(df.head())
   X  Y   Z
0  0  0   0
1  1  2   2
2  1  2 NaN
3  1  3   3
4  1  3 NaN

Answered By: Hennep

Answer 3

I used below code, Here for X and Y method can be different also instead of ffill().

 df1 = df.fillna({
        'X' : df['X'].ffill(),
        'Y' : df['Y'].ffill(),
    })

Answered By: Abhishek Chaurasia

Answer 4

Two columns can be ffill() simultaneously as given below:

df1 = df[['X','Y']].ffill()

Answered By: Souvik Daw

Answer 5

The simplest version I think.

cols = ['X', 'Y']
df[cols] = df[cols].ffill()

Answered By: Bernardo Resolve

Answer 6

Alternatively with the inplace parameter:

df['X'].ffill(inplace=True)
df['Y'].ffill(inplace=True)

And no, you cannot do df[['X','Y]].ffill(inplace=True) as this first creates a slice through the column selection and hence inplace forward fill would create a SettingWithCopyWarning. Of course if you have a list of columns you can do this in a loop:

for col in ['X', 'Y']:
    df[col].ffill(inplace=True)

The point of using inplace is that it avoids copying the column.

Answered By: Uwe Mayer

forward fill specific columns in pandas dataframe

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