Replace a column with another column if another is not null in pandas DataFrame

You should consider using the numpy where function which runs much faster the all the apply methods.

Where is basically if else function for vectors. The first entry is a vector with the condition, the second the value if true and third if false. Here is how it would look:

import numpy as np
df['v_5'] = np.where(~df['pfv'].isnull(),df['pfv'],df['v_5'])

Good luck

Because missing values are strings null, use:

df.loc[df['pfv'].ne('null'), 'v5'] = df["pfv"]
print (df)
      v_4             v5      s_5   vt_5       ex_5            pfv  pfv_cat
0    0-50      StoreSale  Clothes  8-Apr  above 100           null    Shoes
1    0-50  DiscountWorld  Clothes  8-Apr       0-50  DiscountWorld  Clothes
2  51-100    BetterUncle  Clothes  4-Dec     51-100    BetterUncle    Shoes

If missing values are NaN or Nones (not strings) use Series.fillna:

df['v5'] = df['pfv'].fillna(df['v5'])

print (df)
      v_4             v5      s_5   vt_5       ex_5            pfv  pfv_cat
0    0-50      StoreSale  Clothes  8-Apr  above 100            NaN    Shoes
1    0-50  DiscountWorld  Clothes  8-Apr       0-50  DiscountWorld  Clothes
2  51-100    BetterUncle  Clothes  4-Dec     51-100    BetterUncle    Shoes

Late in the game but if truly nulls (not 'null' strings), you could also use

df['v_5'] = df['pfv'].combine_first(df['v_5'])

which is equivalent to COALESCE() in SQL.

You can use Series.update from pandas to replace with non-NA values from passed Series.

Therefore you need first to replace 'null' with NaN.

Example:

import pandas as pd
import numpy as np

 df = pd.DataFrame({'A': ['a', 'b', 'c'],
               'B': ['x', 'null', 'z']})

 # Replace null with nan
 df.loc[:,'B']  =  df.loc[:,'B'].replace('null', np.nan)
 # Update column, only if other column is non-nan (inplace)
 df.A.update(df.B)

Or if you don't want to change null with NaN

import pandas as pd
import numpy as np

 df = pd.DataFrame({'A': ['a', 'b', 'c'],
               'B': ['x', 'null', 'z']})

 # Replace null with nan
 temp  =  df.loc[:,'B'].replace('null', np.nan)
 # Update column, only if other column is non-nan (inplace)
 df.A.update(temp)

My solution is the same of jezrael, but with one more step, based on an essay I made with the null problem. I've added one more row for a pfv without values.

    data = [['0-50','StoreSale','Clothes','8-Apr','above 100','FatimaStore','Shoes'],
    ['0-50','StoreSale','Clothes','8-Apr','0-50','DiscountWorld','Clothes'],
    ['51-100','CleanShop','Clothes','4-Dec','51-100','BetterUncle','Shoes'],
    ['0-50','StoreSale','Clothes','12-Apr','above 100','','Clothes']]

First step is to handle nulls. 'df' is the DataFrame.

    df = df.replace('', np.nan)

          v_4         v5      s_5    vt_5       ex_5            pfv  pfv_cat
    0    0-50  StoreSale  Clothes   8-Apr  above 100    FatimaStore    Shoes
    1    0-50  StoreSale  Clothes   8-Apr       0-50  DiscountWorld  Clothes
    2  51-100  CleanShop  Clothes   4-Dec     51-100    BetterUncle    Shoes
    3    0-50  StoreSale  Clothes  12-Apr  above 100            NaN  Clothes

Now let's update the v5 column. The command says that we will replace v5 for pfv, but if pfv is NaN we will replace with the current value of v5.

    df['v5'] = df['pfv'].fillna(df['v5'])


    print(df)

         v_4             v5      s_5    vt_5       ex_5            pfv  pfv_cat
    0    0-50    FatimaStore  Clothes   8-Apr  above 100    FatimaStore    Shoes
    1    0-50  DiscountWorld  Clothes   8-Apr       0-50  DiscountWorld  Clothes
    2  51-100    BetterUncle  Clothes   4-Dec     51-100    BetterUncle    Shoes
    3    0-50      StoreSale  Clothes  12-Apr  above 100            NaN  Clothes

You can use the apply method from pandas and numpy:

df['v_5'] = df.apply(lambda row: row['pfv'] if row['pfv']==np.NaN else row['v_5'], axis=1)

or without numpy :

df['v_5'] = df.apply(lambda row: row['pfv'] if pd.isnull(row['pfv']) else row['v_5'], axis=1)