Pandas: Replacing NaNs using Median/Mean of the column

Pandas Handling Missing Values: Exercise-14 with Solution

Write a Pandas program to replace NaNs with median or mean of the specified columns in a given DataFrame.

Test Data:

     ord_no  purch_amt  sale_amt    ord_date  customer_id  salesman_id
0   70001.0     150.50     10.50  2012-10-05         3002       5002.0
1       NaN        NaN     20.65  2012-09-10         3001       5003.0
2   70002.0      65.26       NaN         NaN         3001       5001.0
3   70004.0     110.50     11.50  2012-08-17         3003          NaN
4       NaN     948.50     98.50  2012-09-10         3002       5002.0
5   70005.0        NaN       NaN  2012-07-27         3001       5001.0
6       NaN    5760.00     57.00  2012-09-10         3001       5001.0
7   70010.0    1983.43     19.43  2012-10-10         3004          NaN
8   70003.0        NaN       NaN  2012-10-10         3003       5003.0
9   70012.0     250.45     25.45  2012-06-27         3002       5002.0
10      NaN      75.29     75.29  2012-08-17         3001       5003.0
11  70013.0    3045.60     35.60  2012-04-25         3001          NaN

Sample Solution:

Python Code :

import pandas as pd
import numpy as np
pd.set_option('display.max_rows', None)
#pd.set_option('display.max_columns', None)
df = pd.DataFrame({
'ord_no':[70001,np.nan,70002,70004,np.nan,70005,np.nan,70010,70003,70012,np.nan,70013],
'purch_amt':[150.5,np.nan,65.26,110.5,948.5,np.nan,5760,1983.43,np.nan,250.45, 75.29,3045.6],
'sale_amt':[10.5,20.65,np.nan,11.5,98.5,np.nan,57,19.43,np.nan,25.45, 75.29,35.6],
'ord_date': ['2012-10-05','2012-09-10',np.nan,'2012-08-17','2012-09-10','2012-07-27','2012-09-10','2012-10-10','2012-10-10','2012-06-27','2012-08-17','2012-04-25'],
'customer_id':[3002,3001,3001,3003,3002,3001,3001,3004,3003,3002,3001,3001],
'salesman_id':[5002,5003,5001,np.nan,5002,5001,5001,np.nan,5003,5002,5003,np.nan]})
print("Original Orders DataFrame:")
print(df)
print("Using median in purch_amt to replace NaN:")
df['purch_amt'].fillna(df['purch_amt'].median(), inplace=True)
print(df)
print("Using mean to replace NaN:")
df['sale_amt'].fillna(int(df['sale_amt'].mean()), inplace=True)
print(df)

Sample Output:

Original Orders DataFrame:
     ord_no  purch_amt     ...      customer_id salesman_id
0   70001.0     150.50     ...             3002      5002.0
1       NaN        NaN     ...             3001      5003.0
2   70002.0      65.26     ...             3001      5001.0
3   70004.0     110.50     ...             3003         NaN
4       NaN     948.50     ...             3002      5002.0
5   70005.0        NaN     ...             3001      5001.0
6       NaN    5760.00     ...             3001      5001.0
7   70010.0    1983.43     ...             3004         NaN
8   70003.0        NaN     ...             3003      5003.0
9   70012.0     250.45     ...             3002      5002.0
10      NaN      75.29     ...             3001      5003.0
11  70013.0    3045.60     ...             3001         NaN

[12 rows x 6 columns]
Using median in purch_amt to replace NaN:
     ord_no  purch_amt     ...      customer_id salesman_id
0   70001.0     150.50     ...             3002      5002.0
1       NaN     250.45     ...             3001      5003.0
2   70002.0      65.26     ...             3001      5001.0
3   70004.0     110.50     ...             3003         NaN
4       NaN     948.50     ...             3002      5002.0
5   70005.0     250.45     ...             3001      5001.0
6       NaN    5760.00     ...             3001      5001.0
7   70010.0    1983.43     ...             3004         NaN
8   70003.0     250.45     ...             3003      5003.0
9   70012.0     250.45     ...             3002      5002.0
10      NaN      75.29     ...             3001      5003.0
11  70013.0    3045.60     ...             3001         NaN

[12 rows x 6 columns]
Using mean to replace NaN:
     ord_no  purch_amt     ...      customer_id salesman_id
0   70001.0     150.50     ...             3002      5002.0
1       NaN     250.45     ...             3001      5003.0
2   70002.0      65.26     ...             3001      5001.0
3   70004.0     110.50     ...             3003         NaN
4       NaN     948.50     ...             3002      5002.0
5   70005.0     250.45     ...             3001      5001.0
6       NaN    5760.00     ...             3001      5001.0
7   70010.0    1983.43     ...             3004         NaN
8   70003.0     250.45     ...             3003      5003.0
9   70012.0     250.45     ...             3002      5002.0
10      NaN      75.29     ...             3001      5003.0
11  70013.0    3045.60     ...             3001         NaN

[12 rows x 6 columns]

Python Code Editor:

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