[ Python 常見問題 ] Efficiently calculate the difference between two rows in dataframe

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Question 
Consider I have a dataframe as below: 

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1. >>> import pandas as pd  
2. >>> import numpy as np  
3. >>> df = pd.DataFrame([[1, 2], [3, 4]], columns=['f1', 'f2'], index=['r1', 'r2'])  
4. >>> df  
5.     f1  f2  
6. r1   1   2  
7. r2   3   4  

How should I efficiently calculate the absolute different between row r1 and r2 and create another row as r3 to keep the result. That is to say the result will look like: 

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1. >>> for cn in df.columns:  
2. ...     diff_dat.append(abs(df[cn]['r1'] - df[cn]['r2']))  
3. ...   
4. >>> diff_dat  
5. [2, 2]  
6. >>> df.append(pd.DataFrame([diff_dat], index=['r3'], columns=df.columns))  
7.     f1  f2  
8. r1   1   2  
9. r2   3   4  
10. r3   2   2  

How-To 
You can do this: 

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1. In [576]: df.append(df.diff().dropna().abs())  
2. Out[583]:   
3.      f1   f2  
4. r1  1.0  2.0  
5. r2  3.0  4.0  
6. r2  2.0  2.0  

Use loc for select rows, subtract, get abs and last add new row by setting with enlargement: 

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1. df.loc['r3'] = (df.loc['r1'] - df.loc['r2']).abs()  
2. print (df)  
3.     f1  f2  
4. r1   1   2  
5. r2   3   4  
6. r3   2   2  

Performance for 1000 columns: 

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1. np.random.seed(123)  
2. df = pd.DataFrame(np.random.randint(10, size=(2, 1000)), index=['r1', 'r2']).add_prefix('f')-5  
3.   
4. #Mayank Porwal solution  
5. In [40]: %timeit df.append(df.diff().dropna().abs())  
6. 1.51 ms ± 19.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)  
7.   
8. #jezrael solution  
9. In [41]: %timeit df.loc['r3'] = (df.loc['r1'] - df.loc['r2']).abs()  
10. 663 µs ± 54.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)  
11.   
12. #NaT3z solution  
13. In [42]: %timeit df.loc["r3"] = df.apply(lambda c: abs(c["r1"] - c["r2"]), axis=0)  
14. 967 µs ± 80.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)  

For improve performance is possible use numpy: 

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1. In [49]: %timeit df.loc['r3'] = np.abs(df.loc['r1'].values - df.loc['r2'].values)  
2. 414 µs ± 1.68 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)  

Supplement 
* Pandas Doc - Indexing and Selecting Data