Write a program in Python to modify the diagonal of a given DataFrame by 1

In data analysis, we often need to modify specific elements of a DataFrame. One common task is modifying the diagonal elements − where the row index equals the column index (positions [0,0], [1,1], [2,2], etc.).

Understanding the Diagonal

The diagonal of a square DataFrame consists of elements where row index equals column index. For a 3x3 DataFrame, the diagonal elements are at positions (0,0), (1,1), and (2,2).

Method 1: Using Nested Loops

We can iterate through all positions and check if the row index equals column index ?

import pandas as pd

data = [[10, 20, 30], [40, 50, 60], [70, 80, 90]]
df = pd.DataFrame(data)
print("Original DataFrame:")
print(df)

# Modify diagonal elements to 1
for i in range(len(df)):
    for j in range(len(df.columns)):
        if i == j:
            df.iloc[i, j] = 1

print("\nModified DataFrame:")
print(df)

Original DataFrame:
    0   1   2
0  10  20  30
1  40  50  60
2  70  80  90

Modified DataFrame:
   0   1   2
0  1  20  30
1  40   1  60
2  70  80   1

Method 2: Using numpy.fill_diagonal()

NumPy provides a more efficient fill_diagonal() function ?

import pandas as pd
import numpy as np

data = [[10, 20, 30], [40, 50, 60], [70, 80, 90]]
df = pd.DataFrame(data)
print("Original DataFrame:")
print(df)

# Modify diagonal using numpy
np.fill_diagonal(df.values, 1)

print("\nModified DataFrame:")
print(df)

Original DataFrame:
    0   1   2
0  10  20  30
1  40  50  60
2  70  80  90

Modified DataFrame:
   0   1   2
0  1  20  30
1  40   1  60
2  70  80   1

Comparison

Conclusion

Use numpy.fill_diagonal() for efficient diagonal modification in production code. The nested loop approach helps understand the concept but is less efficient for large DataFrames.