Pandas_Challenge_2.py

import numpy as np
import pandas as pd

dfc = pd.DataFrame(np.zeros((1000, 1000), dtype='int'))

There are 2 methods we can put 1's in every diagonal location of the dataframe.
#Method1:
np.fill_diagonal(dfc.values,1)

#Method using for loop and iat
"""
=> for i in range(dfc.shape[0]): it will iterate over all row indices of the DataFrame dfc. Since dfc.shape[0] gives the number of rows in the DataFrame, the loop will go from 0 to dfc.shape[0] - 1, which means it will cover all the row indices.

=> dfc.iat[i,i] = 1: it sets the diagonal element at index (i, i) to 1. In each iteration of the loop, it moves along the diagonal and sets the diagonal elements one by one to 1.
to be precise,in line dfc.iat[i,i],it accesses the DataFrame element at the ith row and ith column, which corresponds to the diagonal position of the DataFrame.
we can use alternative for loop as : for i in range(min(dfc.shape)) and then dfc.iat[i,i] = 1
"""
for i in range(dfc.shape[0]):
dfc.iat[i,i] =1

np.fill_diagonal(dfc.values,1)
dfc.head(10)

#And
for i in range((dfc.shape[0])): #
dfc.iat[i,i] = 1
dfc.head(5)