Python For循环-关于第二个循环的问题

Question

I am just starting to learn how to code in python and I am trying to get around understanding the following code:

import numpy as np
n=4 
matrix=np.zeros((n,n))  
for j in range (0,n):
  for i in range (n-1,n-j-2,-1):  
      matrix[i,j]=2*n-i-j-1 
print (matrix) 

I would greatly appreciate if someone could please help me understand how each line executes and how the code is revaluated with the loop. Especially how can I interpret the second "for" loop regarding "i" Thanks in advance!

Answer 1

Temporarily remove the matrix stuff, add a few print statements, and the code itself will tell you how the loops work!

n=4
for j in range (0,n):
    for i in range (n-1,n-j-2,-1):
        print(j, i)

Answer 2

Not sure if StackOverflow is the right platform for explaining the code. Anyways...

I changed the inner for loop to make it easy to understand

import numpy as np
n=4 

Create an n*n matrix

matrix=np.zeros((n,n))  

For each column in matrix

for j in range (0,n):

For each row in jth column, but from nj-1 to n-1

NOTE: In the original example, -1 at the end indicates reverse order. I reversed the loop order and removed the -1 at the end, to produce the same output. Please check to confirm

nj-1 : This should be understood by examples. For j==0 the value is n-1 -> Last row. For the last column j==n-1 , the value is 0 -> First row. So for each column, starting from the last row, we proceed diagonally upward to the first row.

Simply a logic/equation to move diagonally upward.

NOTE: This is only the starting point for each column.

n - 1 : Last row (though the second number is n, the call range(0, x) or range (x) expands from 0 to x - 1 . Much like array indexing)

    for i in range (n-j-1, n):
        matrix[i,j]=2*n-i-j-1 

print (matrix)