Python For Loop - Question about second loop
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!
python python-2.7 loops for-loop
asked Nov 22 '18 at 1:03
Aylín PérezAylín Pérez
12
add a comment |
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!
python python-2.7 loops for-loop
asked Nov 22 '18 at 1:03
Aylín PérezAylín Pérez
12
add a comment |
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!
python python-2.7 loops for-loop
asked Nov 22 '18 at 1:03
Aylín PérezAylín Pérez
12
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!
python python-2.7 loops for-loop
python python-2.7 loops for-loop
asked Nov 22 '18 at 1:03
Aylín PérezAylín Pérez
12
asked Nov 22 '18 at 1:03
Aylín PérezAylín Pérez
12
asked Nov 22 '18 at 1:03
Aylín PérezAylín Pérez
12
asked Nov 22 '18 at 1:03
Aylín PérezAylín Pérez
12
asked Nov 22 '18 at 1:03
Aylín PérezAylín Pérez
12
12
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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)
answered Nov 22 '18 at 1:33
John GordonJohn Gordon
9,81251729
add a comment |
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 n-j-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
n-j-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)
answered Nov 22 '18 at 1:50
KrishnaKrishna
6021515
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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)
answered Nov 22 '18 at 1:33
John GordonJohn Gordon
9,81251729
add a comment |
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)
answered Nov 22 '18 at 1:33
John GordonJohn Gordon
9,81251729
add a comment |
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)
answered Nov 22 '18 at 1:33
John GordonJohn Gordon
9,81251729
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)
answered Nov 22 '18 at 1:33
John GordonJohn Gordon
9,81251729
answered Nov 22 '18 at 1:33
John GordonJohn Gordon
9,81251729
answered Nov 22 '18 at 1:33
John GordonJohn Gordon
9,81251729
answered Nov 22 '18 at 1:33
John GordonJohn Gordon
9,81251729
9,81251729
add a comment |
add a comment |
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 n-j-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
n-j-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)
answered Nov 22 '18 at 1:50
KrishnaKrishna
6021515
add a comment |
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 n-j-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
n-j-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)
answered Nov 22 '18 at 1:50
KrishnaKrishna
6021515
add a comment |
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 n-j-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
n-j-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)
answered Nov 22 '18 at 1:50
KrishnaKrishna
6021515
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 n-j-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
n-j-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)
answered Nov 22 '18 at 1:50
KrishnaKrishna
6021515
answered Nov 22 '18 at 1:50
KrishnaKrishna
6021515
answered Nov 22 '18 at 1:50
KrishnaKrishna
6021515
answered Nov 22 '18 at 1:50
KrishnaKrishna
6021515
6021515
add a comment |
add a comment |
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