How can I write a python loop to count every other index?

Question

I want to write a python code that outputs the following

1 2
3 4

I tried this:

For I in range(2):
Print(I, I+1)

But this outputs
0 1
1 2

So it does not work for me. How can I fix the loop?

The goal of this loop is to go 1000 times to generate those kind of numbers.

Answer 1

just in case you wanted to get a little fancier.

y = [{a, a+1} for a in range(1, 50, 2)]
print(y)

the code is a list comprehension that creates a set 1, 2 etc. and the range starts at 1 to 50 or any number you want and skips every other number.
you can change the set {} to a list [], tuple (), or just print.

y = [print(a, a+1) for a in range(1, 50, 2)]

Answer 2

Yes loop counter should be incremented by 2 for each iteration 🤭

Answer 3

Andrew your solution will break every third index. If you run the code the answer will be [(1, 2), (3,4), (5,6) ,(8,7)] as you see the third index must be 7,8 not 8,7.

Answer 4

Karzan i didnt run into that problem when i ran the code. have you tried to run the code? if so, do you mind sharing it?

Answer 5

Actually I solved it:

for i in range(2):
if i % 2 ==0:
Print(i, i+1)
else:
print(i+1, i+2)