how to loop step and count in python

Question

I want to be able to count 5 at every step of 2 while condition is less than 1000 for example:

i = 0
j = 2000
k = 3000
while i < 1000:
    i += 2
    for x in range(5):
        print(i)
    j += 2
    for x in range(5):
        print(j)
    k += 2
    for x in range(5):
        print(k)

but the output just print i, j, k 5 times

I want the the result to be: .....

2
3
4
5
6
2002
2003
2004
2005
2006
3002
3003
3004
3005
3006
8      #please note here that 8(i) continue by 2 steps from 6
9
10
etc..........

i will like to know a more simpler and pythonic way to do this. Thanks

There are a gazillion way to get your desired output, but I haven't understood what you are trying to do — Umberto Fontanazza
– Umberto Fontanazza, Commented Aug 13, 2022 at 11:11
There are a gazillion way to skin the cat, I wonder which answer is the most efficient using timeit — leon gxl
– leon gxl, Commented Aug 13, 2022 at 12:16
i want to count from 0 to 1000 but jump 2 steps at every count of 5. e.g. 2,3,4,5,6, 8,9,10,11,12, 14,15,16,17,18...... — 7r0jan005
– 7r0jan005, Commented Aug 13, 2022 at 12:23

Ciro García · Accepted Answer · 2022-08-13 11:41:13Z

2

That's a weird thing you're trying to do, but here is my modified version with your desired output:

a, b, c = 0, 2000, 3000
for i in range(2, 1000, 6):
    for x in range(5):
        print(a+i+x)
    for x in range(5):
        print(b+i+x)
    for x in range(5):
        print(c+i+x)

edited Aug 13, 2022 at 11:41

answered Aug 13, 2022 at 11:17

Ciro García

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1 Comment

Joe Carboni Over a year ago

You can use range in a for-loop to remove defining and stepping i yourself: for i in range(2,1000,6)

threadfin · Accepted Answer · 2022-08-13 12:07:02Z

1

So you're trying to get number progression, here's one way to do it:

progression = [a for a in range(2,7)] \
            + [b for b in range(2002,2007)] \
            + [c for c in range(3002,3007)]
for i in range(0, 1000, 6):
    for p in progression:
        print(p + i)

2
3
4
5
6
2002
2003
2004
2005
2006
3002
3003
3004
3005
3006
8
9
10
11
12
2008
2009
2010
2011
2012
3008
3009
3010
3011
3012
14
15
16
17
18
2014
2015
2016
2017
2018
3014
3015
3016
3017
3018
20
21
22
23
24
2020
2021
2022
2023
2024
3020
3021
3022
3023
3024
26
...<truncated>

edited Aug 13, 2022 at 12:07

answered Aug 13, 2022 at 11:52

threadfin

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Comments

WhoKnowsMe · Accepted Answer · 2022-08-13 11:33:08Z

0

I don't really undestand what you want to do but this output what you want:

i = 0
j = 2000
k = 3000
while i < 1000:
    i += 2
    for x in range(5):
        print(i+x)
    j += 2
    for x in range(5):
        print(j+x)
    k += 2
    for x in range(5):
        print(k+x)
    i += 4

edited Aug 13, 2022 at 11:33

answered Aug 13, 2022 at 11:22

WhoKnowsMe

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1 Comment

7r0jan005 Over a year ago

i want to count from (i,j,k) 0 to 1000 but jump 2 steps at every count of 5. e.g. 2,3,4,5,6, 8,9,10,11,12, 14,15,16,17,18......

Sidharth Mudgil · Accepted Answer · 2022-08-14 05:17:28Z

0

You can do this in other ways but here is my approach

j = 2000
k = 3000
for i in range(2, 1000, 6):
    for x in range(5):
        print(i+x)
    j += 2
    for x in range(5):
        print(j+x)
    k += 2
    for x in range(5):
        print(k+x)

edited Aug 14, 2022 at 5:17

answered Aug 13, 2022 at 11:22

Sidharth Mudgil

1,47112 silver badges30 bronze badges

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