Back and forth loop Python

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up vote
12
down vote

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1

I want to create an infinite loop that counts up and down from 0 to 100 to 0 (and so on) and only stops when some convergence criterion inside the loop is met, so basically something like this:

for i in range(0, infinity):
 for j in range(0, 100, 1):
 print(j) # (in my case 100 lines of code)
 for j in range(100, 0, -1):
 print(j) # (same 100 lines of code as above)

Is there any way to merge the two for loops over j into one so that I don't have write out the same code inside the loops twice?

python loops for-loop infinite-loop

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asked Sep 5 at 14:13

Daniel

708

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  "so that I don't have write out the same code" Sounds like a good usecase for a function
  – DeepSpace
  Sep 5 at 14:16
  
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  @Phylogenesis TypeError: unsupported operand type(s) for +: 'range' and 'range'. This will only work in Python 2
  – DeepSpace
  Sep 5 at 14:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, that's Python 2 only.
  – Phylogenesis
  Sep 5 at 14:19
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Three sidenotes: range(100, 0, -1) does not actually produce range(0, 100, 1) in reverse. If you meant to go from 0 through to 99 included, then from 99 back to 0, use range(99, -1, -1). range(100) is the shorter form for range(0, 100, 1), it is good practice to use that instead. And there is no range(0, infinity) syntax, you'd use for i in itertools.count(): perhaps to create an infinite counter, or while True: to create an endless loop.
  – Martijn Pieters♦
  Sep 5 at 14:27
  

  
  
  
  

add a comment | 

up vote
12
down vote

favorite

1

I want to create an infinite loop that counts up and down from 0 to 100 to 0 (and so on) and only stops when some convergence criterion inside the loop is met, so basically something like this:

for i in range(0, infinity):
 for j in range(0, 100, 1):
 print(j) # (in my case 100 lines of code)
 for j in range(100, 0, -1):
 print(j) # (same 100 lines of code as above)

Is there any way to merge the two for loops over j into one so that I don't have write out the same code inside the loops twice?

python loops for-loop infinite-loop

share|improve this question

asked Sep 5 at 14:13

Daniel

708

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  "so that I don't have write out the same code" Sounds like a good usecase for a function
  – DeepSpace
  Sep 5 at 14:16
  
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  @Phylogenesis TypeError: unsupported operand type(s) for +: 'range' and 'range'. This will only work in Python 2
  – DeepSpace
  Sep 5 at 14:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, that's Python 2 only.
  – Phylogenesis
  Sep 5 at 14:19
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Three sidenotes: range(100, 0, -1) does not actually produce range(0, 100, 1) in reverse. If you meant to go from 0 through to 99 included, then from 99 back to 0, use range(99, -1, -1). range(100) is the shorter form for range(0, 100, 1), it is good practice to use that instead. And there is no range(0, infinity) syntax, you'd use for i in itertools.count(): perhaps to create an infinite counter, or while True: to create an endless loop.
  – Martijn Pieters♦
  Sep 5 at 14:27
  

  
  
  
  

add a comment | 

up vote
12
down vote

favorite

1

up vote
12
down vote

favorite

1

1

I want to create an infinite loop that counts up and down from 0 to 100 to 0 (and so on) and only stops when some convergence criterion inside the loop is met, so basically something like this:

for i in range(0, infinity):
 for j in range(0, 100, 1):
 print(j) # (in my case 100 lines of code)
 for j in range(100, 0, -1):
 print(j) # (same 100 lines of code as above)

Is there any way to merge the two for loops over j into one so that I don't have write out the same code inside the loops twice?

python loops for-loop infinite-loop

share|improve this question

asked Sep 5 at 14:13

Daniel

708

I want to create an infinite loop that counts up and down from 0 to 100 to 0 (and so on) and only stops when some convergence criterion inside the loop is met, so basically something like this:

for i in range(0, infinity):
 for j in range(0, 100, 1):
 print(j) # (in my case 100 lines of code)
 for j in range(100, 0, -1):
 print(j) # (same 100 lines of code as above)

Is there any way to merge the two for loops over j into one so that I don't have write out the same code inside the loops twice?

python loops for-loop infinite-loop

share|improve this question

asked Sep 5 at 14:13

Daniel

708

share|improve this question

share|improve this question

asked Sep 5 at 14:13

Daniel

708

asked Sep 5 at 14:13

Daniel

708

asked Sep 5 at 14:13

Daniel

708

708

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  "so that I don't have write out the same code" Sounds like a good usecase for a function
  – DeepSpace
  Sep 5 at 14:16
  
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  @Phylogenesis TypeError: unsupported operand type(s) for +: 'range' and 'range'. This will only work in Python 2
  – DeepSpace
  Sep 5 at 14:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, that's Python 2 only.
  – Phylogenesis
  Sep 5 at 14:19
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Three sidenotes: range(100, 0, -1) does not actually produce range(0, 100, 1) in reverse. If you meant to go from 0 through to 99 included, then from 99 back to 0, use range(99, -1, -1). range(100) is the shorter form for range(0, 100, 1), it is good practice to use that instead. And there is no range(0, infinity) syntax, you'd use for i in itertools.count(): perhaps to create an infinite counter, or while True: to create an endless loop.
  – Martijn Pieters♦
  Sep 5 at 14:27
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  "so that I don't have write out the same code" Sounds like a good usecase for a function
  – DeepSpace
  Sep 5 at 14:16
  
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  @Phylogenesis TypeError: unsupported operand type(s) for +: 'range' and 'range'. This will only work in Python 2
  – DeepSpace
  Sep 5 at 14:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, that's Python 2 only.
  – Phylogenesis
  Sep 5 at 14:19
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Three sidenotes: range(100, 0, -1) does not actually produce range(0, 100, 1) in reverse. If you meant to go from 0 through to 99 included, then from 99 back to 0, use range(99, -1, -1). range(100) is the shorter form for range(0, 100, 1), it is good practice to use that instead. And there is no range(0, infinity) syntax, you'd use for i in itertools.count(): perhaps to create an infinite counter, or while True: to create an endless loop.
  – Martijn Pieters♦
  Sep 5 at 14:27
  

  
  
  
  

3

3

"so that I don't have write out the same code" Sounds like a good usecase for a function
– DeepSpace
Sep 5 at 14:16

"so that I don't have write out the same code" Sounds like a good usecase for a function
– DeepSpace
Sep 5 at 14:16

6

6

@Phylogenesis TypeError: unsupported operand type(s) for +: 'range' and 'range'. This will only work in Python 2
– DeepSpace
Sep 5 at 14:18

@Phylogenesis TypeError: unsupported operand type(s) for +: 'range' and 'range'. This will only work in Python 2
– DeepSpace
Sep 5 at 14:18

Yeah, that's Python 2 only.
– Phylogenesis
Sep 5 at 14:19

Yeah, that's Python 2 only.
– Phylogenesis
Sep 5 at 14:19

3

3

Three sidenotes: range(100, 0, -1) does not actually produce range(0, 100, 1) in reverse. If you meant to go from 0 through to 99 included, then from 99 back to 0, use range(99, -1, -1). range(100) is the shorter form for range(0, 100, 1), it is good practice to use that instead. And there is no range(0, infinity) syntax, you'd use for i in itertools.count(): perhaps to create an infinite counter, or while True: to create an endless loop.
– Martijn Pieters♦
Sep 5 at 14:27

Three sidenotes: range(100, 0, -1) does not actually produce range(0, 100, 1) in reverse. If you meant to go from 0 through to 99 included, then from 99 back to 0, use range(99, -1, -1). range(100) is the shorter form for range(0, 100, 1), it is good practice to use that instead. And there is no range(0, infinity) syntax, you'd use for i in itertools.count(): perhaps to create an infinite counter, or while True: to create an endless loop.
– Martijn Pieters♦
Sep 5 at 14:27

add a comment | 

9 Answers
9

active

oldest

votes

up vote
34
down vote



accepted

Use the chain method of itertools

import itertools
for i in range(0, infinity):
 for j in itertools.chain(range(0, 100, 1), range(100, 0, -1)):
 print(j) # (in my case 100 lines of code)

As suggested by @Chepner, you can use itertools.cycle() for the infinite loop:

from itertools import cycle, chain

for i in cycle(chain(range(0, 100, 1), range(100, 0, -1))):
 ....

share|improve this answer

edited Sep 6 at 6:01

answered Sep 5 at 14:19

damienfrancois

23.4k43959

• 
  
  
  

  
  12
  

  
  
  
  
  
  

  
  You can add itertools.cycle to this as well, to avoid the nested loop. for i in cycle(chain(range(100), range(100,0,-1)).
  – chepner
  Sep 5 at 14:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As commented by @MartijnPieters it uses too much memory at once
  – N Chauhan
  Sep 5 at 14:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @damienfrancois: I disagree that generic unpacking is better. Try that technique with range(10**6), range(10**6, -1, -1), and watch the memory footprint.
  – Martijn Pieters♦
  Sep 5 at 14:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MartijnPieters Noted. Thanks for the feedback
  – damienfrancois
  Sep 5 at 14:24
  

  
  
  
  

add a comment | 

up vote
14
down vote

As well as the other answers you can use a bit of maths:

while(True):
 for i in range(200):
 if i > 100:
 i = 200 - i

share|improve this answer

edited Sep 5 at 14:24

answered Sep 5 at 14:21

ncfirth

3,206730

add a comment | 

up vote
7
down vote

Here's yet another possibility:

while notConverged:
 for i in xrange(-100, 101):
 print 100 - abs(i)

share|improve this answer

answered Sep 5 at 19:52

twalberg

40.2k76364

• 
  
  
  

  
  

  
  
  
  
  
  

  
  A very interesting approach! +1
  – N Chauhan
  Sep 5 at 20:32
  

  
  
  
  

add a comment | 

up vote
4
down vote

If you've got a repeated set of code, use a function to save space and effort:

def function(x, y, x, num_from_for_loop):
 # 100 lines of code 

while not condition:
 for i in range(1, 101):
 if condition:
 break
 function(x, y, z, i)
 for i in range(100, 0, -1):
 if condition:
 break
 function(x, y, z, i)

You could even use a while True

share|improve this answer

answered Sep 5 at 14:22

N Chauhan

945111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1... I think a lot of the "ooo, I know just the itertools function to help!" answers are neglecting the big picture.
  – Sneftel
  Sep 5 at 20:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I agree, although I honestly don't know anything about itertools save for a few of the simple functions.
  – N Chauhan
  Sep 5 at 20:31
  

  
  
  
  

add a comment | 

up vote
3
down vote

If you're using Python 3.5+, you can using generic unpacking:

for j in (*range(0, 100, 1), *range(100, 0, -1)):

or prior to Python 3.5, you can use itertools.chain:

from itertools import chain

...

for j in chain(range(0, 100, 1), range(100, 0, -1)):

share|improve this answer

answered Sep 5 at 14:19

blhsing

13.8k2628

• 
  
  
  

  
  14
  

  
  
  
  
  
  

  
  Generic unpacking creates a huge tuple with all integers materialised (remember that range() is a very lightweight object). Chaining is better!
  – Martijn Pieters♦
  Sep 5 at 14:22
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

up = True # since we want to go from 0 to 100 first

while True: #for infinite loop

 # For up == True we will print 0-->100 (0,100,1)
 # For up == False we will print 100-->0 (100,0,-1)


 start,stop,step = (0,100,1) if up else (100,0,-1)
 for i in range(start,stop,step):
 print(i)

 up = not up # if we have just printed from 0-->100 (ie up==True), we want to print 100-->0 next so make up False ie up = not up( True) 

 # up will help toggle, between 0-->100 and 100-->0

share|improve this answer

edited Sep 5 at 16:59

answered Sep 5 at 16:10

Tanmay jain

36915

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Pro tip for inverting a boolean: up ^= True
  – wjandrea
  Sep 6 at 0:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @wjandrea Yes thanks for commenting forgot to add as a side note, I was going to use it but then I thought to keep it simple in case OP was beginner.
  – Tanmay jain
  Sep 6 at 3:22
  

  
  
  
  

add a comment | 

up vote
1
down vote

def up_down(lowest_value, highest_value):
 current = lowest_value
 delta = 1
 while True: # Begin infinite loop
 yield current
 current += delta
 if current <= lowest_value or current >= highest_value:
 delta *= -1 # Turn around when either limit is hit

This defines a generator, which will continue to yield values for as long as you need. For example:

>>> u = up_down(0, 10)
>>> count = 0
>>> for j in u:
 print(j) # for demonstration purposes
 count += 1 # your other 100 lines of code here
 if count >= 25: # your ending condition here
 break


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share|improve this answer

answered Sep 6 at 5:23

mathmandan

440713

add a comment | 

up vote
0
down vote

This is more of a partial answer than a direct answer to your question, but you can also use the notion of trigonometric functions and their oscillation to imitate a 'back and forth' loop.

If we have a cos function with an amplitude of 100, shifted left and upwards so that f(x) = 0 and 0 <= f(x) <= 100, we then have the formula f(x) = 50(cos(x-pi)+1) (plot of graph may be found here. The range is what you require, and oscillation occurs so there's no need to negate any values.

>>> from math import cos, pi
>>> f = lambda x: 50*(cos(x-pi)+1)
>>> f(0)
0.0
>>> f(pi/2)
50.0
>>> f(pi)
100.0
>>> f(3*pi/2)
50.0
>>> f(2*pi)
0.0

The issue of course comes in that the function doesn't give integer values so easily, thus it's not that helpful - but this may be useful for future readers where trigonometric functions might be helpful for their case.

share|improve this answer

answered Sep 6 at 5:38

TerryA

41.6k87699

add a comment | 

up vote
0
down vote

I had a similar problem a while ago where I also wanted to create values in the form of an infinite triangle wave, but wanted to step over some values. I ended up using a generator (and the range function as other also have been using):

def tri_wave(min, max, step=1):
 while True:
 yield from range(min, max, step)
 yield from range(max, min, -1 * step)

With carefully selected values on min, max and step (i.e. evenly divisible),

for value in tri_wave(0, 8, 2):
 print(value, end=", ")

I get the min and max value only once, which was my goal:

...0, 2, 4, 6, 8, 6, 4, 2, 0, 2, 4, 6, 8, 6, 4...

I was using Python 3.6 at the time.

share|improve this answer

answered 2 days ago

Amoork

364

add a comment | 

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9 Answers
9

active

oldest

votes

9 Answers
9

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
34
down vote



accepted

Use the chain method of itertools

import itertools
for i in range(0, infinity):
 for j in itertools.chain(range(0, 100, 1), range(100, 0, -1)):
 print(j) # (in my case 100 lines of code)

As suggested by @Chepner, you can use itertools.cycle() for the infinite loop:

from itertools import cycle, chain

for i in cycle(chain(range(0, 100, 1), range(100, 0, -1))):
 ....

share|improve this answer

edited Sep 6 at 6:01

answered Sep 5 at 14:19

damienfrancois

23.4k43959

• 
  
  
  

  
  12
  

  
  
  
  
  
  

  
  You can add itertools.cycle to this as well, to avoid the nested loop. for i in cycle(chain(range(100), range(100,0,-1)).
  – chepner
  Sep 5 at 14:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As commented by @MartijnPieters it uses too much memory at once
  – N Chauhan
  Sep 5 at 14:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @damienfrancois: I disagree that generic unpacking is better. Try that technique with range(10**6), range(10**6, -1, -1), and watch the memory footprint.
  – Martijn Pieters♦
  Sep 5 at 14:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MartijnPieters Noted. Thanks for the feedback
  – damienfrancois
  Sep 5 at 14:24
  

  
  
  
  

add a comment | 

up vote
34
down vote



accepted

Use the chain method of itertools

import itertools
for i in range(0, infinity):
 for j in itertools.chain(range(0, 100, 1), range(100, 0, -1)):
 print(j) # (in my case 100 lines of code)

As suggested by @Chepner, you can use itertools.cycle() for the infinite loop:

from itertools import cycle, chain

for i in cycle(chain(range(0, 100, 1), range(100, 0, -1))):
 ....

share|improve this answer

edited Sep 6 at 6:01

answered Sep 5 at 14:19

damienfrancois

23.4k43959

• 
  
  
  

  
  12
  

  
  
  
  
  
  

  
  You can add itertools.cycle to this as well, to avoid the nested loop. for i in cycle(chain(range(100), range(100,0,-1)).
  – chepner
  Sep 5 at 14:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As commented by @MartijnPieters it uses too much memory at once
  – N Chauhan
  Sep 5 at 14:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @damienfrancois: I disagree that generic unpacking is better. Try that technique with range(10**6), range(10**6, -1, -1), and watch the memory footprint.
  – Martijn Pieters♦
  Sep 5 at 14:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MartijnPieters Noted. Thanks for the feedback
  – damienfrancois
  Sep 5 at 14:24
  

  
  
  
  

add a comment | 

up vote
34
down vote



accepted

up vote
34
down vote



accepted

Use the chain method of itertools

import itertools
for i in range(0, infinity):
 for j in itertools.chain(range(0, 100, 1), range(100, 0, -1)):
 print(j) # (in my case 100 lines of code)

As suggested by @Chepner, you can use itertools.cycle() for the infinite loop:

from itertools import cycle, chain

for i in cycle(chain(range(0, 100, 1), range(100, 0, -1))):
 ....

share|improve this answer

edited Sep 6 at 6:01

answered Sep 5 at 14:19

damienfrancois

23.4k43959

Use the chain method of itertools

import itertools
for i in range(0, infinity):
 for j in itertools.chain(range(0, 100, 1), range(100, 0, -1)):
 print(j) # (in my case 100 lines of code)

As suggested by @Chepner, you can use itertools.cycle() for the infinite loop:

from itertools import cycle, chain

for i in cycle(chain(range(0, 100, 1), range(100, 0, -1))):
 ....

share|improve this answer

edited Sep 6 at 6:01

answered Sep 5 at 14:19

damienfrancois

23.4k43959

share|improve this answer

share|improve this answer

edited Sep 6 at 6:01

edited Sep 6 at 6:01

edited Sep 6 at 6:01

answered Sep 5 at 14:19

damienfrancois

23.4k43959

answered Sep 5 at 14:19

damienfrancois

23.4k43959

answered Sep 5 at 14:19

damienfrancois

23.4k43959

23.4k43959

• 
  
  
  

  
  12
  

  
  
  
  
  
  

  
  You can add itertools.cycle to this as well, to avoid the nested loop. for i in cycle(chain(range(100), range(100,0,-1)).
  – chepner
  Sep 5 at 14:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As commented by @MartijnPieters it uses too much memory at once
  – N Chauhan
  Sep 5 at 14:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @damienfrancois: I disagree that generic unpacking is better. Try that technique with range(10**6), range(10**6, -1, -1), and watch the memory footprint.
  – Martijn Pieters♦
  Sep 5 at 14:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MartijnPieters Noted. Thanks for the feedback
  – damienfrancois
  Sep 5 at 14:24
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  12
  

  
  
  
  
  
  

  
  You can add itertools.cycle to this as well, to avoid the nested loop. for i in cycle(chain(range(100), range(100,0,-1)).
  – chepner
  Sep 5 at 14:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As commented by @MartijnPieters it uses too much memory at once
  – N Chauhan
  Sep 5 at 14:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @damienfrancois: I disagree that generic unpacking is better. Try that technique with range(10**6), range(10**6, -1, -1), and watch the memory footprint.
  – Martijn Pieters♦
  Sep 5 at 14:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MartijnPieters Noted. Thanks for the feedback
  – damienfrancois
  Sep 5 at 14:24
  

  
  
  
  

12

12

You can add itertools.cycle to this as well, to avoid the nested loop. for i in cycle(chain(range(100), range(100,0,-1)).
– chepner
Sep 5 at 14:23

You can add itertools.cycle to this as well, to avoid the nested loop. for i in cycle(chain(range(100), range(100,0,-1)).
– chepner
Sep 5 at 14:23

As commented by @MartijnPieters it uses too much memory at once
– N Chauhan
Sep 5 at 14:23

As commented by @MartijnPieters it uses too much memory at once
– N Chauhan
Sep 5 at 14:23

1

1

@damienfrancois: I disagree that generic unpacking is better. Try that technique with range(10**6), range(10**6, -1, -1), and watch the memory footprint.
– Martijn Pieters♦
Sep 5 at 14:24

@damienfrancois: I disagree that generic unpacking is better. Try that technique with range(10**6), range(10**6, -1, -1), and watch the memory footprint.
– Martijn Pieters♦
Sep 5 at 14:24

@MartijnPieters Noted. Thanks for the feedback
– damienfrancois
Sep 5 at 14:24

@MartijnPieters Noted. Thanks for the feedback
– damienfrancois
Sep 5 at 14:24

add a comment | 

up vote
14
down vote

As well as the other answers you can use a bit of maths:

while(True):
 for i in range(200):
 if i > 100:
 i = 200 - i

share|improve this answer

edited Sep 5 at 14:24

answered Sep 5 at 14:21

ncfirth

3,206730

add a comment | 

up vote
14
down vote

As well as the other answers you can use a bit of maths:

while(True):
 for i in range(200):
 if i > 100:
 i = 200 - i

share|improve this answer

edited Sep 5 at 14:24

answered Sep 5 at 14:21

ncfirth

3,206730

add a comment | 

up vote
14
down vote

up vote
14
down vote

As well as the other answers you can use a bit of maths:

while(True):
 for i in range(200):
 if i > 100:
 i = 200 - i

share|improve this answer

edited Sep 5 at 14:24

answered Sep 5 at 14:21

ncfirth

3,206730

As well as the other answers you can use a bit of maths:

while(True):
 for i in range(200):
 if i > 100:
 i = 200 - i

share|improve this answer

edited Sep 5 at 14:24

answered Sep 5 at 14:21

ncfirth

3,206730

share|improve this answer

share|improve this answer

edited Sep 5 at 14:24

edited Sep 5 at 14:24

edited Sep 5 at 14:24

answered Sep 5 at 14:21

ncfirth

3,206730

answered Sep 5 at 14:21

ncfirth

3,206730

answered Sep 5 at 14:21

ncfirth

3,206730

3,206730

add a comment | 

add a comment | 

up vote
7
down vote

Here's yet another possibility:

while notConverged:
 for i in xrange(-100, 101):
 print 100 - abs(i)

share|improve this answer

answered Sep 5 at 19:52

twalberg

40.2k76364

• 
  
  
  

  
  

  
  
  
  
  
  

  
  A very interesting approach! +1
  – N Chauhan
  Sep 5 at 20:32
  

  
  
  
  

add a comment | 

up vote
7
down vote

Here's yet another possibility:

while notConverged:
 for i in xrange(-100, 101):
 print 100 - abs(i)

share|improve this answer

answered Sep 5 at 19:52

twalberg

40.2k76364

• 
  
  
  

  
  

  
  
  
  
  
  

  
  A very interesting approach! +1
  – N Chauhan
  Sep 5 at 20:32
  

  
  
  
  

add a comment | 

up vote
7
down vote

up vote
7
down vote

Here's yet another possibility:

while notConverged:
 for i in xrange(-100, 101):
 print 100 - abs(i)

share|improve this answer

answered Sep 5 at 19:52

twalberg

40.2k76364

Here's yet another possibility:

while notConverged:
 for i in xrange(-100, 101):
 print 100 - abs(i)

share|improve this answer

answered Sep 5 at 19:52

twalberg

40.2k76364

share|improve this answer

share|improve this answer

answered Sep 5 at 19:52

twalberg

40.2k76364

answered Sep 5 at 19:52

twalberg

40.2k76364

answered Sep 5 at 19:52

twalberg

40.2k76364

40.2k76364

• 
  
  
  

  
  

  
  
  
  
  
  

  
  A very interesting approach! +1
  – N Chauhan
  Sep 5 at 20:32
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  A very interesting approach! +1
  – N Chauhan
  Sep 5 at 20:32
  

  
  
  
  

A very interesting approach! +1
– N Chauhan
Sep 5 at 20:32

A very interesting approach! +1
– N Chauhan
Sep 5 at 20:32

add a comment | 

up vote
4
down vote

If you've got a repeated set of code, use a function to save space and effort:

def function(x, y, x, num_from_for_loop):
 # 100 lines of code 

while not condition:
 for i in range(1, 101):
 if condition:
 break
 function(x, y, z, i)
 for i in range(100, 0, -1):
 if condition:
 break
 function(x, y, z, i)

You could even use a while True

share|improve this answer

answered Sep 5 at 14:22

N Chauhan

945111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1... I think a lot of the "ooo, I know just the itertools function to help!" answers are neglecting the big picture.
  – Sneftel
  Sep 5 at 20:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I agree, although I honestly don't know anything about itertools save for a few of the simple functions.
  – N Chauhan
  Sep 5 at 20:31
  

  
  
  
  

add a comment | 

up vote
4
down vote

If you've got a repeated set of code, use a function to save space and effort:

def function(x, y, x, num_from_for_loop):
 # 100 lines of code 

while not condition:
 for i in range(1, 101):
 if condition:
 break
 function(x, y, z, i)
 for i in range(100, 0, -1):
 if condition:
 break
 function(x, y, z, i)

You could even use a while True

share|improve this answer

answered Sep 5 at 14:22

N Chauhan

945111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1... I think a lot of the "ooo, I know just the itertools function to help!" answers are neglecting the big picture.
  – Sneftel
  Sep 5 at 20:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I agree, although I honestly don't know anything about itertools save for a few of the simple functions.
  – N Chauhan
  Sep 5 at 20:31
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

If you've got a repeated set of code, use a function to save space and effort:

def function(x, y, x, num_from_for_loop):
 # 100 lines of code 

while not condition:
 for i in range(1, 101):
 if condition:
 break
 function(x, y, z, i)
 for i in range(100, 0, -1):
 if condition:
 break
 function(x, y, z, i)

You could even use a while True

share|improve this answer

answered Sep 5 at 14:22

N Chauhan

945111

If you've got a repeated set of code, use a function to save space and effort:

def function(x, y, x, num_from_for_loop):
 # 100 lines of code 

while not condition:
 for i in range(1, 101):
 if condition:
 break
 function(x, y, z, i)
 for i in range(100, 0, -1):
 if condition:
 break
 function(x, y, z, i)

You could even use a while True

share|improve this answer

answered Sep 5 at 14:22

N Chauhan

945111

share|improve this answer

share|improve this answer

answered Sep 5 at 14:22

N Chauhan

945111

answered Sep 5 at 14:22

N Chauhan

945111

answered Sep 5 at 14:22

N Chauhan

945111

945111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1... I think a lot of the "ooo, I know just the itertools function to help!" answers are neglecting the big picture.
  – Sneftel
  Sep 5 at 20:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I agree, although I honestly don't know anything about itertools save for a few of the simple functions.
  – N Chauhan
  Sep 5 at 20:31
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1... I think a lot of the "ooo, I know just the itertools function to help!" answers are neglecting the big picture.
  – Sneftel
  Sep 5 at 20:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I agree, although I honestly don't know anything about itertools save for a few of the simple functions.
  – N Chauhan
  Sep 5 at 20:31
  

  
  
  
  

+1... I think a lot of the "ooo, I know just the itertools function to help!" answers are neglecting the big picture.
– Sneftel
Sep 5 at 20:28

+1... I think a lot of the "ooo, I know just the itertools function to help!" answers are neglecting the big picture.
– Sneftel
Sep 5 at 20:28

I agree, although I honestly don't know anything about itertools save for a few of the simple functions.
– N Chauhan
Sep 5 at 20:31

I agree, although I honestly don't know anything about itertools save for a few of the simple functions.
– N Chauhan
Sep 5 at 20:31

add a comment | 

up vote
3
down vote

If you're using Python 3.5+, you can using generic unpacking:

for j in (*range(0, 100, 1), *range(100, 0, -1)):

or prior to Python 3.5, you can use itertools.chain:

from itertools import chain

...

for j in chain(range(0, 100, 1), range(100, 0, -1)):

share|improve this answer

answered Sep 5 at 14:19

blhsing

13.8k2628

• 
  
  
  

  
  14
  

  
  
  
  
  
  

  
  Generic unpacking creates a huge tuple with all integers materialised (remember that range() is a very lightweight object). Chaining is better!
  – Martijn Pieters♦
  Sep 5 at 14:22
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

If you're using Python 3.5+, you can using generic unpacking:

for j in (*range(0, 100, 1), *range(100, 0, -1)):

or prior to Python 3.5, you can use itertools.chain:

from itertools import chain

...

for j in chain(range(0, 100, 1), range(100, 0, -1)):

share|improve this answer

answered Sep 5 at 14:19

blhsing

13.8k2628

• 
  
  
  

  
  14
  

  
  
  
  
  
  

  
  Generic unpacking creates a huge tuple with all integers materialised (remember that range() is a very lightweight object). Chaining is better!
  – Martijn Pieters♦
  Sep 5 at 14:22
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

If you're using Python 3.5+, you can using generic unpacking:

for j in (*range(0, 100, 1), *range(100, 0, -1)):

or prior to Python 3.5, you can use itertools.chain:

from itertools import chain

...

for j in chain(range(0, 100, 1), range(100, 0, -1)):

share|improve this answer

answered Sep 5 at 14:19

blhsing

13.8k2628

If you're using Python 3.5+, you can using generic unpacking:

for j in (*range(0, 100, 1), *range(100, 0, -1)):

or prior to Python 3.5, you can use itertools.chain:

from itertools import chain

...

for j in chain(range(0, 100, 1), range(100, 0, -1)):

share|improve this answer

answered Sep 5 at 14:19

blhsing

13.8k2628

share|improve this answer

share|improve this answer

answered Sep 5 at 14:19

blhsing

13.8k2628

answered Sep 5 at 14:19

blhsing

13.8k2628

answered Sep 5 at 14:19

blhsing

13.8k2628

13.8k2628

• 
  
  
  

  
  14
  

  
  
  
  
  
  

  
  Generic unpacking creates a huge tuple with all integers materialised (remember that range() is a very lightweight object). Chaining is better!
  – Martijn Pieters♦
  Sep 5 at 14:22
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  14
  

  
  
  
  
  
  

  
  Generic unpacking creates a huge tuple with all integers materialised (remember that range() is a very lightweight object). Chaining is better!
  – Martijn Pieters♦
  Sep 5 at 14:22
  
  

  
  
  
  

14

14

Generic unpacking creates a huge tuple with all integers materialised (remember that range() is a very lightweight object). Chaining is better!
– Martijn Pieters♦
Sep 5 at 14:22

Generic unpacking creates a huge tuple with all integers materialised (remember that range() is a very lightweight object). Chaining is better!
– Martijn Pieters♦
Sep 5 at 14:22

add a comment | 

up vote
3
down vote

up = True # since we want to go from 0 to 100 first

while True: #for infinite loop

 # For up == True we will print 0-->100 (0,100,1)
 # For up == False we will print 100-->0 (100,0,-1)


 start,stop,step = (0,100,1) if up else (100,0,-1)
 for i in range(start,stop,step):
 print(i)

 up = not up # if we have just printed from 0-->100 (ie up==True), we want to print 100-->0 next so make up False ie up = not up( True) 

 # up will help toggle, between 0-->100 and 100-->0

share|improve this answer

edited Sep 5 at 16:59

answered Sep 5 at 16:10

Tanmay jain

36915

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Pro tip for inverting a boolean: up ^= True
  – wjandrea
  Sep 6 at 0:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @wjandrea Yes thanks for commenting forgot to add as a side note, I was going to use it but then I thought to keep it simple in case OP was beginner.
  – Tanmay jain
  Sep 6 at 3:22
  

  
  
  
  

add a comment | 

up vote
3
down vote

up = True # since we want to go from 0 to 100 first

while True: #for infinite loop

 # For up == True we will print 0-->100 (0,100,1)
 # For up == False we will print 100-->0 (100,0,-1)


 start,stop,step = (0,100,1) if up else (100,0,-1)
 for i in range(start,stop,step):
 print(i)

 up = not up # if we have just printed from 0-->100 (ie up==True), we want to print 100-->0 next so make up False ie up = not up( True) 

 # up will help toggle, between 0-->100 and 100-->0

share|improve this answer

edited Sep 5 at 16:59

answered Sep 5 at 16:10

Tanmay jain

36915

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Pro tip for inverting a boolean: up ^= True
  – wjandrea
  Sep 6 at 0:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @wjandrea Yes thanks for commenting forgot to add as a side note, I was going to use it but then I thought to keep it simple in case OP was beginner.
  – Tanmay jain
  Sep 6 at 3:22
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

up = True # since we want to go from 0 to 100 first

while True: #for infinite loop

 # For up == True we will print 0-->100 (0,100,1)
 # For up == False we will print 100-->0 (100,0,-1)


 start,stop,step = (0,100,1) if up else (100,0,-1)
 for i in range(start,stop,step):
 print(i)

 up = not up # if we have just printed from 0-->100 (ie up==True), we want to print 100-->0 next so make up False ie up = not up( True) 

 # up will help toggle, between 0-->100 and 100-->0

share|improve this answer

edited Sep 5 at 16:59

answered Sep 5 at 16:10

Tanmay jain

36915

up = True # since we want to go from 0 to 100 first

while True: #for infinite loop

 # For up == True we will print 0-->100 (0,100,1)
 # For up == False we will print 100-->0 (100,0,-1)


 start,stop,step = (0,100,1) if up else (100,0,-1)
 for i in range(start,stop,step):
 print(i)

 up = not up # if we have just printed from 0-->100 (ie up==True), we want to print 100-->0 next so make up False ie up = not up( True) 

 # up will help toggle, between 0-->100 and 100-->0

share|improve this answer

edited Sep 5 at 16:59

answered Sep 5 at 16:10

Tanmay jain

36915

share|improve this answer

share|improve this answer

edited Sep 5 at 16:59

edited Sep 5 at 16:59

edited Sep 5 at 16:59

answered Sep 5 at 16:10

Tanmay jain

36915

answered Sep 5 at 16:10

Tanmay jain

36915

answered Sep 5 at 16:10

Tanmay jain

36915

36915

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Pro tip for inverting a boolean: up ^= True
  – wjandrea
  Sep 6 at 0:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @wjandrea Yes thanks for commenting forgot to add as a side note, I was going to use it but then I thought to keep it simple in case OP was beginner.
  – Tanmay jain
  Sep 6 at 3:22
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Pro tip for inverting a boolean: up ^= True
  – wjandrea
  Sep 6 at 0:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @wjandrea Yes thanks for commenting forgot to add as a side note, I was going to use it but then I thought to keep it simple in case OP was beginner.
  – Tanmay jain
  Sep 6 at 3:22
  

  
  
  
  

1

1

Pro tip for inverting a boolean: up ^= True
– wjandrea
Sep 6 at 0:58

Pro tip for inverting a boolean: up ^= True
– wjandrea
Sep 6 at 0:58

@wjandrea Yes thanks for commenting forgot to add as a side note, I was going to use it but then I thought to keep it simple in case OP was beginner.
– Tanmay jain
Sep 6 at 3:22

@wjandrea Yes thanks for commenting forgot to add as a side note, I was going to use it but then I thought to keep it simple in case OP was beginner.
– Tanmay jain
Sep 6 at 3:22

add a comment | 

up vote
1
down vote

def up_down(lowest_value, highest_value):
 current = lowest_value
 delta = 1
 while True: # Begin infinite loop
 yield current
 current += delta
 if current <= lowest_value or current >= highest_value:
 delta *= -1 # Turn around when either limit is hit

This defines a generator, which will continue to yield values for as long as you need. For example:

>>> u = up_down(0, 10)
>>> count = 0
>>> for j in u:
 print(j) # for demonstration purposes
 count += 1 # your other 100 lines of code here
 if count >= 25: # your ending condition here
 break


0
1
2
3
4
5
6
7
8
9
10
9
8
7
6
5
4
3
2
1
0
1
2
3
4

share|improve this answer

answered Sep 6 at 5:23

mathmandan

440713

add a comment | 

up vote
1
down vote

def up_down(lowest_value, highest_value):
 current = lowest_value
 delta = 1
 while True: # Begin infinite loop
 yield current
 current += delta
 if current <= lowest_value or current >= highest_value:
 delta *= -1 # Turn around when either limit is hit

This defines a generator, which will continue to yield values for as long as you need. For example:

>>> u = up_down(0, 10)
>>> count = 0
>>> for j in u:
 print(j) # for demonstration purposes
 count += 1 # your other 100 lines of code here
 if count >= 25: # your ending condition here
 break


0
1
2
3
4
5
6
7
8
9
10
9
8
7
6
5
4
3
2
1
0
1
2
3
4

share|improve this answer

answered Sep 6 at 5:23

mathmandan

440713

add a comment | 

up vote
1
down vote

up vote
1
down vote

def up_down(lowest_value, highest_value):
 current = lowest_value
 delta = 1
 while True: # Begin infinite loop
 yield current
 current += delta
 if current <= lowest_value or current >= highest_value:
 delta *= -1 # Turn around when either limit is hit

This defines a generator, which will continue to yield values for as long as you need. For example:

>>> u = up_down(0, 10)
>>> count = 0
>>> for j in u:
 print(j) # for demonstration purposes
 count += 1 # your other 100 lines of code here
 if count >= 25: # your ending condition here
 break


0
1
2
3
4
5
6
7
8
9
10
9
8
7
6
5
4
3
2
1
0
1
2
3
4

share|improve this answer

answered Sep 6 at 5:23

mathmandan

440713

def up_down(lowest_value, highest_value):
 current = lowest_value
 delta = 1
 while True: # Begin infinite loop
 yield current
 current += delta
 if current <= lowest_value or current >= highest_value:
 delta *= -1 # Turn around when either limit is hit

This defines a generator, which will continue to yield values for as long as you need. For example:

>>> u = up_down(0, 10)
>>> count = 0
>>> for j in u:
 print(j) # for demonstration purposes
 count += 1 # your other 100 lines of code here
 if count >= 25: # your ending condition here
 break


0
1
2
3
4
5
6
7
8
9
10
9
8
7
6
5
4
3
2
1
0
1
2
3
4

share|improve this answer

answered Sep 6 at 5:23

mathmandan

440713

share|improve this answer

share|improve this answer

answered Sep 6 at 5:23

mathmandan

440713

answered Sep 6 at 5:23

mathmandan

440713

answered Sep 6 at 5:23

mathmandan

440713

440713

add a comment | 

add a comment | 

up vote
0
down vote

This is more of a partial answer than a direct answer to your question, but you can also use the notion of trigonometric functions and their oscillation to imitate a 'back and forth' loop.

If we have a cos function with an amplitude of 100, shifted left and upwards so that f(x) = 0 and 0 <= f(x) <= 100, we then have the formula f(x) = 50(cos(x-pi)+1) (plot of graph may be found here. The range is what you require, and oscillation occurs so there's no need to negate any values.

>>> from math import cos, pi
>>> f = lambda x: 50*(cos(x-pi)+1)
>>> f(0)
0.0
>>> f(pi/2)
50.0
>>> f(pi)
100.0
>>> f(3*pi/2)
50.0
>>> f(2*pi)
0.0

The issue of course comes in that the function doesn't give integer values so easily, thus it's not that helpful - but this may be useful for future readers where trigonometric functions might be helpful for their case.

share|improve this answer

answered Sep 6 at 5:38

TerryA

41.6k87699

add a comment | 

up vote
0
down vote

This is more of a partial answer than a direct answer to your question, but you can also use the notion of trigonometric functions and their oscillation to imitate a 'back and forth' loop.

If we have a cos function with an amplitude of 100, shifted left and upwards so that f(x) = 0 and 0 <= f(x) <= 100, we then have the formula f(x) = 50(cos(x-pi)+1) (plot of graph may be found here. The range is what you require, and oscillation occurs so there's no need to negate any values.

>>> from math import cos, pi
>>> f = lambda x: 50*(cos(x-pi)+1)
>>> f(0)
0.0
>>> f(pi/2)
50.0
>>> f(pi)
100.0
>>> f(3*pi/2)
50.0
>>> f(2*pi)
0.0

The issue of course comes in that the function doesn't give integer values so easily, thus it's not that helpful - but this may be useful for future readers where trigonometric functions might be helpful for their case.

share|improve this answer

answered Sep 6 at 5:38

TerryA

41.6k87699

add a comment | 

up vote
0
down vote

up vote
0
down vote

This is more of a partial answer than a direct answer to your question, but you can also use the notion of trigonometric functions and their oscillation to imitate a 'back and forth' loop.

If we have a cos function with an amplitude of 100, shifted left and upwards so that f(x) = 0 and 0 <= f(x) <= 100, we then have the formula f(x) = 50(cos(x-pi)+1) (plot of graph may be found here. The range is what you require, and oscillation occurs so there's no need to negate any values.

>>> from math import cos, pi
>>> f = lambda x: 50*(cos(x-pi)+1)
>>> f(0)
0.0
>>> f(pi/2)
50.0
>>> f(pi)
100.0
>>> f(3*pi/2)
50.0
>>> f(2*pi)
0.0

The issue of course comes in that the function doesn't give integer values so easily, thus it's not that helpful - but this may be useful for future readers where trigonometric functions might be helpful for their case.

share|improve this answer

answered Sep 6 at 5:38

TerryA

41.6k87699

This is more of a partial answer than a direct answer to your question, but you can also use the notion of trigonometric functions and their oscillation to imitate a 'back and forth' loop.

If we have a cos function with an amplitude of 100, shifted left and upwards so that f(x) = 0 and 0 <= f(x) <= 100, we then have the formula f(x) = 50(cos(x-pi)+1) (plot of graph may be found here. The range is what you require, and oscillation occurs so there's no need to negate any values.

>>> from math import cos, pi
>>> f = lambda x: 50*(cos(x-pi)+1)
>>> f(0)
0.0
>>> f(pi/2)
50.0
>>> f(pi)
100.0
>>> f(3*pi/2)
50.0
>>> f(2*pi)
0.0

The issue of course comes in that the function doesn't give integer values so easily, thus it's not that helpful - but this may be useful for future readers where trigonometric functions might be helpful for their case.

share|improve this answer

answered Sep 6 at 5:38

TerryA

41.6k87699

share|improve this answer

share|improve this answer

answered Sep 6 at 5:38

TerryA

41.6k87699

answered Sep 6 at 5:38

TerryA

41.6k87699

answered Sep 6 at 5:38

TerryA

41.6k87699

41.6k87699

add a comment | 

add a comment | 

up vote
0
down vote

I had a similar problem a while ago where I also wanted to create values in the form of an infinite triangle wave, but wanted to step over some values. I ended up using a generator (and the range function as other also have been using):

def tri_wave(min, max, step=1):
 while True:
 yield from range(min, max, step)
 yield from range(max, min, -1 * step)

With carefully selected values on min, max and step (i.e. evenly divisible),

for value in tri_wave(0, 8, 2):
 print(value, end=", ")

I get the min and max value only once, which was my goal:

...0, 2, 4, 6, 8, 6, 4, 2, 0, 2, 4, 6, 8, 6, 4...

I was using Python 3.6 at the time.

share|improve this answer

answered 2 days ago

Amoork

364

add a comment | 

up vote
0
down vote

I had a similar problem a while ago where I also wanted to create values in the form of an infinite triangle wave, but wanted to step over some values. I ended up using a generator (and the range function as other also have been using):

def tri_wave(min, max, step=1):
 while True:
 yield from range(min, max, step)
 yield from range(max, min, -1 * step)

With carefully selected values on min, max and step (i.e. evenly divisible),

for value in tri_wave(0, 8, 2):
 print(value, end=", ")

I get the min and max value only once, which was my goal:

...0, 2, 4, 6, 8, 6, 4, 2, 0, 2, 4, 6, 8, 6, 4...

I was using Python 3.6 at the time.

share|improve this answer

answered 2 days ago

Amoork

364

add a comment | 

up vote
0
down vote

up vote
0
down vote

I had a similar problem a while ago where I also wanted to create values in the form of an infinite triangle wave, but wanted to step over some values. I ended up using a generator (and the range function as other also have been using):

def tri_wave(min, max, step=1):
 while True:
 yield from range(min, max, step)
 yield from range(max, min, -1 * step)

With carefully selected values on min, max and step (i.e. evenly divisible),

for value in tri_wave(0, 8, 2):
 print(value, end=", ")

I get the min and max value only once, which was my goal:

...0, 2, 4, 6, 8, 6, 4, 2, 0, 2, 4, 6, 8, 6, 4...

I was using Python 3.6 at the time.

share|improve this answer

answered 2 days ago

Amoork

364

I had a similar problem a while ago where I also wanted to create values in the form of an infinite triangle wave, but wanted to step over some values. I ended up using a generator (and the range function as other also have been using):

def tri_wave(min, max, step=1):
 while True:
 yield from range(min, max, step)
 yield from range(max, min, -1 * step)

With carefully selected values on min, max and step (i.e. evenly divisible),

for value in tri_wave(0, 8, 2):
 print(value, end=", ")

I get the min and max value only once, which was my goal:

...0, 2, 4, 6, 8, 6, 4, 2, 0, 2, 4, 6, 8, 6, 4...

I was using Python 3.6 at the time.

share|improve this answer

answered 2 days ago

Amoork

364

share|improve this answer

share|improve this answer

answered 2 days ago

Amoork

364

answered 2 days ago

Amoork

364

answered 2 days ago

Amoork

364

364

add a comment | 

add a comment | 

 

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