Back and forth loop Python

Question

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?

Wenzel

"so that I don't have write out the same code" Sounds like a good usecase for a function – Motile

@Phylogenesis TypeError: unsupported operand type(s) for +: 'range' and 'range'. This will only work in Python 2 – Motile

Yeah, that's Python 2 only. – Fitter

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. – Malversation

Answer 1

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))):
    ....

Josh

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)). – Inference

As commented by @MartijnPieters it uses too much memory at once – Wetmore

@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. – Malversation

Thanks. I didn't knew about itertools before. They are great! – Luxuriance

Answer 2

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

Boeotia

Answer 3

Here's yet another possibility:

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

Oersted

Answer 4

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

Wetmore

+1... I think a lot of the "ooo, I know just the itertools function to help!" answers are neglecting the big picture. – Metalliferous

I agree, although I honestly don't know anything about itertools save for a few of the simple functions. – Wetmore

Answer 5

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)):

Mechellemechlin

Generic unpacking creates a huge tuple with all integers materialised (remember that range() is a very lightweight object). Chaining is better! – Malversation

Answer 6

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

Nicely

Pro tip for inverting a boolean: up ^= True – Ravelin

@Ravelin 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. – Nicely

Answer 7

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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Hoofbound

Answer 8

I became curious if it's possible to implement such kind of triangle oscillator without conditions and enumerations. Well, one option is the following:

def oscillator(magnitude):
   i = 0
   x = y = -1
   double_magnitude = magnitude + magnitude

   while True:
       yield i
       x = (x + 1) * (1 - (x // (double_magnitude - 1)))  # instead of (x + 1) % double_magnitude
       y = (y + 1) * (1 - (y // (magnitude - 1)))         # instead of (y + 1) % magnitude
       difference = x - y                                 # difference ∈ {0, magnitude}
       derivative = (-1 * (difference > 0) + 1 * (difference == 0))
       i += derivative

The idea behind this is to take 2 sawtooth waves with different periods and subtract one from another. The result will be a square wave with values in {0, magnitude}. Then we just substitute {0, magnitude} with {-1, +1} respectively to get derivative values for our target signal.

Let's look at example with magnitude = 5:

o = oscillator(5)
[next(o) for _ in range(21)]

This outputs [0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0].

If abs() is allowed, it can be used for simplicity. For example, the following code gives the same output as above:

[abs(5 - ((x + 5) % 10)) for x in range(21)]

Panpsychist

Answer 9

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.

Stefanistefania

Answer 10

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.

Insolence

Answer 11

Here is a simple and straightforward solution that does not require any imports:

index = 0
constant = 100

while True:
    print(index)
    if index == constant:
        step = -1
    elif index == 0:
        step = 1

    index += step

Pongee