Incrementing in a for loop

Question

So my issue is with this not incrementing correctly… I tried to uses an int “step” to + 1 every time this loop is ran but it doesn’t do anything. Why is that? Also when I print(step) it only adds up to 337. It does not go the full 1000 like I had thought I asked it too. How do I do this correctly?

lockers = []

step = 3

locker = 0

while len(lockers) <= 1000:
     lockers.append(1)

for i in range(0, len(lockers)):
     lockers[i] = 0

for i in range(0, len(lockers), 2):
     lockers[i] = 1

for i in range(0, len(lockers), step):
     if lockers[i] == 0:
          lockers [i] = 1
     else:
          lockers[i] = 0

     step += 1


print(lockers)

Asked By: Casey

||

Source

Answer 1

range gives you an iterable object:

>>> range(10,20 , 2)
range(10, 20, 2)
>>> list(range(10,20 , 2))
[10, 12, 14, 16, 18]

The values in it are fully decided as soon as the call returns, and aren’t re-evaluated each time around the loop. Your step only goes up to 337 because you are incrementing it once for each element in the object range(0, 1000, 3), which has 334 items, not 1000:

>>> len(range(0,1000,3))
334

To get something that works like range but advances the step, you would need to write your own generator:

def advancing_range(start, stop, step):
    ''' Like range(start, stop, step) except that step is incremented 
        between each value 
    '''
    while start < stop:
        yield start
        start += step
        step += 1

You can then do for i in advancing_range(0, 1000, 3): and it will work as you intend.

But this is a very strange thing to want to do. Judging by your variable names, I would guess you’re coding the locker problem, which says:

A new high school has just been completed. There are 1,000 lockers
in the school and they have been numbered from 1 through 1,000.
During recess (remember this is a fictional problem), the students
decide to try an experiment. When recess is over each student will
walk into the school one at a time. The first student will open all
of the locker doors. The second student will close all of the locker
doors with even numbers. The third student will change all of the
locker doors that are multiples of 3 (change means closing lockers
that are open, and opening lockers that are closed.) The fourth
student will change the position of all locker doors numbered with
multiples of four and so on. After 1,000 students have entered the
school, which locker doors will be open, and why?

But the advancing range logic says something more like "the first student opens the first locker, then the second opens the second locker after that, then the third student opens the third locker after that …". You want to affect multiple lockers each time, but further spaced out. Essentially, you want to copy and paste your first two loops another 998 times with a one higher step each time. Of course, you can do better than copy and paste, and this seems like you want two nested loops, where the outer one advances the step that the inner one uses. That would look like this:

for step in range(1, len(lockers)):
    for i in range(step, len(lockers), step):

Simplifying your other logic by using booleans instead of 1 and 0, the whole program looks like this:

lockers = [True] * 1000

for step in range(1, len(lockers)):
    for i in range(step, len(lockers), step):
        lockers[i] = not lockers[i]

print(sum(lockers))

It prints that the number of open lockers is 969.

Answered By: lvc

Answer 2

If you want to adjust the step size while iterating, you can have an own range object:

class AdjustableRange(object):
    def __init__(self, start, stop, step):
        self.start = start
        self.stop = stop
        self.step = step
        self.value = None
    def __iter__(self):
        if self.value is None:
            self.value = start
        while self.value < self.stop:
            yield self.value
            self.value += self.step

This (untested) one you can use for iterting like

rg = AdjustableRange(0, len(lockers), step):
for i in rg:
    if lockers[i] == 0:
        lockers [i] = 1
    else:
        lockers[i] = 0
    rg.step += 1 # this influences the iteration

But, as was already said, there are better ways to solve your “real” problem.

Answered By: glglgl

Incrementing in a for loop

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