Getting next element while cycling through a list

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li = [0, 1, 2, 3]

running = True
while running:
    for elem in li:
        thiselem = elem
        nextelem = li[li.index(elem)+1]

When this reaches the last element, an IndexError is raised (as is the case for any list, tuple, dictionary, or string that is iterated). I actually want at that point for nextelem to equal li[0]. My rather cumbersome solution to this was

while running:
    for elem in li:
        thiselem = elem
        nextelem = li[li.index(elem)-len(li)+1]   # negative index

Is there a better way of doing this?

Answer

After thinking this through carefully, I think this is the best way. It lets you step off in the middle easily without using break, which I think is important, and it requires minimal computation, so I think it’s the fastest. It also doesn’t require that li be a list or tuple. It could be any iterator.

from itertools import cycle

li = [0, 1, 2, 3]

running = True
licycle = cycle(li)
# Prime the pump
nextelem = next(licycle)
while running:
    thiselem, nextelem = nextelem, next(licycle)

I’m leaving the other solutions here for posterity.

All of that fancy iterator stuff has its place, but not here. Use the % operator.

li = [0, 1, 2, 3]

running = True
while running:
    for idx, elem in enumerate(li):
        thiselem = elem
        nextelem = li[(idx + 1) % len(li)]

Now, if you intend to infinitely cycle through a list, then just do this:

li = [0, 1, 2, 3]

running = True
idx = 0
while running:
    thiselem = li[idx]
    idx = (idx + 1) % len(li)
    nextelem = li[idx]

I think that’s easier to understand than the other solution involving tee, and probably faster too. If you’re sure the list won’t change size, you can squirrel away a copy of len(li) and use that.

This also lets you easily step off the ferris wheel in the middle instead of having to wait for the bucket to come down to the bottom again. The other solutions (including yours) require you check running in the middle of the for loop and then break.



Source: stackoverflow