# Getting next element while cycling through a list

Each Answer to this Q is separated by one/two green lines.

``````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?

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`.

``````while running:
for elem,next_elem in zip(li, li[1:]+[li[0]]):
...
``````

You can use a pairwise cyclic iterator:

``````from itertools import izip, cycle, tee

def pairwise(seq):
a, b = tee(seq)
next(b)
return izip(a, b)

for elem, next_elem in pairwise(cycle(li)):
...
``````

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

Use the zip method in Python. This function returns a list of tuples, where the i-th tuple contains the i-th element from each of the argument sequences or iterables

``````    while running:
for thiselem,nextelem in zip(li, li[1 : ] + li[ : 1]):
#Do whatever you want with thiselem and nextelem
``````

A rather different way to solve this:

``````   li = [0,1,2,3]

for i in range(len(li)):

if i < len(li)-1:

# until end is reached
print 'this', li[i]
print 'next', li[i+1]

else:

# end
print 'this', li[i]
``````

As simple as this:

``````li = [0, 1, 2, 3]
while 1:
for i, item in enumerate(x):
k = i + 1 if i != len(x) - 1 else 0
print('Current index {} : {}'.format(i,li[k]))
``````

The simple solution is to remove IndexError by incorporating the condition:

``````if(index<(len(li)-1))
``````

The error ‘index out of range’ will not occur now as the last index will not be reached. The idea is to access the next element while iterating. On reaching the penultimate element, you can access the last element.

Use enumerate method to add index or counter to an iterable(list, tuple, etc.). Now using the index+1, we can access the next element while iterating through the list.

``````li = [0, 1, 2, 3]

running = True
while running:
for index, elem in enumerate(li):
if(index<(len(li)-1)):
thiselem = elem
nextelem = li[index+1]
``````

``````        li = [0, 1, 2, 3]
for elem in li:
if (li.index(elem))+1 != len(li):
thiselem = elem
nextelem = li[li.index(elem)+1]
print 'thiselem',thiselem
print 'nextel',nextelem
else:
print 'thiselem',li[li.index(elem)]
print 'nextel',li[li.index(elem)]
``````

``````   while running:
lenli = len(li)
for i, elem in enumerate(li):
thiselem = elem
nextelem = li[(i+1)%lenli] # This line is vital
``````

I’ve used enumeration to handle this problem.

``````storage=""
for num, value in enumerate(result, start=0):
content = value
if 'A' == content:
storage = result[num + 1]
``````

I’ve used num as Index here, when it finds the correct value it adds up one to the current index of actual list. Which allows me to maneuver to the next index.

I hope this helps your purpose.

For strings list from 1(or whatever > 0) until end.

``````itens = ['car', 'house', 'moon', 'sun']

v = 0
for item in itens:
b = itens[1 + v]
print(b)
print('any other command')
if b == itens[-1]:
print('End')
break
v += 1
``````

The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .