CODEWITHSUNDEEP
pythonlistiteration
ignoramus
ignoramus

Reputation:

Getting next element while cycling through a list

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?

Upvotes: 77

Views: 302193

Answers (12)

Bipin
Bipin

Reputation: 483

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]

Upvotes: 1

Marcos Paolo
Marcos Paolo

Reputation: 371

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

Upvotes: 0

Mehdi Khlifi
Mehdi Khlifi

Reputation: 415

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]))

Upvotes: 1

vijayraj34
vijayraj34

Reputation: 2415

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.

Upvotes: 0

T Joseph
T Joseph

Reputation: 1

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

Upvotes: 0

Omnifarious
Omnifarious

Reputation: 56088

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.

Upvotes: 124

Satheesh Alathiyur
Satheesh Alathiyur

Reputation: 111

        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)]

Upvotes: 0

Nirmal
Nirmal

Reputation: 81

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

Upvotes: 6

Timothy Dalton
Timothy Dalton

Reputation: 1390

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]

Upvotes: 6

Ned Batchelder
Ned Batchelder

Reputation: 375992

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

Upvotes: 7

Ants Aasma
Ants Aasma

Reputation: 54935

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)):
    ...

Upvotes: 7

John La Rooy
John La Rooy

Reputation: 304463

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

Upvotes: 24

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