CODEWITHSUNDEEP
pythonfor-loopiteration
pyramidface
pyramidface

Reputation: 1277

Is there an easy way to get next and prev values in a for-loop?

So...is there an easy way to get next and previous values while iterating with a for-loop in Python?

I know this can be easily done if you do something like:

a = [3,4,5,6,3,4,5]
for x in range(len(a)):
    next = a[x+1]

But what about:

for x in a:
    x.next??

Upvotes: 4

Views: 152

Answers (6)

Stuart
Stuart

Reputation: 9858

Probably overkill but I sometimes use the following more general generator for this, which yields a sequence of 'windows' of any size on a list or other iterable. (The window size must be less than the length of the iterable.)

def sliding_window(iterable, size):
    try:                    # indexed iterable
        for i in range(len(iterable) - size + 1):
            yield iterable[i:i+size]
    except TypeError:       # iterator
        window = [next(iterable) for _ in range(size)]
        yield window
        for item in iterable:
            window = window[1:] + [item]
            yield window

a = [3,4,5,6,3,4,5]
for current, following in sliding_window(a, 2):
    print(current, following)

Upvotes: 1

user2555451
user2555451

Reputation:

You could always convert a into an iterator with iter and then iterate over that. This will allow you to use next inside the loop to advance the iterator that you are iterting over:

>>> a = [3,4,5,6,3,4,5]
>>> it = iter(a)
>>> for i in it:
...     j = next(it, None)
...     print('i={}, j={}'.format(i, j))
...
i=3, j=4
i=5, j=6
i=3, j=4
i=5, j=None
>>>

Also, the None in there is the default value to return if there is no next item. You can set it to whatever value you want though. Omitting the argument will cause a StopIteration exception to be raised:

>>> a = [1, 2, 3, 4, 5]
>>> it = iter(a)
>>> for i in it:
...     j = next(it)
...     print('i={}, j={}'.format(i, j))
...
i=1, j=2
i=3, j=4
Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
StopIteration
>>>

Upvotes: 3

Hackaholic
Hackaholic

Reputation: 19733

this is easy too:

>>> a = [3,4,5,6,3,4,5]
>>> for i in range(1,len(a)):
...     print a[i-1],a[i]
... 
3 4
4 5
5 6
6 3
3 4
4 5

Upvotes: 1

kums
kums

Reputation: 2691

If you want both the previous and the next element in a circular sequence for each iteration:

a = [3,4,5,6,3,4,5]
l = len(a)

for k, v in enumerate(a):
    print a[(k-1)%l], v, a[(k+1)%l] #prints previous, current, next elements

Upvotes: 1

FogleBird
FogleBird

Reputation: 76762

Here is a common pattern that I use to iterate over pairs of items in a sequence:

>>> a = range(10)
>>> for i, j in zip(a, a[1:]):
...  print i, j
... 
0 1
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9

If you want three items (prev, item, next) you can do this:

>>> for i, j, k in zip(a, a[1:], a[2:]):
...  print i, j, k
... 
0 1 2
1 2 3
2 3 4
3 4 5
4 5 6
5 6 7
6 7 8
7 8 9

i is the previous element, j is the current element, k is the next element.

Of course, this "starts" at 1 and "ends" at 8. What should you receive as prev/next at the ends? Perhaps None? Probably easiest to just do this:

>>> a = [None] + a + [None]
>>> for i, j, k in zip(a, a[1:], a[2:]):
...  print i, j, k
... 
None 0 1
0 1 2
1 2 3
2 3 4
3 4 5
4 5 6
5 6 7
6 7 8
7 8 9
8 9 None

Upvotes: 4

Jon Kiparsky
Jon Kiparsky

Reputation: 7743

easiest way I know of is 

for x,next in zip (a, a[1:]):
  # now you have x and next available

Upvotes: 3

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