Iterate Over All Pairs of Consecutive Items in a List

Iterate over all pairs of consecutive items in a list

Just use zip

>>> l = [1, 7, 3, 5]
>>> for first, second in zip(l, l[1:]):
...     print first, second
...
1 7
7 3
3 5

If you use Python 2 (not suggested) you might consider using the izip function in itertools for very long lists where you don't want to create a new list.

import itertools

for first, second in itertools.izip(l, l[1:]):
    ...

R iterate over consecutive pairs in a list

We can remove the first and last elements and concatenate in Map

Map(c,  v1[-length(v1)], v1[-1])
#[[1]]
#[1] 1 2

#[[2]]
#[1] 2 3

#[[3]]
#[1] 3 4

---

Or rbind and use asplit

asplit(rbind(v1[-length(v1)], v1[-1]), 2)

python - iterate over all pairs of consecutive items with gaps

One fairly easy way to do it is to think about it in terms of the index rather than the list items itself. Start with:

import itertools
list(itertools.combinations(range(len(mylist)), 3)

This gets you all the possible index triple combinations in a list with the length of your list. Now you want to filter them to exclude any where the last index is 4 or more away from the first:

list(filter(lambda seq: (seq[-1] - seq[0]) <= 4, itertools.combinations(l, 3)))

This gets you the indeces you want. So now you can get the triples you need based on the indeces:

[[mylist[i] for i in seq] for seq in filter(lambda seq: (seq[-1] - seq[0]) < 4, itertools.combinations(l, 3))]

which produces:

[[1, 6, 4], [1, 6, 9], [1, 4, 9], [6, 4, 9], [6, 4, 2], [6, 9, 2], [4, 9, 2]]

Iterate over all pairs (or triples) of consecutive elements with wrap-around

Probably you are looking for something like this:

mylist = [1,2,3,4,5,6,7,8,9,10]
for previous, current, following in zip([mylist[-1]] + mylist[:-1], mylist, mylist[1:] + [mylist[0]]):
    print(previous, current, following)

Here I simply used zip and list concatenation.

Iterate consecutive elements in a list in Python such that the last element combines with first

You can simply extend the second list in zip() with a list with only the first item, something like:

for first, second in zip(L, L[1:] + L[0:1]):  # or simply zip(L, L[1:] + L[:1])
    print([first, second])

Iterating over every two elements in a list

You need a pairwise() (or grouped()) implementation.

def pairwise(iterable):
    "s -> (s0, s1), (s2, s3), (s4, s5), ..."
    a = iter(iterable)
    return zip(a, a)

for x, y in pairwise(l):
   print("%d + %d = %d" % (x, y, x + y))

Or, more generally:

def grouped(iterable, n):
    "s -> (s0,s1,s2,...sn-1), (sn,sn+1,sn+2,...s2n-1), (s2n,s2n+1,s2n+2,...s3n-1), ..."
    return zip(*[iter(iterable)]*n)

for x, y in grouped(l, 2):
   print("%d + %d = %d" % (x, y, x + y))

In Python 2, you should import izip as a replacement for Python 3's built-in zip() function.

All credit to martineau for his answer to my question, I have found this to be very efficient as it only iterates once over the list and does not create any unnecessary lists in the process.

N.B: This should not be confused with the pairwise recipe in Python's own itertools documentation, which yields s -> (s0, s1), (s1, s2), (s2, s3), ..., as pointed out by @lazyr in the comments.

Little addition for those who would like to do type checking with mypy on Python 3:

from typing import Iterable, Tuple, TypeVar

T = TypeVar("T")

def grouped(iterable: Iterable[T], n=2) -> Iterable[Tuple[T, ...]]:
    """s -> (s0,s1,s2,...sn-1), (sn,sn+1,sn+2,...s2n-1), ..."""
    return zip(*[iter(iterable)] * n)

How can I iterate over overlapping (current, next) pairs of values from a list?

Here's a relevant example from the itertools module docs:

import itertools
def pairwise(iterable):
    "s -> (s0, s1), (s1, s2), (s2, s3), ..."
    a, b = itertools.tee(iterable)
    next(b, None)
    return zip(a, b)   

For Python 2, you need itertools.izip instead of zip:

import itertools
def pairwise(iterable):
    "s -> (s0, s1), (s1, s2), (s2, s3), ..."
    a, b = itertools.tee(iterable)
    next(b, None)
    return itertools.izip(a, b)

How this works:

First, two parallel iterators, a and b are created (the tee() call), both pointing to the first element of the original iterable. The second iterator, b is moved 1 step forward (the next(b, None)) call). At this point a points to s0 and b points to s1. Both a and b can traverse the original iterator independently - the izip function takes the two iterators and makes pairs of the returned elements, advancing both iterators at the same pace.

One caveat: the tee() function produces two iterators that can advance independently of each other, but it comes at a cost. If one of the iterators advances further than the other, then tee() needs to keep the consumed elements in memory until the second iterator comsumes them too (it cannot 'rewind' the original iterator). Here it doesn't matter because one iterator is only 1 step ahead of the other, but in general it's easy to use a lot of memory this way.

And since tee() can take an n parameter, this can also be used for more than two parallel iterators:

def threes(iterator):
    "s -> (s0, s1, s2), (s1, s2, s3), (s2, s3, 4), ..."
    a, b, c = itertools.tee(iterator, 3)
    next(b, None)
    next(c, None)
    next(c, None)
    return zip(a, b, c)

Python group consecutive elements of a list into two's without repetition

What we do is we first flatten the list and then we index the flattened list to the 2d list

l1 = [[0,1],[0,1,2,3,4,5,6,7],[0,1,2,3]]
def fix(l1):
    l1 = [item for sublist in l1 for item in sublist]
    l1 = [l1[i:i+2] for i in range(0, len(l1), 2)]
    return l1
l1 = fix(l1)
print(l1)

Output:

[[0, 1], [0, 1], [2, 3], [4, 5], [6, 7], [0, 1], [2, 3]]

Scala iterate over two consecutive elements of a list

If I understand correctly you probably want to iterate over a sliding window.

list.sliding(2).map{
  case List(a, b) => a.diff(b)
  case List(a) => a
}.toList

Alternatively you might also want grouped(2) which partitions the list into groups instead.

---

Related Topics