Iterating over every two elements in a list [duplicate] Iterating over every two elements in a list [duplicate] python python

Iterating over every two elements in a list [duplicate]


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, TypeVarT = 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)


Well you need tuple of 2 elements, so

data = [1,2,3,4,5,6]for i,k in zip(data[0::2], data[1::2]):    print str(i), '+', str(k), '=', str(i+k)

Where:

  • data[0::2] means create subset collection of elements that (index % 2 == 0)
  • zip(x,y) creates a tuple collection from x and y collections same index elements.


>>> l = [1,2,3,4,5,6]>>> zip(l,l[1:])[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]>>> zip(l,l[1:])[::2][(1, 2), (3, 4), (5, 6)]>>> [a+b for a,b in zip(l,l[1:])[::2]][3, 7, 11]>>> ["%d + %d = %d" % (a,b,a+b) for a,b in zip(l,l[1:])[::2]]['1 + 2 = 3', '3 + 4 = 7', '5 + 6 = 11']


matomo