## Question

Asked By – Apalala

Often enough, I’ve found the need to process a list by pairs. I was wondering which would be the pythonic and efficient way to do it, and found this on Google:

```
pairs = zip(t[::2], t[1::2])
```

I thought that was pythonic enough, but after a recent discussion involving idioms versus efficiency, I decided to do some tests:

```
import time
from itertools import islice, izip
def pairs_1(t):
return zip(t[::2], t[1::2])
def pairs_2(t):
return izip(t[::2], t[1::2])
def pairs_3(t):
return izip(islice(t,None,None,2), islice(t,1,None,2))
A = range(10000)
B = xrange(len(A))
def pairs_4(t):
# ignore value of t!
t = B
return izip(islice(t,None,None,2), islice(t,1,None,2))
for f in pairs_1, pairs_2, pairs_3, pairs_4:
# time the pairing
s = time.time()
for i in range(1000):
p = f(A)
t1 = time.time() - s
# time using the pairs
s = time.time()
for i in range(1000):
p = f(A)
for a, b in p:
pass
t2 = time.time() - s
print t1, t2, t2-t1
```

These were the results on my computer:

```
1.48668909073 2.63187503815 1.14518594742
0.105381965637 1.35109519958 1.24571323395
0.00257992744446 1.46182489395 1.45924496651
0.00251388549805 1.70076990128 1.69825601578
```

If I’m interpreting them correctly, that should mean that the implementation of lists, list indexing, and list slicing in Python is very efficient. It’s a result both comforting and unexpected.

**Is there another, “better” way of traversing a list in pairs?**

Note that if the list has an odd number of elements then the last one will not be in any of the pairs.

**Which would be the right way to ensure that all elements are included?**

I added these two suggestions from the answers to the tests:

```
def pairwise(t):
it = iter(t)
return izip(it, it)
def chunkwise(t, size=2):
it = iter(t)
return izip(*[it]*size)
```

These are the results:

```
0.00159502029419 1.25745987892 1.25586485863
0.00222492218018 1.23795199394 1.23572707176
```

## Results so far

Most pythonic and very efficient:

```
pairs = izip(t[::2], t[1::2])
```

Most efficient and very pythonic:

```
pairs = izip(*[iter(t)]*2)
```

It took me a moment to grok that the first answer uses two iterators while the second uses a single one.

To deal with sequences with an odd number of elements, the suggestion has been to augment the original sequence adding one element (`None`

) that gets paired with the previous last element, something that can be achieved with `itertools.izip_longest()`

.

## Finally

Note that, in Python 3.x, `zip()`

behaves as `itertools.izip()`

, and `itertools.izip()`

is gone.

**Now we will see solution for issue: Pairs from single list [duplicate] **

## Answer

My favorite way to do it:

```
def pairwise(t):
it = iter(t)
return zip(it,it)
# for "pairs" of any length
def chunkwise(t, size=2):
it = iter(t)
return zip(*[it]*size)
```

When you want to pair all elements you obviously might need a fillvalue:

```
from itertools import izip_longest
def blockwise(t, size=2, fillvalue=None):
it = iter(t)
return izip_longest(*[it]*size, fillvalue=fillvalue)
```

With Python 3, `itertools.izip`

is now simply `zip`

.. to work with an older Python, use

```
from itertools import izip as zip
```

This question is answered By – Jochen Ritzel

**This answer is collected from stackoverflow and reviewed by FixPython community admins, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 **