This has been one of the processes I’ve normally not solved in a clean or readable way.
In []: a = [1,2,3,4,5,6,7,8]
Taking the top output I want to return something similar to below.
Out[]: [(1, 2), (3, 4), (5, 6), (7, 8)]
Using a Pythonic approach this task isn’t to difficult, but there is some explaining to do.
First lets look at the code:
In []: a = iter([1,2,3,4,5,6,7,8])
In []: [ i for i in zip(a, a) ]
Out[]: [(1, 2), (3, 4), (5, 6), (7, 8)]
First we need to pass our list to the iter function, this turns our list into a iterator object:
In []: a
Out[]: <listiterator at 0x1037c3110>
This iterator can be used like any other iterator as it has a next method:
In []: a = iter([1,2,3,4,5,6,7,8])
In []: a.next()
Out[]: 1
In []: a.next()
Out[]: 2
In []: a.next()
Out[]: 3
The next bit to understand is the zip function:
In []: a = [1,2,3,4]
In []: b = [5,6,7,8]
In []: zip(a, b)
Out[]: [(1, 5), (2, 6), (3, 7), (4, 8)]
What zip does is takes the 1st object ( next method ) from each iterator-able object passed in and packs them together in a tuple, it then continues till one or all objects throws a StopIteration :
In []: a = iter([1,2])
In []: a.next()
Out[]: 1
In []: a.next()
Out[]: 2
In []: a.next()
---------------------------------------------------------------------------
StopIteration Traceback (most recent call last)
<ipython-input-52-aa817a57a973> in <module>()
----> 1 a.next()
StopIteration:
So to put it all together zip takes our ‘a’ variable (an iter object of our list) as a argument
twice
.
[ i for i in zip(**a**, **a**) ]
Zip then grabs the first item from our first argument
using the next method,
and the first item from our second argument
using the next method
and bundles them in a tuple. If we were to do this by hand it would look like this:
In []: a = iter([1,2,3,4,5,6,7,8])
In []: first = a.next()
In []: second = a.next()
In []: first, second
Out[]: (1, 2)
Then if we were to do it again:
In []: first = a.next()
In []: second = a.next()
In []: first, second
Out[]: (3, 4)
Hopefully this makes sense now.