EliseB -4 years ago 158
Python Question

# Better way to swap elements in a list?

I have a bunch of lists that look like this one:

``````l = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
``````

I want to swap elements as follows:

``````final_l = [2, 1, 4, 3, 6, 5, 8, 7, 10, 9]
``````

The size of the lists may vary, but they will always contain an even number of elements.

I'm fairly new to Python and am currently doing it like this:

``````l =  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
final_l = []
for i in range(0, len(l)/2):
final_l.append(l[2*i+1])
final_l.append(l[2*i])
``````

I know this isn't really Pythonic and would like to use something more efficient. Maybe a list comprehension?

No need for complicated logic, simply rearrange the list with slicing and step:

``````In [1]: l = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

In [2]: l[::2], l[1::2] = l[1::2], l[::2]

In [3]: l
Out[3]: [2, 1, 4, 3, 6, 5, 8, 7, 10, 9]
``````

## TLDR;

Edited with explanation

I believe most viewers are already familiar with list slicing and multiple assignment. In case you don't I will try my best to explain what's going on (hope I do not make it worse).

To understand list slicing, here already has an excellent answer and explanation of list slice notation. Simply put:

``````a[start:end] # items start through end-1
a[start:]    # items start through the rest of the array
a[:end]      # items from the beginning through end-1
a[:]         # a copy of the whole array

There is also the step value, which can be used with any of the above:

a[start:end:step] # start through not past end, by step
``````

Let's look at OP's requirements:

`````` [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]  # list l
^  ^  ^  ^  ^  ^  ^  ^  ^  ^
0  1  2  3  4  5  6  7  8  9    # respective index of the elements
l[0]  l[2]  l[4]  l[6]  l[8]      # first tier : start=0, step=2
l[1]  l[3]  l[5]  l[7]  l[9]   # second tier: start=1, step=2
-----------------------------------------------------------------------
l[1]  l[3]  l[5]  l[7]  l[9]
l[0]  l[2]  l[4]  l[6]  l[8]   # desired output
``````

First tier will be: `l[::2] = [1, 3, 5, 7, 9]` Second tier will be: `l[1::2] = [2, 4, 6, 8, 10]`

As we want to re-assign `first = second` & `second = first`, we can use multiple assignment, and update the original list in place:

``````first , second  = second , first
``````

that is:

``````l[::2], l[1::2] = l[1::2], l[::2]
``````

As a side note, to get a new list but not altering original `l`, we can assign a new list from `l`, and perform above, that is:

``````n = l[:]  # assign n as a copy of l (without [:], n still points to l)
n[::2], n[1::2] = n[1::2], n[::2]
``````

Hopefully I do not confuse any of you with this added explanation. If it does, please help update mine and make it better :-)

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