Better way to swap elements in a list?

Question

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:

fi         


        

Answer 1

An(other) alternative:

final_l = list() # make an empty list
for i in range(len(l)): # for as many items there are in the original list
    if i % 2 == 0: # if the item is even
        final_l.append(l[i+1]) # make this item in the new list equal to the next in the original list
    else: # else, so when the item is uneven
        final_l.append(l[i-1]) # make this item in the new list equal to the previous in the original list

This assumes that the original list has an even number of items. If not, a try-except can be added:

final_l = list()
for i in range(len(l)):
    if i % 2 == 0:
        try: # try if we can add the next item
            final_l.append(l[i+1])
        except:  # if we can't (because i+1 doesnt exist), add the current item
            final_l.append(l[i])
    else:
            final_l.append(l[i-1])

Answer 2

newList = [(x[2*i+1], x[2*i]) for i in range(0, len(x)/2)]

Now find a way to unzip the tuples. I won't do all of your homework.

Answer 3

Here a single list comprehension that does the trick:

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

In [2]: [l[i^1] for i in range(len(l))]
Out[2]: [2, 1, 4, 3, 6, 5, 8, 7, 10, 9]

The key to understanding it is the following demonstration of how it permutes the list indices:

In [3]: [i^1 for i in range(10)]
Out[3]: [1, 0, 3, 2, 5, 4, 7, 6, 9, 8]

The ^ is the exclusive or operator. All that i^1 does is flip the least-significant bit of i, effectively swapping 0 with 1, 2 with 3 and so on.

Answer 4

You can use the pairwise iteration and chaining to flatten the list:

>>> from itertools import chain
>>>
>>> list(chain(*zip(l[1::2], l[0::2])))
[2, 1, 4, 3, 6, 5, 8, 7, 10, 9]

Or, you can use the itertools.chain.from_iterable() to avoid the extra unpacking:

>>> list(chain.from_iterable(zip(l[1::2], l[0::2])))
[2, 1, 4, 3, 6, 5, 8, 7, 10, 9]

Answer 5

One of the possible answer using chain and list comprehension

>>> l = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> list(chain([(l[2*i+1], l[2*i]) for i in range(0, len(l)/2)]))
[(2, 1), (4, 3), (6, 5), (8, 7), (10, 9)]

Answer 6

New to stack overflow. Please free to leave a comment or feedback on this solution. swap = [2, 1, 4, 3, 5]

lst = []
for index in range(len(swap)):
        if index%2 == 0 and index < len(swap)-1:
            swap[index],swap[index+1]  = swap[index+1],swap[index]
        lst.append(swap[index])
print(lst)


out = [1, 2, 3, 4, 5]