How to perform a pairwise swap of a list?

Question

I want to write a small code in python that Swap Elements in a list this program will accept a list, and will return a list that exchanges the positions of each pair of adjacent elements: positions 0 and 1, positions 2 and 3, and so on. If the list has an odd number of elements, then the element in the last position stays “in place”.

Before: [1,2,3,4,5]
After: [2,1,4,3,5]

This looks unpythonic. What is the Python way to do it?

Answer 1

Most pythonic way:

def swappairwise(a):
  l = len(a)&~1
  a[1:l:2],a[:l:2] = a[:l:2],a[1:l:2]

Answer 2

Here is a neat one, if you are always guaranteed to have even numbers:

 nums = [1,2,3,4,5,6]
 print([nums[i^1] for i in range(len(nums))])
 >>[2, 1, 4, 3, 6, 5] 

Explanation:

print (0^1)    #1
print (1^1)    #0
print (2^1)    #3
print (3^1)    #2
print (4^1)    #5
print (5^1)    #4

As a refresher, the XOR has the following effect:

A   B | Output
---------------
0   0     0
0   1     1
1   0     1
1   1     0

And the official description: Each bit of the output is the same as the corresponding bit in x if that bit in y is 0, and it's the complement of the bit in x if that bit in y is 1.

Answer 3

Building on the answer above from @Arpegius, here a, hopefully, somewhat more readable solution. Uses the same approach.

def swap_list_pairwise(lis):
    """Pairwise swap of all elements in a list.

    If the number of elements is odd, the leftover element
    stays at its place.
    """
    length = len(lis)  
    # Stop at second last if length is odd, otherwise use full list. 
    end = length - 1 if length % 2 else length
    lis[1:end:2], lis[:end:2] = lis[:end:2], lis[1:end:2]

Answer 4

If you want "pythonic", try "How do you split a list into evenly sized chunks in Python?", followed by a map() that reverses every chunk. May not be too performant, though.

(Oh, forgot the flattening of the list at the end)

Answer 5

Here is a way:

def pairwise_swap(iterable):
    for i, value in enumerate(iterable):
        if i % 2 == 0:
            saved = value
        else:
            yield value
            yield saved

>>> list(pairwise_swap(range(10)))
[1, 0, 3, 2, 5, 4, 7, 6, 9, 8]

Answer 6

Nice approach by @Alok above. This should fix the missing last element if the number of elements is odd.

def pairwise_swap(iterable):
    """Pairwise swap of all elements in an iterable.

    If the number of elements is odd, the leftover element
    stays at its place.
    """
    for i, value in enumerate(iterable):
        if i % 2 == 0:
            saved = value
        else:
            yield value
            yield saved
    # Don't forget the last element if `iterable` has an odd 
    # number of elements. Since counting starts a zero, we need 
    # to test if `i` is even. 
    if iterable and i % 2 == 0:
        yield value

Answer 7

How about trying this one out?

>>> l = [1,2,3,4,5,6]
>>> pl = [l[i:i+2] for i in range(0,len(l),2)]
>>> pl
[[1, 2], [3, 4], [5, 6]]
>>> for i in pl:
...     i[0],i[1] = i[1],i[0]
...     print i
... 
[2, 1]
[4, 3]
[6, 5]
>>> pl
[[2, 1], [4, 3], [6, 5]]
>>> 
>>> zpl = [i for sublist in pl for i in sublist]
>>> zpl
[2, 1, 4, 3, 6, 5]
>>> 

I tried to resolve in the easiest way out.

Answer 8

It will work for both even and odd elements. if list elements are even 
first part will work if not else will do his task.

a = list(input("Put your list here: "))
len_a = len(a)
last_element = len_a-1

result = 0
if len_a % 2==0:
        for i in range(0, len(a), 2):
                a[i], a[i + 1] = a[i + 1], a[i]
        print("its if",a)

else:
        a_n = a.pop()
        for i in range(0, len(a), 2):
                a[i], a[i + 1] = a[i + 1], a[i]
        a.insert(0,a_n)
        # a.insert(last_element,a_n) if you want last element remain unmodified
        print("its else:",a)

Answer 9

This is what worked for me. Hopefully, it can help others. I think about using boxes. The value in the first position needs a temporary place to be before it can be swapped, so I assign it to x and then make the swap. Incrementing by 2 will skip over the last element, so it's okay with odd-numbered lists as well.

a = [int(s) for s in input().split()]
i = 0
while i in range(0, len(a) - 1):
    x = a[i]
    a[i] = a[i + 1]
    a[i + 1] = x
    i += 2
print(a)

Answer 10

Found this to be much simpler solution and handels the list with odd elements

elements = list(map(int,input().split()))
swap = []

i = 0
while i < len(elements):
    if i+1 < len(elements):
        swap.append(elements[i+1])
        swap.append(elements[i]) 
    else:
        swap.append(elements[-1])
    i = i+2
print(swap)