I have a list of 4 items like this:
a, b, c, d = [1, 2, 3, 4]
I'm reordering the list, flipping each pair:
[b, a, d, c]
Is there a way to do this in one expression? I've tried using list comprehension and unpacking, but can't seem to get it right.
I have [1, 2, 3, 4]. I'm trying to get [2, 1, 4, 3].
More generically, if you're looking to flip pairs of numbers in a list:
>>> L = [1, 2, 3, 4, 5, 6]
>>> from itertools import chain
>>> list(chain.from_iterable(zip(L[1::2], L[::2])))
[2, 1, 4, 3, 6, 5]
Look at this:
>>> lst = [1, 2, 3, 4]
>>> [y for x in zip(*[iter(lst)]*2) for y in x[::-1]]
[2, 1, 4, 3]
>>>
>>> lst = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> [y for x in zip(*[iter(lst)]*2) for y in x[::-1]]
[2, 1, 4, 3, 6, 5, 8, 7, 10, 9]
>>>
If this is only about 4 member lists - this would suffice:
list = [1, 2, 3, 4]
reordered_list = [list[1], list[0], list[3],list[2]]
Because absolutely nobody has given an answer that works on generic iterables,
from itertools import chain
items = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
zip(*[iter(items)]*2)
#>>> <zip object at 0x7fd673afd050>
[itms for itms in zip(*[iter(items)]*2)]
#>>> [(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)]
So zip(*[iter(x)]*2) means ix = iter(x); zip(ix, ix) which pairs each element.
Then you can reverse:
[(y, x) for (x, y) in zip(*[iter(items)]*2)]
#>>> [(2, 1), (4, 3), (6, 5), (8, 7), (10, 9)]
Putting it all together and flattening:
[itm for (x, y) in zip(*[iter(items)]*2) for itm in (y, x)]
#>>> [2, 1, 4, 3, 6, 5, 8, 7, 10, 9]
It's generic and short!
If you want something faster at the expense of genericism, you'll be hard pressed to better this:
new = list(items)
new[::2], new[1::2] = new[1::2], new[::2]
new
#>>> [2, 1, 4, 3, 6, 5, 8, 7, 10, 9]
Note that this still works on arbitrary iterables, but there are fewer layers of abstraction; you can't bump up the size of the flipped sub-lists as easily and can't output iterables, etc.
Do you mean this:
>>> a, b, c, d = [1, 2, 3, 4]
>>> b, a, d, c = a, b, c, d
>>> a
2
>>> b
1
>>> c
4
>>> d
3
?
Try this list comprenhension solution:
a = [1,2,3,4,5,6] # Any list with even number of elements
b = [a[e+1] if (e%2 == 0) else a[e-1] for e in range(len(a))]
This just works if the list a have an even number of elements.
In [1]: l = [1, 2, 3, 4]
In [2]: list(chain(*map(reversed, zip(l[::2], l[1::2]))))
Out[2]: [2, 1, 4, 3]
Am I missing something? Reorder given_list with a loop:
rez = []
for i in range(len(given_list)-1, -1, -1):
rez.append(given_list[i])
return rez
