I have a list of tuples like this:
[('foo','bar'),('foo1','bar1'),('foofoo','barbar')]
What is the fastest way in python (running on a very low cpu/ram machine) to swap values like this...
[('bar','foo'),('bar1','foo1'),('barbar','foofoo')]
I am currently using:
for x in mylist:
self.my_new_list.append(((x[1]),(x[0])))
Is there a better or faster way???
You could use map:
map (lambda t: (t[1], t[0]), mylist)
Or list comprehension:
[(t[1], t[0]) for t in mylist]
List comprehensions are preferred and supposedly much faster than map when lambda is needed, however note that list comprehension has a strict evaluation, that is it will be evaluated as soon as it gets bound to variable, if you're worried about memory consumption use a generator instead:
g = ((t[1], t[0]) for t in mylist)
#call when you need a value
g.next()
There are some more details here: Python List Comprehension Vs. Map
You can use reversed like this:
tuple(reversed((1, 2)) == (2, 1)
To apply it to a list, you can use map or a list/generator comprehension:
map(tuple, map(reversed, tuples)) # map
[tuple(reversed(x)) for x in tuples] # list comprehension
(tuple(reversed(x)) for x in tuples) # generator comprehension
If you're interested primarily in runtime speed, I can only recommend that you profile the various approaches and pick the fastest.
A fancy way:
[t[::-1] for t in mylist]
Using a list comprehension I find more elegant and understandable to use separate variables instead of indices for a single variable as in the solution provided by @iabdalkader:
[(b, a) for a, b in mylist]
To modify the current list in-place, the most efficient way is:
my_list[:] = [(y, x) for x, y in my_list]
It is assigning to the list slice, which covers the entire list, without creating an extra duplicate of the list in memory. See also this answer.