best way to do this

[TP]

Hi everybody,

c=[(5,3), (6,8)]

From c, I want to obtain a list with 5,3,6, and 8, in any order.
I do this:

[i for (i,j) in c] + [ j for (i,j) in c]

[5, 6, 3, 8]

Is there a quicker way to do this?

Thanks

Julien
--
python -c "print ''.join([chr(154 - ord(c)) for c in '*9(9&(18%.9&1+,\'Z
(55l4('])"

"When a distinguished but elderly scientist states that something is
possible, he is almost certainly right. When he states that something is
impossible, he is very probably wrong." (first law of AC Clarke)

 

[Diez B. Roggisch]

Hi everybody,

c=[(5,3), (6,8)]

From c, I want to obtain a list with 5,3,6, and 8, in any order.
I do this:

[i for (i,j) in c] + [ j for (i,j) in c]

[5, 6, 3, 8]

Is there a quicker way to do this?

dunno if it's faster, but less cluttered:

list(sum(c, ()))

Diez

 

[Hrvoje Niksic]

Hi everybody,

c=[(5,3), (6,8)]

From c, I want to obtain a list with 5,3,6, and 8, in any order.
I do this:

[i for (i,j) in c] + [ j for (i,j) in c]

[5, 6, 3, 8]

Is there a quicker way to do this?

Quicker? Hard to say. Using itertools elegantly? Definitely:

list(chain(*c))

As an added benefit, it works regardless of the number of elements in
the tuple. (Also assumes from itertools import chain for effect.)

 

[Arnaud Delobelle]

Hi everybody,

c=[(5,3), (6,8)]

From c, I want to obtain a list with 5,3,6, and 8, in any order.
I do this:

[i for (i,j) in c] + [ j for (i,j) in c]

[5, 6, 3, 8]

Is there a quicker way to do this?

One list comprehension is enough:

c=[(5,3), (6,8)]
[x for t in c for x in t]

[5, 3, 6, 8]

HTH

 

[thor]

Hi everybody,

c=[(5,3), (6,8)]

From c, I want to obtain a list with 5,3,6, and 8, in any order.
I do this:

[i for (i,j) in c] + [ j for (i,j) in c]

[5, 6, 3, 8]

Is there a quicker way to do this?

c = [(5, 3), (6, 8)]
[x for t in zip(*c) for x in t]

[5, 6, 3, 8]

 

[alex23]

c = [(5, 3), (6, 8)]
[x for t in zip(*c) for x in t]

[5, 6, 3, 8]

The zip here is superfluous.

c = [(5, 3), (6, 8)]
zip(*c)

[(5, 6), (3, 8)]

Unless you're -only- using it to ensure your result is the same order
as the OP...but he also stressed that order was unimportant.

Leaving out the zip call gives you Arnaud's solution, which seems the
clearest.