I have this list comprehension:
[[x,x] for x in range(3)]
which results in this list:
[[0, 0], [1, 1], [2, 2]]
but what I want is this list:
[0, 0, 1, 1, 2, 2]
What's the easiest to way to generate this list?
views:
113answers:
6If one wants to generatlize it one could write `[x]*2` instead of `[x,x]`.
phimuemue
2010-10-13 16:01:05
@phimuemue: which would be even slower
SilentGhost
2010-10-13 16:18:19
Maybe using the integer division operator with `x // 2` would be better than `int(x / 2)`? In Python 2.7, `timeit` shows `//` to be a little more than twice as fast, In Python 3.1, nearly three times, for large ranges.
gotgenes
2010-10-13 19:26:37
A:
You might get away with this:
[floor(x/2) for x in range(6)]
edit1
[int(x/2) for x in range(6)]
is the more portable solution in the same vein. Although the other presented answers seem better.
JoshD
2010-10-13 15:48:01
`math.floor()` returns a `float`, so this is not quite the same.
Ignacio Vazquez-Abrams
2010-10-13 16:12:43
+1
A:
>>> [i for i in range(3) for _ in range(2)]
[0, 0, 1, 1, 2, 2]
SilentGhost
2010-10-13 15:50:13
+2
A:
a general solution;
m = 3 #the list of integers
n = 2 # of repetitions
[x for x in range(m) for y in range(n)]
joaquin
2010-10-13 15:52:51
A:
[x/2 for x in range(6)]
update:
[x//2 for x in range(6)] #ok now ?
singularity
2010-10-13 16:41:26
@Ant: change to py3k syntax is trivial. However the same solution was given by Ned
SilentGhost
2010-10-14 14:57:47
@silent ghost and why it's trivial? write x//2 it's better, in my opinion...
Ant
2010-10-15 11:33:58