On 11/07/2011 01:01 PM, JoeM wrote:
Thanks guys, I was just looking for a one line solution instead of a
for loop if possible. Why do you consider
[x.remove(x[0]) for x in [a,b,c]]
cheating? It seems compact and elegant enough for me.
Cheers
Are you considering the possibility that two of these names might
reference the same list?
a = [42, 44, 6, 19, 48]
b = a
c = b
for x in [a,b,c]:
x.remove(x[0])
now a will have [19,48] as its content.
--
DaveA
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