karma wrote:
I was playing around with eliminating duplicates in a list not using
groupby. From the two solutions below, which is more "pythonic".
Alternative solutions would be welcome.
Thanks
x=[1,1,1,3,2,2,2,2,4,4]
[v for i,v in enumerate(x) if x[i]!=x[i-1] or i==0]
[x[i] for i in range(len(x)-1) if i==0 or x[i]!=x[i-1]]
output:
[1, 3, 2, 4]
I'd vote for the first form, though I might make a generator out of it.
gen = (v for i,v in enumerate(x) if x[i]!=x[i-1] or i==0)
print list(gen)
Of course, that depends on the lifetime of the various objects.
Interestingly, if you modify x after defining gen, but before using it,
gen will use the new contents of x.
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