On Thu, Jun 18, 2009 at 9:15 AM, 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.
But why not use groupby()? That seems much clearer to me:
In [1]: from
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(l
Whoops. That's called assuming what I read is really what I see. A good
lesson in reading questions twice.
I remember this from a post some time back and I remember having been
intrigued by it. I used Google, and since I tend to keep extensive
notes, the solution I found is not uniquely mine, but
On 6/18/2009 7:23 AM karma said...
Hi Robert,
Thanks for the link. However, I am looking for eliminating consecutive
duplicates rather than all duplicates - my example wasn't clear,
apologies for that.
x=[1,1,1,3,2,2,2,4,4,2,2]
[1 ,3 ,2 ,4 ,2 ]
Something like
[ ii for ii,jj in zip(x,x[1:]+[
Hi Robert,
Thanks for the link. However, I am looking for eliminating consecutive
duplicates rather than all duplicates - my example wasn't clear,
apologies for that.
x=[1,1,1,3,2,2,2,4,4,2,2]
[1 ,3 ,2 ,4 ,2 ]
2009/6/18 Robert Berman :
> This might help: http://code.activestate.com/recipes/525
This might help: http://code.activestate.com/recipes/52560/
Robert
On Thu, 2009-06-18 at 15:15 +0200, 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.
>
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
