On 31/01/2015 02:38, Chris Angelico wrote:
On Sat, Jan 31, 2015 at 1:27 PM, <rajanb...@gmail.com> wrote:
l1 = ["a","b","c","d","e","f","g","h","i","j"]
l2 = ["aR","bR","cR"]
l2 will always be smaller or equal to l1
numL1PerL2 = len(l1)/len(l2)
I want to create a dictionary that has key from l1 and value from l2 based on
numL1PerL2
So
{
a:aR,
b:aR,
c:aR,
d:bR,
e:bR,
f:bR,
g:cR,
h:cR,
i:cR,
j:cR
}
So last item from l2 is key for remaining items from l1
So the Nth element of l1 will always be paired with the
(N/numL1PerL2)th element of l2 (with the check at the end)? Seems easy
enough.
dups = len(l1)/len(l2)
l2.append(l2[-1])
result = {x:l2[i/dups] for i,x in enumerate(l1)}
This mutates l2 for convenience, but you could also adjust the index
to take care of the excess. As a one-liner:
result = {x:l2[min(i/(len(l1)/len(l2)),len(l2)-1)] for i,x in enumerate(l1)}
But the one-liner is not better code :)
ChrisA
The one-liner might not be better code, but it must be better speed wise
precisely because it's on one line, right? :)
--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.
Mark Lawrence
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