Comments on: Obtaining overrepresented motifs in DNA sequences, part 12.5 http://python.genedrift.org/2008/08/15/obtaining-overrepresented-motifs-in-dna-sequences-part-12/ a step-by-step guide to create Python applications in bioinformatics Mon, 22 Feb 2010 18:22:18 +0000 http://wordpress.org/?v=3.0-alpha hourly 1 By: Paulo Nuin http://python.genedrift.org/2008/08/15/obtaining-overrepresented-motifs-in-dna-sequences-part-12/comment-page-1/#comment-17648 Paulo Nuin Fri, 15 Aug 2008 13:12:52 +0000 http://python.genedrift.org/?p=137#comment-17648 Thanks Andrew, I will change the code and post the results. Thanks Andrew, I will change the code and post the results.

]]> By: Andrew Dalke http://python.genedrift.org/2008/08/15/obtaining-overrepresented-motifs-in-dna-sequences-part-12/comment-page-1/#comment-17631 Andrew Dalke Fri, 15 Aug 2008 06:24:04 +0000 http://python.genedrift.org/?p=137#comment-17631 I sent a couple of links to binomial function implementations without reviewing the code myself. You used the second of those links, which happens to be amazingly slow. It's actually computing all of the values of the distribution for a given size n, then returning the one requested. Choosing one of the implementations from the first link (the "non-clever" one at http://groups.google.com/group/comp.lang.python/msg/6e7c3358b086ff9c?dmode=source): def choose(n, k): if 0 <= k <= n: ntok = 1 ktok = 1 for t in xrange(1, min(k, n - k) + 1): ntok *= n ktok *= t n -= 1 return ntok // ktok else: return 0 I get as output fac: 0.01 choose: 0.00 Since it doesn't actually take 0.0 time I changed the repeat count from 10 to 10,000 and removed the "/0". This will give you an idea of the performance difference: fac: 140.19 choose: 0.05 or roughly a factor of 3,000 times faster. (BTW, I had a hard time writing "binomial" - my fingers keep wanting to type "bionomial" :) I sent a couple of links to binomial function implementations without reviewing the code myself. You used the second of those links, which happens to be amazingly slow. It’s actually computing all of the values of the distribution for a given size n, then returning the one requested.

Choosing one of the implementations from the first link (the “non-clever” one at http://groups.google.com/group/comp.lang.python/msg/6e7c3358b086ff9c?dmode=source):

def choose(n, k):
if 0 <= k <= n:
ntok = 1
ktok = 1
for t in xrange(1, min(k, n – k) + 1):
ntok *= n
ktok *= t
n -= 1
return ntok // ktok
else:
return 0

I get as output

fac: 0.01
choose: 0.00

Since it doesn’t actually take 0.0 time I changed the repeat count from 10 to 10,000 and removed the “/0″. This will give you an idea of the performance difference:

fac: 140.19
choose: 0.05

or roughly a factor of 3,000 times faster.

(BTW, I had a hard time writing “binomial” – my fingers keep wanting to type “bionomial” :)

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