Comments on: Obtaining overrepresented motifs in DNA sequences, part 4

By: Paulo Nuin

Paulo Nuin — Fri, 16 May 2008 12:55:14 +0000

Thanks Mike. I am away until Tuesday and I am going to test too. Would you mind sending me the code you have?

Cheers

By: Mike

Mike — Thu, 15 May 2008 00:34:33 +0000

I coded the routine 8 different ways. The code in comment number 1 was fastest. A routine similar to the code in the original post was second fastest. The difference in run time among the 4 fastest routines was not significant (only about 2 or 3%). The slowest routine was about 20% slower than the fastest routine.

By: Paulo Nuin

Paulo Nuin — Tue, 13 May 2008 02:08:28 +0000

Hi Mike

I don’t know if I can thank you enough, your comments always make the topic better.

I was aware of the iteritems method but forgot to use in this case. Thanks for pointing out the enumerate. I am wondering about a possible speed increase with the solution you suggested.

Cheers

By: Mike

Mike — Tue, 13 May 2008 00:02:30 +0000

The “enumerate” function is handy when you want to number items in a list (or tuple, generator, etc.) as you loop over them with a ‘for’ loop. You pass the iterable into enumerate and it returns two-tuples a sequence number and an item from the iterator.

The built-in “set” class is handy when you want to keep track of a group of items, but don’t want duplicates. The .add() method adds an element to the set, unless it is already in the set.

Dictionary methods items() and iteritems() return tuples of key-value pairs

(Leading “.”s represent spaces for indentation)

quorum = defaultdict(set)

for seq_number,seq in enumerate(seqs):
…for n in range(len(i.sequence) – length):
……quorum[i.sequence[n:n+length]].add(seq_number)

for motif,seq_numbers in quorum.iteritems():
…print motif.upper(), len(seq_numbers)