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Commit bdf447e8 authored by Blaise Li's avatar Blaise Li
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Fix print, comment solution for kmer counting.

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......@@ -391,7 +391,7 @@ Compute the 6 mers occurences of the sequence above, and print each 6mer and it'
>>> s = s.replace('\n', '')
>>> kmers = get_kmer_occurences(s, 6)
>>> for kmer in kmers:
>>> print kmer[0], '..', kmer[1]
>>> print(kmer[0], '..', kmer[1])
gcagag .. 2
aacttc .. 1
gcaact .. 1
......
def get_kmer_occurences(seq, kmer_len):
"""
return a list of tuple
return a list of tuple
each tuple contains a kmers present in seq and its occurence
"""
# Counter dictionary (keys will be kmers, values will be counts)
kmers = {}
stop = len(seq) - kmer_len
for i in range(stop + 1):
# Extract kmer starting at position i
kmer = seq[i: i + kmer_len]
# Increment the count for kmer
# or create the entry the first time this kmer is seen
# using default value 0, plus 1
kmers[kmer] = kmers.get(kmer, 0) + 1
return kmers.items()
\ No newline at end of file
# pairs (kmer, count)
return kmers.items()
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