<p>如果我正确地理解了你的问题,你想确定基因组中得分最高的k-mers。k-mer的得分是它在基因组中出现的次数加上汉明距离小于<code>m</code>的任何k-mer在基因组中出现的次数。请注意,这假设您只对出现在您的基因组中的k-mer感兴趣(正如@j_random_hacker所指出的)。在</p>
<p>您可以通过四个基本步骤来解决此问题:</p>
<ol>
<li>确定基因组中的所有k-mer。在</li>
<li>计算每个k-mer在<code>G</code>中出现的次数。在</li>
<li>对于每对k-mers(<code>K1</code>,<code>K2</code>),如果<code>K1</code>和{<cd6>}的汉明距离小于<code>m</code>,则增加它们的计数。在</li>
<li>找到<code>max</code>k-mer及其计数。在</li>
</ol>
<p>下面是Python代码示例:</p>
<pre><code>from itertools import combinations
from collections import Counter
# Hamming distance function
def hamming_distance(s, t):
if len(s) != len(t):
raise ValueError("Hamming distance is undefined for sequences of different lengths.")
return sum( s[i] != t[i] for i in range(len(s)) )
# Main function
# - k is for k-mer
# - m is max hamming distance
def hamming_kmer(genome, k, m):
# Enumerate all k-mers
kmers = [ genome[i:i+k] for i in range(len(genome)-k + 1) ]
# Compute initial counts
initcount = Counter(kmers)
kmer2count = dict(initcount)
# Compare all pairs of k-mers
for kmer1, kmer2 in combinations(set(kmers), 2):
# Check if the hamming distance is small enough
if hamming_distance(kmer1, kmer2) <= m:
# Increase the count by the number of times the other
# k-mer appeared originally
kmer2count[kmer1] += initcount[kmer2]
kmer2count[kmer2] += initcount[kmer1]
return kmer2count
# Count the occurrences of each mismatched k-mer
genome = 'ACGTTGCATGTCGCATGATGCATGAGAGCT'
kmer2count = hamming_kmer(genome, 4, 1)
# Print the max k-mer and its count
print max(kmer2count.items(), key=lambda (k,v ): v )
# Output => ('ATGC', 5)
</code></pre>