我试图在一个字符串上找到最频繁的子字符串(具有一定长度L),但是允许某个阈值或错误T(在子字符串的amy位置),显然不超过L本身。然而,到目前为止,我还没有取得任何成功。我如何完成代码?在
stn='KLHLHLHKPLHLHLPHHKLH'
L = 4 #(lenght of the pattern)
T = 1 #(maximum tolerance or permitted error in any position of the query pattern)
pattcount = {}
for n in range(len(stn)- L+1):
patt = stn[n:n+L]
s_ = stn[i:i+len(patt)]
if LevenshteinDistance(s_, patt) == T:
pattcount[patt] = pattcount[patt] + 1 if pattcount.has_key(patt) else 1
max = 0
max_patt = []
for p,v in pattcount.iteritems():
if v > max:
max_patt = [p]
max = v
elif v == max:
max_patt += [p]
print (" ".join(max_patt))
因此,例如,如果频率最高的是KLH,那么HLH、PLH、KLP、KPH的频率如何膨胀KLH的频率(以便报告)?在
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