Python 如何扩展模糊dna序列
假设你有这样的DNA序列:Python 如何扩展模糊dna序列,python,biopython,dna-sequence,Python,Biopython,Dna Sequence,假设你有这样的DNA序列: AATCRVTAA 其中R和V是DNA核苷酸的模糊值,其中R表示A或G,V表示A,C或G 是否有一种Biopython方法来生成可以由上述不明确序列表示的所有不同序列组合 例如,这里的输出是: AATCAATAA AATCACTAA AATCAGTAA AATCGATAA AATCGCTAA AATCGGTAA 我最终编写了自己的函数: from Bio import Seq from itertools import product def extend_am
AATCRVTAA
其中R
和V
是DNA核苷酸的模糊值,其中R
表示A
或G
,V
表示A
,C
或G
是否有一种Biopython方法来生成可以由上述不明确序列表示的所有不同序列组合
例如,这里的输出是:
AATCAATAA
AATCACTAA
AATCAGTAA
AATCGATAA
AATCGCTAA
AATCGGTAA
我最终编写了自己的函数:
from Bio import Seq
from itertools import product
def extend_ambiguous_dna(seq):
"""return list of all possible sequences given an ambiguous DNA input"""
d = Seq.IUPAC.IUPACData.ambiguous_dna_values
r = []
for i in product(*[d[j] for j in seq]):
r.append("".join(i))
return r
In [1]: extend_ambiguous_dna("AV")
Out[1]: ['AA', 'AC', 'AG']
它允许您使用
In [2]: extend_ambiguous_dna("NN")
Out[2]: ['GG', 'GA', 'GT', 'GC',
'AG', 'AA', 'AT', 'AC',
'TG', 'TA', 'TT', 'TC',
'CG', 'CA', 'CT', 'CC']
希望这能为其他人节省时间 我不确定是否有一种biopython方法可以做到这一点,但这里有一种使用itertools的方法:
s = "AATCRVTAA"
ambig = {"R": ["A", "G"], "V":["A", "C", "G"]}
groups = itertools.groupby(s, lambda char:char not in ambig)
splits = []
for b,group in groups:
if b:
splits.extend([[g] for g in group])
else:
for nuc in group:
splits.append(ambig[nuc])
answer = [''.join(p) for p in itertools.product(*splits)]
输出:
In [189]: answer
Out[189]: ['AATCAATAA', 'AATCACTAA', 'AATCAGTAA', 'AATCGATAA', 'AATCGCTAA', 'AATCGGTAA']
也许是一种更短更快的方法,因为很可能这个函数将用于非常大的数据:
from Bio import Seq
from itertools import product
def extend_ambiguous_dna(seq):
"""return list of all possible sequences given an ambiguous DNA input"""
d = Seq.IUPAC.IUPACData.ambiguous_dna_values
return [ list(map("".join, product(*map(d.get, seq)))) ]
使用map
可以使用C语言而不是Python语言执行循环。事实证明,这比使用普通循环甚至列表理解要快得多
现场试验
用一个简单的dict作为d
而不是不明确的值返回的dict
产出:
# len(seq) = 2:
List delay: 0.02 ms
Map delay: 0.01 ms
# len(seq) = 3:
List delay: 0.04 ms
Map delay: 0.02 ms
# len(seq) = 4
List delay: 0.08 ms
Map delay: 0.06 ms
# len(seq) = 5
List delay: 0.43 ms
Map delay: 0.17 ms
# len(seq) = 10
List delay: 126.68 ms
Map delay: 77.15 ms
# len(seq) = 12
List delay: 1887.53 ms
Map delay: 1320.49 ms
显然,map
更好,但只是2或3倍。可以肯定的是,它可以进一步优化。还有一个itertools解决方案:
from itertools import product
import re
lu = {'R':'AG', 'V':'ACG'}
def get_seqs(seq):
seqs = []
nrepl = seq.count('R') + seq.count('V')
sp_seq = [a for a in re.split(r'(R|V)', seq) if a]
pr_terms = [lu[a] for a in sp_seq if a in 'RV']
for cmb in product(*pr_terms):
seqs.append(''.join(sp_seq).replace('R', '%s').replace('V', '%s') % cmb)
return seqs
seq = 'AATCRVTAA'
print 'seq: ', seq
print '\n'.join(get_seqs(seq))
seq1 = 'RAATCRVTAAR'
print 'seq: ', seq1
print '\n'.join(get_seqs(seq1))
输出:
特殊情况下的错误输出,其中有两个或多个相同的相邻不明确代码,如“RRATCGGTAAA”
from itertools import product
import re
lu = {'R':'AG', 'V':'ACG'}
def get_seqs(seq):
seqs = []
nrepl = seq.count('R') + seq.count('V')
sp_seq = [a for a in re.split(r'(R|V)', seq) if a]
pr_terms = [lu[a] for a in sp_seq if a in 'RV']
for cmb in product(*pr_terms):
seqs.append(''.join(sp_seq).replace('R', '%s').replace('V', '%s') % cmb)
return seqs
seq = 'AATCRVTAA'
print 'seq: ', seq
print '\n'.join(get_seqs(seq))
seq1 = 'RAATCRVTAAR'
print 'seq: ', seq1
print '\n'.join(get_seqs(seq1))
seq: AATCRVTAA
AATCAATAA
AATCACTAA
AATCAGTAA
AATCGATAA
AATCGCTAA
AATCGGTAA
seq: RAATCRVTAAR
AAATCAATAAA
AAATCAATAAG
AAATCACTAAA
AAATCACTAAG
AAATCAGTAAA
AAATCAGTAAG
AAATCGATAAA
AAATCGATAAG
AAATCGCTAAA
AAATCGCTAAG
AAATCGGTAAA
AAATCGGTAAG
GAATCAATAAA
GAATCAATAAG
GAATCACTAAA
GAATCACTAAG
GAATCAGTAAA
GAATCAGTAAG
GAATCGATAAA
GAATCGATAAG
GAATCGCTAAA
GAATCGCTAAG
GAATCGGTAAA
GAATCGGTAAG