Python:如何计算三核苷酸的频率

Question

因此，我的翻译工作正常，但是当我通过断言检查运行时，它不会以错误的方式传递：它应该是字符串而不是元组。我遇到了问题，但我只是不知道该如何解决。

AssertionError:
     <class 'tuple'> != <class 'str'>

---

def frequency(dna_sequence):
    '''
    takes a DNA sequence (in string format) as input, parses it into codons using parse_sequence(),
    counts each type of codon and returns the codons' frequency as a dictionary of counts;
    the keys of the dictionary must be in string format
    '''
    codon_freq = dict()
    # split string with parse_sequence()
    parsed = parse_sequence(dna_sequence) # it's a function made previously, which actually makes a sequence of string to one-element tuple.  
    # count each type of codons in DNA sequence
    from collections import Counter
    codon_freq = Counter(parsed)
    return codon_freq
codon_freq1 = codon_usage(dna_sequence1)
print("Sequence 1 Codon Frequency:n{0}".format(codon_freq1))
codon_freq2 = codon_usage(dna_sequence2)
print("nSequence 2 Codon Frequency:n{0}".format(codon_freq2))

断言检查

assert_equal(codon_usage('ATATTAAAGAATAATTTTATAAAAATATGT'), 
             {'AAA': 1, 'AAG': 1, 'AAT': 2, 'ATA': 3, 'TGT': 1, 'TTA': 1, 'TTT': 1})
assert_equal(type((list(codon_frequency1.keys()))[0]), str)

关于parse_sequence：

def parse_sequence(dna_sequence):
    codons = []
    if len(dna_sequence) % 3 == 0:
        for i in range(0,len(dna_sequence),3):
            codons.append((dna_sequence[i:i + 3],))
    return codons

Answer 1

您可能会发现直接使用理解的计数器更容易。例如

>>> s = 'ATATTAAAGAATAATTTTATAAAAATATGT'
>>> [s[3*i:3*i+3] for i in xrange(0, len(s)/3)]
['ATA', 'TTA', 'AAG', 'AAT', 'AAT', 'TTT', 'ATA', 'AAA', 'ATA', 'TGT']
>>> from collections import Counter
>>> Counter([s[3*i:3*i+3] for i in xrange(0, len(s)/3)])
Counter({'ATA': 3, 'AAT': 2, 'AAG': 1, 'AAA': 1, 'TGT': 1, 'TTT': 1, 'TTA': 1})

Answer 2

您正确解析了，但是结果是元组而不是所需的字符串，例如

>>> s = "ATATTAAAGAATAATTTTATAAAAATATGT"
>>> parse_sequence(s)
[('ATA',),
 ('TTA',),
 ('AAG',),
 ('AAT',),
 ('AAT',),
 ('TTT',),
 ('ATA',),
 ('AAA',),
 ('ATA',),
 ('TGT',)]

只需从此线中删除尾随逗号：

    ...
    codons.append((dna_sequence[i:i + 3],))
    ...

---

fyi，滑动窗口是一种可以应用于密码子匹配的技术。这是使用more_itertools.windowed（第三方工具）的完整，简化的示例：

import collections as ct
import more_itertools as mit

def parse_sequence(dna_sequence):
    """Return a generator of codons."""
    return ("".join(codon) for codon in mit.windowed(dna_sequence, 3, step=3))
def frequency(dna_sequence):
    """Return a Counter of codon frequency."""
    parsed = parse_sequence(dna_sequence)
    return ct.Counter(parsed)

测试

s = "ATATTAAAGAATAATTTTATAAAAATATGT"
expected = {'AAA': 1, 'AAG': 1, 'AAT': 2, 'ATA': 3, 'TGT': 1, 'TTA': 1, 'TTT': 1}
assert frequency(s) == expected