2015-01-05 62 views
1

我目前正在通过麻省理工学院开放式课件上的其中一个问题集,其任务是在DNA序列中查找匹配的子串。Python中的生成器函数

我在努力编写一个返回长度为k的子序列的函数。当使用字符串时,我可以使它工作,但使用迭代器设置问题时,使用迭代器时,函数似乎每次都会重置,而不是使用yield返回到原始位置。

这是一个正确的功能,我写了使用字符串:

def subs(seq, k): 
    subseq = '' 
    pos = 0 
    while pos < len(seq): 
     while len(subseq) < k: 
      subseq += seq[pos] 
      pos += 1 
     yield subseq, pos - k 
     subseq = subseq[1:] 

一个正确的答案:

>>> a = 'hello' 
>>> b = subs(a,2) 
>>> b.next() 
('he', 0) 
>>> b.next() 
('el', 1) 
>>> b.next() 
('ll', 2) 
>>> b.next() 
('lo', 3) 
>>> b.next() 

Traceback (most recent call last): 
    File "<pyshell#13>", line 1, in <module> 
    b.next() 
StopIteration 

我的问题

的任务设置使用类创建一个长串字符串序列的迭代器,我不会在这里进入它,但测试给出也创建一个迭代器从一个字符串

# This test case may break once you add the argument m (skipping). 
class TestExactSubmatches(dnaseq.unittest.TestCase): 
    def test_one(self): 
     foo = 'yabcabcabcz' 
     bar = 'xxabcxxxx' 
     matches = list(dnaseq.getExactSubmatches(iter(foo), iter(bar), 3, 1)) 
     correct = [(1,2), (4,2), (7,2)] 
     self.assertTrue(len(matches) == len(correct)) 
     for x in correct: 
      self.assertTrue(x in matches) 

和我目前的解决方案:

def subsequenceHashes(seq, k): 
    subseq = '' 
    pos = 0 
    print 'Start of subseqHashes' 
    try: 
     while True: 
      while len(subseq) < k: 
       subseq += seq.next() 
       pos += 1 
      print subseq, pos - k 
      yield hash(subseq), pos - k 
      subseq = subseq[1:] 
    except StopIteration: 
     return 

函数调用它得到的子序列的散列,使他们与其中的子序列开始到字典(类multidict)的位置沿与将子字符串与相同的散列进行比较,看它们是否相同。然后它应该返回两个相同的子串中的位置对。我没有设法调试这个函数的大部分,因为我在启动时遇到了问题。

def getExactSubmatches(a, b, k, m): 
    # a and b are the strings compared, k is the length of substring, parameter m is unused, need it for later on in the problem set 
    ahash, apos = subsequenceHashes(a, k).next() 
    bhash, bpos = subsequenceHashes(b, k).next() 
    multidict = Multidict() 
    print 'starting' 
    while ahash: 
     print 'iterate' 
     multidict.put(ahash, ('a', apos)) 
     ahash, apos = subsequenceHashes(a, k).next() 
     print apos 
    while bhash: 
     multidict.put(bhash, ('b', bpos)) 
     bhash, bpos = subsequenceHashes(b, k).next() 
    for key in multidict.mydict: 
     if len(multidict.get(key)) > 1: 
      for t in multidict.get(key): 
       if t[0] == 'a': 
        for s in multidict.get(key): 
         if s[0] == 'b': 
          if a[apos:apos+k] == b[bpos:bpos+k]: 
           print apos, bpos 
           yield apos, bpos 

当我运行测试,会发生什么:

Start of subseqHashes 


yab 0 
Start of subseqHashes 


xxa 0 
starting 
iterate 
Start of subseqHashes 


cab 0 
0 
iterate 
Start of subseqHashes 


cab 0 
0 
iterate 
Start of subseqHashes 


F.. 
====================================================================== 
FAIL: test_one (__main__.TestExactSubmatches) 
---------------------------------------------------------------------- 
Traceback (most recent call last): 
    File "C:\Users\Alex\Desktop\Pythonwork\6.006\ps4\dist\test_dnaseq.py", line 32, in test_one 
    self.assertTrue(len(matches) == len(correct)) 
AssertionError: False is not true 

什么,似乎是想错了是subsequenceHashes被重置每次我使用的.next(时间),当其拿到在它的身上迭代器作为反对在使用字符串时留在循环中。

+1

每次打电话给'子序列哈希(b,k)'它会重新开始。你应该在函数的开始创建一次。 – jonrsharpe

+0

我将比较的DNA序列长度为数千万个核苷酸,问题集建议生成发生器函数。 http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-006-introduction-to-algorithms-fall-2011/assignments/MIT6_006F11_ps4.pdf – Az00123

+0

是的,但你应该*只打电话发电机功能一次*。在那之后,你只想*重复它*,而不是重新启动它。从'gen_a = subsequenceHashes(a,k)'开始,并从那里开始。 – jonrsharpe

回答

0

正如@jonrsharpe提到的,我的错误是多次调用生成器函数,而不是实际迭代它。