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。
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
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
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