重叠正则表达式匹配

Question

我正在尝试创建以下正则表达式：从以下RNA字符串返回AUG和（ UAG或UGA或UAA ）之间的字符串： AGCCAUGUAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAAUGAUCCGAGUAGCAUCUCAG ，以便找到所有匹配项，包括重叠项。

我已经尝试了几个正则表达式，结果是这样的：

matches = re.findall('(?=AUG)(\w+)(?=UAG|UGA|UAA)',"AGCCAUGUAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAAUGAUCCGAGUAGCAUCUCAG")

你能告诉我正则表达式模式中的错误吗？

Answer 1

使用一个正则表达式执行此操作实际上非常困难，因为大多数使用特别不希望重叠匹配。 但是，您可以通过一些简单的迭代来完成此操作：

regex = re.compile('(?=AUG)(\w+)(?=UAG|UGA|UAA)');
RNA = 'AGCCAUGUAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAAUGAUCCGAGUAGCAUCUCAG'
matches = []
tmp = RNA
while (match = regex.search(tmp)):
    matches.append(match)
    tmp = tmp[match.start()-2:]  #Back up two to get the UG portion.  Shouldn't matter, but safer.

for m in matches:
    print m.group(0)

虽然，这有一些问题。 对于AUGUAGUGAUAA你期望回报是什么？ 有两个字符串要返回吗？ 还是只有一个？ 现在，你的正则表达式甚至无法捕获UAG ，因为它继续匹配UAGUGA并在UAA被切断。 为了解决这个问题，你可能希望使用? 运算符使您的运算符变得懒惰 - 这种方法将无法捕获更长的子字符串。

也许在字符串上迭代两次就是答案，但是如果您的RNA序列包含AUGAUGUAGUGAUAA ？ 那里的正确行为是什么？

我可能喜欢正则表达式自由方法，通过遍历字符串及其子串：

RNA = 'AGCCAUGUAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAAUGAUCCGAGUAGCAUCUCAG'
candidates = []
start = 0

while (RNA.find('AUG', start) > -1):
    start = RNA.find('AUG', start) #Confound python and its lack of assignment returns
    candidates.append(RNA[start+3:])
    start += 1

matches = []

for candidate in candidates:
    for terminator in ['UAG', 'UGA', 'UAA']:
        end = 1;
        while(candidate.find(terminator, end) > -1):
            end = candidate.find(terminator, end)
            matches.append(candidate[:end])
            end += 1

for match in matches:
    print match

这样，无论如何，你肯定会得到所有的比赛。

如果需要跟踪每个匹配的位置，可以修改候选数据结构以使用保持起始位置的元组：

RNA = 'AGCCAUGUAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAAUGAUCCGAGUAGCAUCUCAG'
candidates = []
start = 0

while (RNA.find('AUG', start) > -1):
    start = RNA.find('AUG', start) #Confound python and its lack of assignment returns
    candidates.append((RNA[start+3:], start+3))
    start += 1

matches = []

for candidate in candidates:
    for terminator in ['UAG', 'UGA', 'UAA']:
        end = 1;
        while(candidate[0].find(terminator, end) > -1):
            end = candidate[0].find(terminator, end)
            matches.append((candidate[1], candidate[1] + end, candidate[0][:end]))
            end += 1

for match in matches:
    print "%d - %d: %s" % match

打印：

7 - 49: UAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAU
7 - 85: UAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAAUGAUCCGAG
7 - 31: UAGCUAACUCAGGUUACAUGGGGA
7 - 72: UAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCC
7 - 76: UAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAA
7 - 11: UAGC
7 - 66: UAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAA
27 - 49: GGGAUGACCCCGCGACUUGGAU
27 - 85: GGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAAUGAUCCGAG
27 - 31: GGGA
27 - 72: GGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCC
27 - 76: GGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAA
27 - 66: GGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAA
33 - 49: ACCCCGCGACUUGGAU
33 - 85: ACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAAUGAUCCGAG
33 - 72: ACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCC
33 - 76: ACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAA
33 - 66: ACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAA
78 - 85: AUCCGAG

地狱，还有三行，你甚至可以根据它们落入RNA序列的位置对匹配进行排序：

from operator import itemgetter
matches.sort(key=itemgetter(1))
matches.sort(key=itemgetter(0)) 

在最终印刷网之前放置你：

007 - 011: UAGC
007 - 031: UAGCUAACUCAGGUUACAUGGGGA
007 - 049: UAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAU
007 - 066: UAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAA
007 - 072: UAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCC
007 - 076: UAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAA
007 - 085: UAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAAUGAUCCGAG
027 - 031: GGGA
027 - 049: GGGAUGACCCCGCGACUUGGAU
027 - 066: GGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAA
027 - 072: GGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCC
027 - 076: GGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAA
027 - 085: GGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAAUGAUCCGAG
033 - 049: ACCCCGCGACUUGGAU
033 - 066: ACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAA
033 - 072: ACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCC
033 - 076: ACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAA
033 - 085: ACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAAUGAUCCGAG
078 - 085: AUCCGAG

Answer 2

不幸的是， re模块目前不支持重叠匹配，但您可以轻松地将解决方案分解为：

'AGCCAUGUAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAAUGAUCCGAGUAGCAUCUCAG'
matches = []

for m in re.finditer('AUG', str):
    for n in re.finditer('(UAG)|(UGA)|(UAA)', str[m.start():]):
        matches.append(str[m.start()+3:m.start()+n.end()-3]

print matches

Answer 3

如果你不考虑“匹配”，而是考虑“间隔”，我认为你会发现它更容易。 这就是@ ionut-hulub所做的。 您可以在下面演示的单次传递中执行此操作，但是您应该使用更简单的finditer（）方法，除非您有足够的RNA字符串（或者它们足够长），您需要避免字符串上的冗余传递。

s = 'AGCCAUGUAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAAUGAUCCGAGUAGCAUCUCAG'

def intervals(s):
    state = []
    i = 0
    max = len(s) - 2
    while i < max:
        if s[i] == 'A' and s[i+1] == 'U' and s[i+2] == 'G':
            state.append(i)
        if s[i] == 'U' and (s[i+1] == 'A' and s[i+2] == 'G') or (s[i+1] == 'G' and s[i+2] == 'A') or (s[i+1] == 'A' and s[i+2] == 'A'):
            for b in state:
                yield (b, i)
        i += 1

for interval in intervals(s):
    print interval