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:{ {1}},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:
AGCCAUGUAGCUAACUCAGGUUACAUGGGGAUGACCCCGCGACUUGGAUUAGAGUCUCUUUUGGAAUAAGCCUGAAUGAUCCGAGUAGCAUCUCAG
你能告诉我正则表达式模式中的错误吗?
答案 0 :(得分:3)
使用一个正则表达式执行此操作实际上非常困难,因为大多数使用不需要重叠匹配。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。在AUGUAGUGAUAA
的情况下,您期望得到什么回报?有两个字符串要返回吗?还是只有一个?现在,您的正则表达式甚至无法捕获UAG
,因为它继续匹配UAGUGA
并在UAA
处被截断。为了解决这个问题,您可能希望使用?
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,但是如果你的RNA序列包含AUGAUGUAGUGAUAA
怎么办?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,你一定会得到所有的比赛。
如果您需要跟踪每个匹配的位置,您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,你还可以根据它们落入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))
在最终打印之前放置你:
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
答案 1 :(得分:0)
不幸的是re
模块目前不支持重叠匹配,但您可以轻松地解决问题:
'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
答案 2 :(得分:0)
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),您需要避免在字符串上进行冗余传递。
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