匹配python中可能最长的字符集

Question

有一个字符为0123456789AB的字符串。 我有一个regexp：

([^1368A]+|[^2479B]+|[^0358A]+|[^1469B]+|[^0257A]+|[^1368B]+|[^02479]+|[^1358A]+|[^2469B]+|[^0357A]+|[^1468B]+|[^02579]+)

首先匹配而不是最长匹配的问题。如何使它与python中最长的匹配？我不希望在regexp中有这种可能性。 编辑：我需要找到所有匹配项。最好有成功模式的索引。 输入示例：

^{pr2}$

另一个例子：

^{3}$

输出示例：

'470470574704705747047057570570574704705727027B5747047057570570574704705727027B5747047057570570574704705727027B5747047057570570574704705727027B5747047057570570574704705727027B5747047057570570574704705727027B5747047057570570574704705727027B5747047057570570574704705727027B5747047057570570574704705727027B57470470574704705747047057B2727875377AA0577AA0577AA0577AA0577AA0577AA059959959959952257777225'
('1368A','470470574704705747047057570570574704705727027B5747047057570570574704705727027B5747047057570570574704705727027B5747047057570570574704705727027B5747047057570570574704705727027B5747047057570570574704705727027B5747047057570570574704705727027B5747047057570570574704705727027B5747047057570570574704705727027B57470470574704705747047057B2727'),('','8'),('1468B','75377AA0577AA0577AA0577AA0577AA0577AA059959959959952257777225')

添加：当前我使用此代码：

import sys,re
from midplay import MidiFile,NoteOn
from collections import deque
notes=("C","C#","D","Eb","E","F","F#","G","G#","A","Bb","B")
noteshex=('0','1','2','3','4','5','6','7','8','9','A','B')
major=lambda x:((x)%12,(x+2)%12,(x+4)%12,(x+5)%12,(x+7)%12,(x+9)%12,(x+11)%12,)
minor=lambda x:((x)%12,(x+2)%12,(x+3)%12,(x+5)%12,(x+7)%12,(x+8)%12,(x+10)%12,)
nomajor=lambda x:{(x+1)%12,(x+3)%12,(x+6)%12,(x+8)%12,(x+10)%12}
nominor=lambda x:{(x+1)%12,(x+4)%12,(x+6)%12,(x+9)%12,(x+11)%12}
nomajortonelist=[re.compile('([^'+''.join([noteshex[note] for note in nomajor(tonality)])+']+)') for tonality in range(12)]
nominortonelist=nomajortonelist[3:]+nomajortonelist[:3]
if len(sys.argv)!=2:
    sys.exit('usage: py tonalitydetect.py [C:\path]filename.mid')
midi=MidiFile(sys.argv[1])
for num, track in enumerate(midi):
    print('Track:',num,'messages:',len(track))
    channelnotes=['','','','','','','','','','','','','','','','']
    channeltonality=[deque(),deque(),deque(),deque(),deque(),deque(),deque(),deque(),deque(),deque(),deque(),deque(),deque(),deque(),deque(),deque()]
    for msg in track:
        if isinstance(msg,NoteOn):
            channelnotes[msg.channel]+=(noteshex[msg.note%12])
    for chnum,channel in enumerate(channelnotes):
        tomatch=[channel]
        matches=[]
        while ''.join(tomatch)!='':
            curchanmaxmatch=deque()
            for string in tomatch:
                for exp in nomajortonelist:
                    curchanmaxmatch.append((exp,max(exp.findall(string)+[''], key=len)))
            matches.append(max(curchanmaxmatch+deque([('','',)]), key=lambda x:len(x[1])))
            newmatch=[]
            found=0
            for x in tomatch:
                if not found:
                    match=x.split(matches[-1][1],1)
                    if len(match)>1:
                        found=1
                    newmatch.extend(match)
                else:
                    newmatch.append(x)
            tomatch=[x for x in newmatch if x!='']
        matches=sorted(matches, key=lambda x:len(x[1]))
        toseek=channel
        while len(matches):
            for num,match in enumerate(matches):
                if not toseek.find(match[1]):
                    channeltonality[chnum].append(match)
                    toseek=toseek[len(match[1]):]
                    del matches[num]
                    break
    for chnum,channel in enumerate(channeltonality):
        print('Channel',chnum,':',[notes[nomajortonelist.index(x[0])]+' major, '+notes[nominortonelist.index(x[0])]+' minor' for x in channel])

Answer 1

编辑：有关显示最长匹配位置的解决方案，请参见下文。

最适合您的问题的内置工具是^{}:“以字符串列表的形式返回字符串中模式的所有非重叠匹配。'

对于您的情况，一个问题是不同的匹配可以重叠-但是findall只返回不重叠的匹配。例如，输入字符串2B001AA包含两个不同的匹配项：2B00和{}。re.findall函数将查找并返回第一个匹配项2B00。然后，它继续执行它停止的地方-只返回1AA作为下一个匹配。在

您可以通过将regexp分解为一个接一个匹配的片段来解决此问题：

import re
patterns=[
    r'[^1368A]+', r'[^2479B]+', r'[^0358A]+', r'[^1469B]+',
    r'[^0257A]+', r'[^1368B]+', r'[^02479]+', r'[^1358A]+',
    r'[^2469B]+', r'[^0357A]+', r'[^1468B]+', r'[^02579]+'
]

def match_patterns(string):
    for pattern in patterns:
        for match in re.findall(pattern,string):
            yield match

函数match_pattern返回所有匹配项（但不总是按顺序）。在python3中，可以将此函数写得更短：

^{pr2}$

在任何情况下，都可以使用内置函数max提取最长匹配：

def find_longest_match(string):
    return max(match_patterns(string), key=len)

print(find_longest_match('12A34B32A43')) # prints: A34B3

---

如果您还想获得最长匹配的位置，请使用 ^{}:'返回一个迭代器，该迭代器对字符串中的RE模式的所有非重叠匹配产生match objects。'对于每个返回的match，match.start()给出匹配的开始位置和文本。在

import re
patterns=[
    r'[^1368A]+', r'[^2479B]+', r'[^0358A]+', r'[^1469B]+',
    r'[^0257A]+', r'[^1368B]+', r'[^02479]+', r'[^1358A]+',
    r'[^2469B]+', r'[^0357A]+', r'[^1468B]+', r'[^02579]+'
]

def match_patterns(string):
    for pattern in patterns:
        yield from re.finditer(pattern, string)

def find_longest_match(string):
    match=max(match_patterns(string), key=lambda m: len(m.group(0)))
    if match:
        return match.start(), match.group(0)
    else:
        return None

print(find_longest_match('12A34B32A43')) # prints: (2, 'A34B3')