（學習）python非貪婪、多行匹配Regex例子

一些regular的tips：

1 非貪婪flag >>> re . findall ( r " a( \d +?) " , " a23b " ) # 非貪婪模式
         [ ' 2 ' ]
>>> re . findall ( r " a( \d +) " , " a23b " )
         [ ' 23 ' ]

注意比較這種情況： >>> re . findall ( r " a( \d +)b " , " a23b " )
         [ ' 23 ' ]
>>> re . findall ( r " a( \d +?)b " , " a23b " ) #如果前後均有限定條件，則非匹配模式失效
         [ ' 23 ' ]

2 如果你要多行匹配，那麼加上re.S和re.M標誌
re.S：.將會匹配分行符號，預設.逗號不會匹配分行符號 >>> re . findall ( r " a( \d +)b.+a( \d +)b " , " a23b \n a34b " )
         []
>>> re . findall ( r " a( \d +)b.+a( \d +)b " , " a23b \n a34b " , re . S )
         [( ' 23 ' , ' 34 ' )]
>>>

re.M：^$標誌將會匹配每一行，預設^只會匹配符合正則的第一行；預設$只會匹配符合正則的末行 >>> re . findall ( r " ^a( \d +)b " , " a23b \n a34b " )
         [ ' 23 ' ]
>>> re . findall ( r " ^a( \d +)b " , " a23b \n a34b " , re . M )
         [ ' 23 ' , ' 34 ' ]

但是，如果沒有^標誌， >>> re . findall ( r " a( \d +)b " , " a23b \n a34b " )
         [ ' 23 ' , ' 43 ' ]

可見，是無需re.M

import re

n='''12 drummers drumming,
11 pipers piping, 10 lords a-leaping'''

p=re.compile('^\d+')
p_multi=re.compile('^\d+',re.MULTILINE) #設定 MULTILINE 標誌
print re.findall(p,n)  #['12']
print re.findall(p_multi,n) # ['12', '11']

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import re
a = 'a23b'
print re.findall('a(\d+?)',a)  #['2']
print re.findall('a(\d+)',a) #['23']
print re.findall(r'a(\d+)b',a) #['23']
print re.findall(r'a(\d+?)b',a) # ['23']
============================
b='a23b\na34b'
''' . 匹配非分行符號的任意一個字元'''

print re.findall(r'a(\d+)b.+a(\d+)b',b) #[]

print re.findall(r'a(\d+)b',b,re.M) # ['23', '34']

print re.findall(r'^a(\d+)b',b,re.M) # ['23', '34']

print re.findall(r'a(\d+)b',b) #['23','34'] 可以匹配多行

print re.findall(r'^a(\d+)b',b) # ['23'] 預設^只會匹配符合正則的第一行

print re.findall(r'a(\d+)b$',b) # ['34'] 預設$只會匹配符合正則的末行

print re.findall(r'a(\d+)b',b,re.M) #['23', '34']

print re.findall(r'a(\d+)b.?',b,re.M)  # ['23', '34']

print re.findall(r"a(\d+)b", "a23b\na34b")  # ['23', '34']

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