使用python re模块在两个字符（a * b）之间的字符串中查找子字符串的数量

Question

给出字符串S作为输入。程序必须找到匹配a * b的模式数量。 *代表1个或多个字母。

import re
s = input()
matches = re.findall(r'MAGIC',s)
print(len(matches))


'''
i/p - aghba34bayyb
o/p - 2
(i.e aghb,ayyb)
It should not take a34b in count.

i/p - aabb
o/p - 3
(i.e aab abb aabb)

i/p : adsbab 
o/p : 2 
(i.e adsb ab)'''

Answer 1

您可以使用

a[a-zA-Z]+?b

---

import re
s = input()
matches = re.findall(r'a[a-zA-Z]+?b',s)
print(len(matches))

Python Demo

Answer 2

使用 <cffunction name="getRegularInvoice" access="remote" output="false" returnformat="json"> <cfargument name="cusno" type="any" required="true"> <cfquery name="getitems" datasource="#datasrc#"> SELECT * FROM reg_invoice_items WHERE cusno = '#arguments.cusno#' ORDER BY code asc </cfquery> <cfset returnArray = arrayNew(1)> <cfloop query="getitems"> <cfset data = structNew() /> <cfset data['refid'] = #getitems.refid# /> <cfset data['code'] = #getitems.code# /> <cfset data['description'] = #getitems.description# /> <cfset data['qty'] = #getitems.qty# /> <cfset data['price'] = #getitems.price# /> <cfset data['total'] = #getitems.total# /> <cfset data['originalprice'] = #getitems.originalprice# /> <cfset data['disc'] = #getitems.disc# /> <cfset data['exvat'] = #getitems.exvat# /> <cfset data['location'] = #getitems.location# /> <cfset data['costnon'] = #getitems.costnon# /> <cfset data['ajustnon'] = #getitems.ajustnon# /> <cfset data['regmonth'] = #getitems.regmonth# /> <cfset data['regday'] = #getitems.regday# /> <cfset data['onceoff'] = "#getitems.once_off#" /> <cfset data['dep'] = #getitems.dep# /> <cfset data['payment_type'] = #getitems.payment_type# /> <cfset data['currency'] = #getitems.currency# /> <cfset arrayAppend(returnArray,data) /> </cfloop> <cfreturn returnArray /> 来匹配所有子字符串：

re.finditer

打印：

inputs = ['aghba34bayyb',
'aabb',
'adsbab']

import re

def all_substrings(s):
    length, seen = len(s), set()
    for i in range(length):
        for j in range(i + 1, length + 1):
            for g in re.finditer(r'(a[^\d]+b)', s[i:j]):
                if (i+g.start(), i+g.end()) in seen:
                    continue
                seen.add((i+g.start(), i+g.end()))
                yield g.groups()[0]

for i in inputs:
    print('Input="{}" Matches:'.format(i))
    for s in all_substrings(i):
        print(' "{}"'.format(s))

Answer 3

您可以在单词中找到a和b的位置，找到所有可能的子字符串，然后过滤仅在其中包含一个或多个字符的子字符串

from itertools import product

words = ['aghba34bayyb', 'aabb', 'adsbab']

for word in words:
    a_pos = [i for i,c in enumerate(word) if c=='a']
    b_pos = [i for i,c in enumerate(word) if c=='b']
    all_substrings = [word[s:e+1] for s,e in product(a_pos, b_pos) if e>s]
    substrings = [s for s in all_substrings if re.match(r'a[a-zA-Z]+b$', s)]
    print (word, substrings)

输出

aghba34bayyb ['aghb', 'ayyb']
aabb ['aab', 'aabb', 'abb']
adsbab ['adsb', 'adsbab']

Answer 4

re.findall(r'a[A-Za-z]+?b',s)

哪里

• [A-Za-z]匹配字母字符，
• +是一个或多个字符
• ?告诉它不贪心

Answer 5

您可以匹配a，后跟1个字符a-z，然后使用匹配0+次a或c-z的字符类，然后匹配第一个{{1} }

b

Regex demo

如果您要匹配以下所有b以匹配a[a-z][ac-z]*b 而不是aabb，则可以使用

aab

Regex demo | Python demo

a[a-z][ac-z]*b+