如何在两个标记之间提取 substring？

Question

Let's say I have a string 'gfgfdAAA1234ZZZuijjk' and I want to extract just the '1234' part.

I only know what will be the few characters directly before AAA , and after ZZZ the part I am interested in 1234 .

With sed it is possible to do something like this with a string:

echo "$STRING" | sed -e "s|.*AAA\(.*\)ZZZ.*|\1|"

And this will give me 1234 as a result.

How to do the same thing in Python?

Answer 1

Using regular expressions - documentation for further reference

import re

text = 'gfgfdAAA1234ZZZuijjk'

m = re.search('AAA(.+?)ZZZ', text)
if m:
    found = m.group(1)

# found: 1234

or:

import re

text = 'gfgfdAAA1234ZZZuijjk'

try:
    found = re.search('AAA(.+?)ZZZ', text).group(1)
except AttributeError:
    # AAA, ZZZ not found in the original string
    found = '' # apply your error handling

# found: 1234

Answer 2

>>> s = 'gfgfdAAA1234ZZZuijjk'
>>> start = s.find('AAA') + 3
>>> end = s.find('ZZZ', start)
>>> s[start:end]
'1234'

Answer 3

regular expression<\/h3>

import re re.search(r"(?<=AAA).*?(?=ZZZ)", your_text).group(0)<\/code><\/pre>
 The above as-is will fail with an AttributeError<\/code> if there are no "AAA" and "ZZZ" in your_text<\/code>

Answer 4

Surprised that nobody has mentioned this which is my quick version for one-off scripts:

>>> x = 'gfgfdAAA1234ZZZuijjk'
>>> x.split('AAA')[1].split('ZZZ')[0]
'1234'

Answer 5

you can do using just one line of code

>>> import re

>>> re.findall(r'\d{1,5}','gfgfdAAA1234ZZZuijjk')

>>> ['1234']

Answer 6

import re
print re.search('AAA(.*?)ZZZ', 'gfgfdAAA1234ZZZuijjk').group(1)

Answer 7

You can use re module for that:

>>> import re
>>> re.compile(".*AAA(.*)ZZZ.*").match("gfgfdAAA1234ZZZuijjk").groups()
('1234,)

Answer 8

In python, extracting substring form string can be done using findall method in regular expression ( re ) module.

>>> import re
>>> s = 'gfgfdAAA1234ZZZuijjk'
>>> ss = re.findall('AAA(.+)ZZZ', s)
>>> print ss
['1234']

Answer 9

With sed it is possible to do something like this with a string:

echo "$STRING" | sed -e "s|.*AAA\\(.*\\)ZZZ.*|\\1|"

And this will give me 1234 as a result.

You could do the same with re.sub function using the same regex.

>>> re.sub(r'.*AAA(.*)ZZZ.*', r'\1', 'gfgfdAAA1234ZZZuijjk')
'1234'

In basic sed, capturing group are represented by \\(..\\) , but in python it was represented by (..) .

Answer 10

text = 'I want to find a string between two substrings'
left = 'find a '
right = 'between two'

print(text[text.index(left)+len(left):text.index(right)])

Answer 11

>>> s = '/tmp/10508.constantstring'
>>> s.split('/tmp/')[1].split('constantstring')[0].strip('.')

Answer 12

You can find first substring with this function in your code (by character index). Also, you can find what is after a substring.

def FindSubString(strText, strSubString, Offset=None):
    try:
        Start = strText.find(strSubString)
        if Start == -1:
            return -1 # Not Found
        else:
            if Offset == None:
                Result = strText[Start+len(strSubString):]
            elif Offset == 0:
                return Start
            else:
                AfterSubString = Start+len(strSubString)
                Result = strText[AfterSubString:AfterSubString + int(Offset)]
            return Result
    except:
        return -1

# Example:

Text = "Thanks for contributing an answer to Stack Overflow!"
subText = "to"

print("Start of first substring in a text:")
start = FindSubString(Text, subText, 0)
print(start); print("")

print("Exact substring in a text:")
print(Text[start:start+len(subText)]); print("")

print("What is after substring \"%s\"?" %(subText))
print(FindSubString(Text, subText))

# Your answer:

Text = "gfgfdAAA1234ZZZuijjk"
subText1 = "AAA"
subText2 = "ZZZ"

AfterText1 = FindSubString(Text, subText1, 0) + len(subText1)
BeforText2 = FindSubString(Text, subText2, 0) 

print("\nYour answer:\n%s" %(Text[AfterText1:BeforText2]))

Answer 13

Just in case somebody will have to do the same thing that I did. I had to extract everything inside parenthesis in a line. For example, if I have a line like 'US president (Barack Obama) met with ...' and I want to get only 'Barack Obama' this is solution:

regex = '.*\((.*?)\).*'
matches = re.search(regex, line)
line = matches.group(1) + '\n'

Ie you need to block parenthesis with slash \\ sign. Though it is a problem about more regular expressions that Python.

Also, in some cases you may see 'r' symbols before regex definition. If there is no r prefix, you need to use escape characters like in C. Here is more discussion on that.

Answer 14

Using PyParsing

import pyparsing as pp

word = pp.Word(pp.alphanums)

s = 'gfgfdAAA1234ZZZuijjk'
rule = pp.nestedExpr('AAA', 'ZZZ')
for match in rule.searchString(s):
    print(match)

which yields:

[['1234']]

Answer 15

一个带有 Python 3.8 的班轮：

text[text.find(start:='AAA')+len(start):text.find('ZZZ')]

Answer 16

Here's a solution without regex that also accounts for scenarios where the first substring contains the second substring. This function will only find a substring if the second marker is after the first marker.

def find_substring(string, start, end):
    len_until_end_of_first_match = string.find(start) + len(start)
    after_start = string[len_until_end_of_first_match:]
    return string[string.find(start) + len(start):len_until_end_of_first_match + after_start.find(end)]

Answer 17

Another way of doing it is using lists (supposing the substring you are looking for is made of numbers, only) :

string = 'gfgfdAAA1234ZZZuijjk'
numbersList = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
output = []

for char in string:
    if char in numbersList: output.append(char)

print(f"output: {''.join(output)}")
### output: 1234

Answer 18

Typescript. Gets string in between two other strings.

<\/blockquote>

Answer 19

also, you can find all combinations in the bellow function

s = 'Part 1. Part 2. Part 3 then more text'
def find_all_places(text,word):
    word_places = []
    i=0
    while True:
        word_place = text.find(word,i)
        i+=len(word)+word_place
        if i>=len(text):
            break
        if word_place<0:
            break
        word_places.append(word_place)
    return word_places
def find_all_combination(text,start,end):
    start_places = find_all_places(text,start)
    end_places = find_all_places(text,end)
    combination_list = []
    for start_place in start_places:
        for end_place in end_places:
            print(start_place)
            print(end_place)
            if start_place>=end_place:
                continue
            combination_list.append(text[start_place:end_place])
    return combination_list
find_all_combination(s,"Part","Part")

result:

['Part 1. ', 'Part 1. Part 2. ', 'Part 2. ']

Answer 20

In case you want to look for multiple occurences.

content ="Prefix_helloworld_Suffix_stuff_Prefix_42_Suffix_andsoon"
strings = []
for c in content.split('Prefix_'):
    spos = c.find('_Suffix')
    if spos!=-1:
        strings.append( c[:spos])
print( strings )

Or more quickly:

strings = [ c[:c.find('_Suffix')] for c in content.split('Prefix_') if c.find('_Suffix')!=-1 ]

Answer 21

One liners that return other string if there was no match. Edit: improved version uses next function, replace "not-found" with something else if needed:

import re
res = next( (m.group(1) for m in [re.search("AAA(.*?)ZZZ", "gfgfdAAA1234ZZZuijjk" ),] if m), "not-found" )

My other method to do this, less optimal, uses regex 2nd time, still didn't found a shorter way:

import re
res = ( ( re.search("AAA(.*?)ZZZ", "gfgfdAAA1234ZZZuijjk") or re.search("()","") ).group(1) )