如何提取两个标记之间的子字符串?

问题:如何提取两个标记之间的子字符串?

假设我有一个字符串,'gfgfdAAA1234ZZZuijjk'而我只想提取'1234'一部分。

我只知道我感兴趣的部分之前AAA和之后ZZZ的几个字符1234

使用sed字符串可以执行以下操作:

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

结果,这会给我1234

如何在Python中做同样的事情?

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?


回答 0

使用正则表达式- 文档以供进一步参考

import re

text = 'gfgfdAAA1234ZZZuijjk'

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

# found: 1234

要么:

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

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

回答 1

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

然后,您也可以在re模块中使用正则表达式,如果需要的话,但这不是必需的。

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

Then you can use regexps with the re module as well, if you want, but that’s not necessary in your case.


回答 2

正则表达式

import re

re.search(r"(?<=AAA).*?(?=ZZZ)", your_text).group(0)

如果原样AttributeError中没有“ AAA”和“ ZZZ”,则上述原样会失败your_text

字符串方法

your_text.partition("AAA")[2].partition("ZZZ")[0]

如果中不存在“ AAA”或“ ZZZ”,则上面的内容将返回一个空字符串your_text

PS Python挑战?

regular expression

import re

re.search(r"(?<=AAA).*?(?=ZZZ)", your_text).group(0)

The above as-is will fail with an AttributeError if there are no “AAA” and “ZZZ” in your_text

string methods

your_text.partition("AAA")[2].partition("ZZZ")[0]

The above will return an empty string if either “AAA” or “ZZZ” don’t exist in your_text.

PS Python Challenge?


回答 3

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

回答 4

惊讶的是没有人提到这是我一次性脚本的快速版本:

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

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'

回答 5

您可以只使用一行代码

>>> import re

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

>>> ['1234']

结果将收到清单…

you can do using just one line of code

>>> import re

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

>>> ['1234']

result will receive list…


回答 6

您可以使用re模块:

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

You can use re module for that:

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

回答 7

使用sed可以用字符串执行以下操作:

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

结果是我会得到1234。

您可以re.sub使用相同的正则表达式对函数执行相同的操作。

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

在基本sed中,捕获组由表示\(..\),但是在python中,捕获组由表示(..)

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 (..).


回答 8

在python中,可以使用findall正则表达式(re)模块中的方法来提取子字符串形式的字符串。

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

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']

回答 9

您可以在代码中找到此功能的第一个子字符串(按字符索引)。另外,您可以找到子字符串之后的内容。

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]))

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]))

回答 10

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

回答 11

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)])

string
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)])

Gives

string

回答 12

以防万一某人必须做与我相同的事情。我必须在一行中提取括号内的所有内容。例如,如果我有一条类似“美国总统(巴拉克·奥巴马)与…会面……”这样的句子,而我只想获得“巴拉克·奥巴马”,这就是解决方案:

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

即您需要用slash \符号来阻止括号。虽然这是关于Python的更多正则表达式的问题。

另外,在某些情况下,您可能会在正则表达式定义之前看到“ r”符号。如果没有r前缀,则需要像C中那样使用转义符。这里有更多讨论。

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'

I.e. 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.


回答 13

使用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)

生成:

[['1234']]

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']]


回答 14

这是一个不使用正则表达式的解决方案,它也解决了第一个子字符串包含第二个子字符串的情况。仅当第二个标记在第一个标记之后时,此函数才会找到子字符串。

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)]

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)]

回答 15

另一种方法是使用列表(假设您要查找的子字符串仅由数字组成):

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

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

回答 16

如果没有匹配项,一个衬里返回其他字符串。编辑:改进的版本使用next功能,"not-found"如果需要,请替换为其他内容:

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

我执行此操作的另一种方法(不太理想)第二次使用正则表达式,但仍未找到更短的方法:

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

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) )