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)])
给
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) )
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