Let's say I have a string 'gfgfdAAA1234ZZZuijjk'
and I want to extract just the '1234'
part.
假设我有一个字符串“gfgfdAAA1234ZZZuijjk”,我想提取“1234”部分。
I only know what will be the few characters directly before AAA
, and after ZZZ
the part I am interested in 1234
.
我只知道在AAA之前会有哪些角色,在ZZZ之后我对1234感兴趣。
With sed
it is possible to do something like this with a string:
有了sed,有可能用字符串来做这样的事情:
echo "$STRING" | sed -e "s|.*AAA\(.*\)ZZZ.*|\1|"
And this will give me 1234
as a result.
结果是1234。
How to do the same thing in Python?
如何在Python中做同样的事情?
12 个解决方案
#1
319
Using regular expressions - documentation for further reference
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
or:
或者:
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
#2
79
>>> 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.
然后,如果您愿意,也可以在re模块中使用regexp,但在您的情况下,这不是必需的。
#3
25
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
如果在your_text中没有“AAA”和“ZZZ”,那么上面的as-is将在AttributeError中失败
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
.
如果在your_text中不存在“AAA”或“ZZZ”,上面的语句将返回一个空字符串。
PS Python Challenge?
PS Python挑战?
#4
13
import re
print re.search('AAA(.*?)ZZZ', 'gfgfdAAA1234ZZZuijjk').group(1)
#5
6
You can use re module for that:
可以使用re模块实现:
>>> import re
>>> re.compile(".*AAA(.*)ZZZ.*").match("gfgfdAAA1234ZZZuijjk").groups()
('1234,)
#6
5
With sed it is possible to do something like this with a string:
有了sed,有可能用字符串来做这样的事情:
echo "$STRING" | sed -e "s|.*AAA\(.*\)ZZZ.*|\1|"
echo "$STRING" | sed -e "s|.*AAA\(.*\)ZZZ.*|\1|"
And this will give me 1234 as a result.
结果是1234。
You could do the same with re.sub
function using the same regex.
您可以使用相同的regex对re.sub函数执行相同的操作。
>>> re.sub(r'.*AAA(.*)ZZZ.*', r'\1', 'gfgfdAAA1234ZZZuijjk')
'1234'
In basic sed, capturing group are represented by \(..\)
, but in python it was represented by (..)
.
在基本sed中,捕获组由\(..\)表示,但在python中,它由(..)表示。
#7
4
You can find first substring with this function in your code (by character index). Also, you can find what is after a substring.
您可以在代码中(通过字符索引)找到第一个带有此函数的子字符串。此外,您还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]))
#8
2
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:
以防万一有人要做和我一样的事情。我必须把括号里的所有东西都提取出来。例如,如果我说“美国总统(巴拉克·奥巴马)与……”我只希望得到“巴拉克•奥巴马”这就是解决方案:
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.
也就是说,你需要用斜杠来阻止括号。尽管Python有一个关于正则表达式的问题。
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.
此外,在某些情况下,您可能会在regex定义之前看到“r”符号。如果没有r前缀,则需要使用转义字符,如c。
#9
1
>>> s = '/tmp/10508.constantstring'
>>> s.split('/tmp/')[1].split('constantstring')[0].strip('.')
#10
1
you can do using just one line of code
您可以只使用一行代码
>>> import re
>>> re.findall(r'\d{1,5}','gfgfdAAA1234ZZZuijjk')
>>> ['1234']
result will receive list...
结果将得到清单……
#11
0
In python, extracting substring form string can be done using findall
method in regular expression (re
) module.
在python中,可以在正则表达式(re)模块中使用findall方法提取子字符串表单字符串。
>>> import re
>>> s = 'gfgfdAAA1234ZZZuijjk'
>>> ss = re.findall('AAA(.+)ZZZ', s)
>>> print ss
['1234']
#12
0
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:
如果没有匹配,返回另一个字符串的衬垫。编辑:改进后的版本使用下一个功能,如果需要,可以用其他功能替换“not-found”:
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:
我的另一个方法,不是最优的,使用regex第二次,仍然没有找到更短的方法:
import re
res = ( ( re.search("AAA(.*?)ZZZ", "gfgfdAAA1234ZZZuijjk") or re.search("()","") ).group(1) )
#1
319
Using regular expressions - documentation for further reference
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
or:
或者:
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
#2
79
>>> 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.
然后,如果您愿意,也可以在re模块中使用regexp,但在您的情况下,这不是必需的。
#3
25
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
如果在your_text中没有“AAA”和“ZZZ”,那么上面的as-is将在AttributeError中失败
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
.
如果在your_text中不存在“AAA”或“ZZZ”,上面的语句将返回一个空字符串。
PS Python Challenge?
PS Python挑战?
#4
13
import re
print re.search('AAA(.*?)ZZZ', 'gfgfdAAA1234ZZZuijjk').group(1)
#5
6
You can use re module for that:
可以使用re模块实现:
>>> import re
>>> re.compile(".*AAA(.*)ZZZ.*").match("gfgfdAAA1234ZZZuijjk").groups()
('1234,)
#6
5
With sed it is possible to do something like this with a string:
有了sed,有可能用字符串来做这样的事情:
echo "$STRING" | sed -e "s|.*AAA\(.*\)ZZZ.*|\1|"
echo "$STRING" | sed -e "s|.*AAA\(.*\)ZZZ.*|\1|"
And this will give me 1234 as a result.
结果是1234。
You could do the same with re.sub
function using the same regex.
您可以使用相同的regex对re.sub函数执行相同的操作。
>>> re.sub(r'.*AAA(.*)ZZZ.*', r'\1', 'gfgfdAAA1234ZZZuijjk')
'1234'
In basic sed, capturing group are represented by \(..\)
, but in python it was represented by (..)
.
在基本sed中,捕获组由\(..\)表示,但在python中,它由(..)表示。
#7
4
You can find first substring with this function in your code (by character index). Also, you can find what is after a substring.
您可以在代码中(通过字符索引)找到第一个带有此函数的子字符串。此外,您还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]))
#8
2
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:
以防万一有人要做和我一样的事情。我必须把括号里的所有东西都提取出来。例如,如果我说“美国总统(巴拉克·奥巴马)与……”我只希望得到“巴拉克•奥巴马”这就是解决方案:
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.
也就是说,你需要用斜杠来阻止括号。尽管Python有一个关于正则表达式的问题。
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.
此外,在某些情况下,您可能会在regex定义之前看到“r”符号。如果没有r前缀,则需要使用转义字符,如c。
#9
1
>>> s = '/tmp/10508.constantstring'
>>> s.split('/tmp/')[1].split('constantstring')[0].strip('.')
#10
1
you can do using just one line of code
您可以只使用一行代码
>>> import re
>>> re.findall(r'\d{1,5}','gfgfdAAA1234ZZZuijjk')
>>> ['1234']
result will receive list...
结果将得到清单……
#11
0
In python, extracting substring form string can be done using findall
method in regular expression (re
) module.
在python中,可以在正则表达式(re)模块中使用findall方法提取子字符串表单字符串。
>>> import re
>>> s = 'gfgfdAAA1234ZZZuijjk'
>>> ss = re.findall('AAA(.+)ZZZ', s)
>>> print ss
['1234']
#12
0
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:
如果没有匹配,返回另一个字符串的衬垫。编辑:改进后的版本使用下一个功能,如果需要,可以用其他功能替换“not-found”:
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:
我的另一个方法,不是最优的,使用regex第二次,仍然没有找到更短的方法:
import re
res = ( ( re.search("AAA(.*?)ZZZ", "gfgfdAAA1234ZZZuijjk") or re.search("()","") ).group(1) )