I try find in text using regex the elements like this: abs=abs , 1=1 etc.
i wrote this i this way:
opis="Some text abs=abs sfsdvc"
wyn=re.search('([\w]*)=\1',opis)
print(wyn.group(0))
And this find nothing, when i tried this code in the websites like www.regexr.com it was working correctly.
Am I doing something wrong in python re ?
You must specify the regex as raw string r'..'
>>> opis="Some text abs=abs sfsdvc"
>>> wyn=re.search(r'([\w]*)=\1',opis)
>>> print wyn.group(0)
abs=abs
From re documentation
Raw string notation (r"text") keeps regular expressions sane. Without it, every backslash ('\') in a regular expression would have to be prefixed with another one to escape it. For example, the two following lines of code are functionally identical:
Meaning, if you are not planing to use raw string, then all the \ in the string must be escaped as
>>> opis="Some text abs=abs sfsdvc"
>>> wyn=re.search('([\\w]*)=\\1',opis)
>>> print wyn.group(0)
abs=abs
Change your regex to:
re.search(r'(\w+)=\1', opis).group()
↑
Note that you don't really need character class here, the [ and ] are redundant, also it's better to have \w+ if you don't want to match the string "=" (lonely equal sign).
Related
I'm trying to unescape the escaped regex pattern to apply it to a string.
It's actually dynamic I don't exactly know what it would look like, but throughout my testing I encountered one problem, the string with escaped regex pattern looks like this:
\\d{4}
I've written a simple regex which replaces every single combination of backslash and a character with just a character
And I'm applying it this way:
sub(r"\\(.)", "\\1", escaped_pattern)
But what it gives me afterwards is d{4} not \d{4} as I expect.
I've tried using raw strings for repl, escape\unescape it, it still doesnt return what I expect it to return. Would appreciate any help.
EDIT
escaped_pattern = settings.reg_exp
regexp = sub(r"\\(.)", "\\1", escaped_pattern)
search(regexp, string_to_regexp).group()[0]
Based on you update I'm pretty sure that you would get exactly your desired output if you just stopped trying to unescape it.
import re
s1 = "1234astring"
matches = re.search("\\d{4}", s1)
matches.group(0)
"1234"
matches.group()[0]
"1"
Try r"\\\\(.)" in search pattern and '\\\1' in substitution pattern.
works OK here: https://regex101.com/r/M3ikqj/1
My string will contain () in it. What I need to do is to change the text between the brackets.
Example string: "B.TECH(CS,IT)".
In my string I need to change the content present inside the brackets to something like this.. B.TECH(ECE,EEE)
What I tried to resolve this problem is as follows..
reg = r'(()([\s\S]*?)())'
a = 'B.TECH(CS,IT)'
re.sub(reg,"(ECE,EEE)",a)
But I got output like this..
'(ECE,EEE)B(ECE,EEE).(ECE,EEE)T(ECE,EEE)E(ECE,EEE)C(ECE,EEE)H(ECE,EEE)((ECE,EEE)C(ECE,EEE)S(ECE,EEE),(ECE,EEE)I(ECE,EEE)T(ECE,EEE))(ECE,EEE)'
Valid output should be like this..
B.TECH(CS,IT)
Where I am missing and how to correctly replace the text.
The problem is that you're using parentheses, which have another meaning in RegEx. They're used as grouping characters, to catch output.
You need to escape the () where you want them as literal tokens. You can escape characters using the backslash character: \(.
Here is an example:
reg = r'\([\s\S]*\)'
a = 'B.TECH(CS,IT)'
re.sub(reg, '(ECE,EEE)', a)
# == 'B.TECH(ECE,EEE)'
The reason your regex does not work is because you are trying to match parentheses, which are considered meta characters in regex. () actually captures a null string, and will attempt to replace it. That's why you get the output that you see.
To fix this, you'll need to escape those parens – something along the lines of
\(...\)
For your particular use case, might I suggest a simpler pattern?
In [268]: re.sub(r'\(.*?\)', '(ECE,EEE)', 'B.TECH(CS,IT)')
Out[268]: 'B.TECH(ECE,EEE)'
I'm trying to get a python regex sub function to work but I'm having a bit of trouble. Below is the code that I'm using.
string = 'á:tdfrec'
newString = re.sub(ur"([aeioäëöáéíóàèìò])([aeioäëöáéíóúàèìò]):", ur"\1:\2", string)
#newString = re.sub(ur"([a|e|i|o|ä|ë|ö|á|é|í|ó|à|è|ì|ò])([a|e|i|o|ä|ë|ö|á|é|í|ó|ú|à|è|ì|ò]):", ur"\1:\2", string)
print newString
# a:́tdfrec is printed
So the the above code is not working the way that I intend. It's not displaying correctly but the string printed has the accute accent over the :. The regex statement is moving the accute accent from over the a to over the :. For the string that I'm declaring this regex is not suppose be applied. My intention for this regex statement is to only be applied for the following examples:
aä:dtcbd becomes a:ädtcbd
adfseì:gh becomes adfse:ìgh
éò:fdbh becomes é:òfdbh
but my regex statement is being applied and I don't want it to be. I think my problem is the second character set followed by the : (ie á:) is what's causing the regex statement to be applied. I've been staring at this for a while and tried a few other things and I feel like this should work but I'm missing something. Any help is appreciated!
The follow code with re.UNICODE flag also doesn't achieve the desired output:
>>> import re
>>> original = u'á:tdfrec'
>>> pattern = re.compile(ur"([aeioäëöáéíóàèìò])([aeioäëöáéíóúàèìò]):", re.UNICODE)
>>> print pattern.sub(ur'\1:\2', string)
á:tdfrec
Is it because of the diacritic and the tony the pony example for les misérable? The diacritic is on the wrong character after reversing it:
>>> original = u'les misérable'
>>> print ''.join([i for i in reversed(original)])
elbarésim sel
edit: Definitely an issue with the combining diacritics, you need to normalize both the regular expression and the strings you are trying to match. For example:
import unicodedata
regex = unicodedata.normalize('NFC', ur'([aeioäëöáéíóàèìò])([aeioäëöáéíóúàèìò]):')
string = unicodedata.normalize('NFC', u'aä:dtcbd')
newString = re.sub(regex, ur'\1:\2', string)
Here is an example that shows why you might hit an issue without the normalization. The string u'á' could either be the single code point LATIN SMALL LETTER A WITH ACCUTE (U+00E1) or it could be two code points, LATIN SMALL LETTER A (U+0061) followed by COMBINING ACUTE ACCENT (U+0301). These will probably look the same, but they will have very different behaviors in a regex because you can match the combining accent as its own character. That is what is happening here with the string 'á:tdfrec', a regular 'a' is captured in group 1, and the combining diacritic is captured in group 2.
By normalizing both the regex and the string you are matching you ensure this doesn't happen, because the NFC normalization will replace the diacritic and the character before it with a single equivalent character.
Original answer below.
I think your issue here is that the string you are attempting to do the replacement on is a byte string, not a Unicode string.
If these are string literals make sure you are using the u prefix, e.g. string = u'aä:dtcbd'. If they are not literals you will need to decode them, e.g. string = string.decode('utf-8') (although you may need to use a different codec).
You should probably also normalize your string, because part of the issue may have something to do with combining diacritics.
Note that in this case the re.UNICODE flag will not make a difference, because that only changes the meaning of character class shorthands like \w and \d. The important thing here is that if you are using a Unicode regular expression, it should probably be applied to a Unicode string.
In Python, I am extracting emails from a string like so:
split = re.split(" ", string)
emails = []
pattern = re.compile("^[a-zA-Z0-9_\.-]+#[a-zA-Z0-9-]+.[a-zA-Z0-9-\.]+$");
for bit in split:
result = pattern.match(bit)
if(result != None):
emails.append(bit)
And this works, as long as there is a space in between the emails. But this might not always be the case. For example:
Hello, foo#foo.com
would return:
foo#foo.com
but, take the following string:
I know my best friend mailto:foo#foo.com!
This would return null. So the question is: how can I make it so that a regex is the delimiter to split? I would want to get
foo#foo.com
in all cases, regardless of punctuation next to it. Is this possible in Python?
By "splitting by regex" I mean that if the program encounters the pattern in a string, it will extract that part and put it into a list.
I'd say you're looking for re.findall:
>>> email_reg = re.compile(r'[a-zA-Z0-9_.-]+#[a-zA-Z0-9-]+\.[a-zA-Z0-9-.]+')
>>> email_reg.findall('I know my best friend mailto:foo#foo.com!')
['foo#foo.com']
Notice that findall can handle more than one email address:
>>> email_reg.findall('Text text foo#foo.com, text text, baz#baz.com!')
['foo#foo.com', 'baz#baz.com']
Use re.search or re.findall.
You also need to escape your expression properly (. needs to be escaped outside of character classes, not inside) and remove/replace the anchors ^ and $ (for example with \b), eg:
r"\b[a-zA-Z0-9_.+-]+#[a-zA-Z0-9-]+\.[a-zA-Z0-9-.]+\b"
The problem I see in your regex is your use of ^ which matches the start of a string and $ which matches the end of your string. If you remove it and then run it with your sample test case it will work
>>> re.findall("[A-Za-z0-9\._-]+#[A-Za-z0-9-]+.[A-Za-z0-9-\.]+","I know my best friend mailto:foo#foo.com!")
['foo#foo.com']
>>> re.findall("[A-Za-z0-9\._-]+#[A-Za-z0-9-]+.[A-Za-z0-9-\.]+","Hello, foo#foo.com")
['foo#foo.com']
>>>
I'm searching for strings within strings using Regex. The pattern is a string literal that ends in (, e.g.
# pattern
" before the bracket ("
# string
this text is before the bracket (and this text is inside) and this text is after the bracket
I know the pattern will work if I escape the character with a backslash, i.e.:
# pattern
" before the bracket \\("
But the pattern strings are coming from another search and I can not control what characters will be or where. Is there a way of escaping an entire string literal so that anything between markers is treated as a string? For example:
# pattern
\" before the ("
The only other option I have is to do a substitute adding escapes for every protected character.
re.escape is exactly what I need. I'm using regexp in Access VBA which doens't have that method. I only have replace, execute or test methods.
Is there a way to escape everything within a string in VBA?
Thanks
You didn't specify the language, but it looks like Python, so if you have a string in Python whose special regex characters you need to escape, use re.escape():
>>> import re
>>> re.escape("Wow. This (really) is *cool*")
'Wow\\.\\ This\\ \\(really\\)\\ is\\ \\*cool\\*'
Note that spaces are escaped, too (probably to ensure that they still work in a re.VERBOSE regex).
Maybe write your own VBA escape function:
Function EscapeRegEx(text As String) As String
Dim regEx As RegExp
Set regEx = New RegExp
regEx.Global = True
regEx.Pattern = "(\[|\\|\^|\$|\.|\||\?|\*|\+|\(|\)|\{|\})"
EscapeRegEx = regEx.Replace(text, "\$1")
End Function
I'm pretty sure that with the limitations of the RegExp abilities in VBA/VBScript, you are going to have to replace the special characters in your pattern before using it. There doesn't seem to be anything built into it like there is in Python.
The following regex will capture everything from the beginning of the string to the first (. The first captured group $1 will contain the portion before (.
^([^(]+)\(
Depending on your language, you might have to escape it as:
"^([^(]+)\\("