Develop Reference

How to substitute a regex with another regex in a string

This question showed how to replace a regex with another regex like this
$string = '"SIP/1037-00000014","SIP/CL-00000015","Dial","SIP/CL/61436523277,45"';
$$pattern = '["SIP/CL/(\d*),(\d*)",]';
$replacement = '"SIP/CL/\1|\2",';
$string = preg_replace($pattern, $replacement, $string);
print($string);
However, I couldn't adapt that pattern to solve my case where I want to remove the full stop that lies between 2 words but not between a word and a number:
text = 'this . is bad. Not . 820'
regex1 = r'(\w+)(\s\.\s)(\D+)'
regex2 = r'(\w+)(\s)(\D+)'
re.sub(regex1, regex2, text)
# Desired outcome:
'this is bad. Not . 820'
Basically I like to remove the . between the two alphabet words. Could someone please help me with this problem? Thank you in advance.

These expressions might be close to what you might have in mind:
\s[.](?=\s\D)
or
(?<=\s)[.](?=\s\D)
Test
import re
regex = r"\s[.](?=\s\D)"
test_str = "this . is bad. Not . 820"
print(re.sub(regex, "", test_str))
Output
this is bad. Not . 820
If you wish to explore/simplify/modify the expression, it's been
explained on the top right panel of
regex101.com. If you'd like, you
can also watch in this
link, how it would match
against some sample inputs.

Firstly, you can't really take PHP and apply it directly to Python, for obvious reasons.
Secondly, it always helps to specify which version of Python you're using as APIs change. Luckily in this instance, the API of re.sub has remained the same between Python 2.x and Python 3.
Onto your issue.
The second argument to re.sub is either a string or a function. If you pass in regex2 it'll just replace regex1 with the string contents of regex2, it won't apply regex2 as a regex.
If you want to use groups derived from the first regex (similar to your example, which is using \1 and \2 to extract the first and second matching group from the first regex), then you'd want to use a function, which takes a match object as its sole argument, which you could then use to extract matching groups and return them as part of the replacement string.

python regex and replace

I am trying to learn python and regex at the same time and I am having some trouble in finding how to match till end of string and make a replacement on the fly.
So, I have a string like so:
ss="this_is_my_awesome_string/mysuperid=687y98jhAlsji"
What I'd want is to first find 687y98jhAlsji (I do not know this content before hand) and then replace it to myreplacedstuff like so:
ss="this_is_my_awesome_string/mysuperid=myreplacedstuff"
Ideally, I'd want to do a regex and replace by first finding the contents after mysuperid= (till the end of string) and then perform a .replace or .sub if this makes sense.
I would appreciate any guidance on this.

You can try this:
re.sub(r'[^=]+$', 'myreplacedstuff', ss)
The idea is to use a character class that exclude the delimiter (here =) and to anchor the pattern with $
explanation:
[^=] is a character class and means all characters that are not =
[^=]+ one or more characters from this class
$ end of the string
Since the regex engine works from the left to the right, only characters that are not an = at the end of the string are matched.

You can use regular expressions:
>>> import re
>>> mymatch = re.search(r'mysuperid=(.*)', ss)
>>> ss.replace(mymatch.group(1), 'replacing_stuff')
'this_is_my_awesome_string/mysuperid=replacing_stuff'
You should probably use #Casimir's answer though. It looks cleaner, and I'm not that good at regex :p.

Python split by regular expression

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']
>>>

python regular expression replacing part of a matched string

i got an string that might look like this
"myFunc('element','node','elementVersion','ext',12,0,0)"
i'm currently checking for validity using, which works fine
myFunc\((.+?)\,(.+?)\,(.+?)\,(.+?)\,(.+?)\,(.+?)\,(.+?)\)
now i'd like to replace whatever string is at the 3rd parameter.
unfortunately i cant just use a stringreplace on whatever sub-string on the 3rd position since the same 'sub-string' could be anywhere else in that string.
with this and a re.findall,
myFunc\(.+?\,.+?\,(.+?)\,.+?\,.+?\,.+?\,.+?\)
i was able to get the contents of the substring on the 3rd position, but re.sub does not replace the string it just returns me the string i want to replace with :/
here's my code
myRe = re.compile(r"myFunc\(.+?\,.+?\,(.+?)\,.+?\,.+?\,.+?\,.+?\)")
val = "myFunc('element','node','elementVersion','ext',12,0,0)"
print myRe.findall(val)
print myRe.sub("noVersion",val)
any idea what i've missed ?
thanks!
Seb

In re.sub, you need to specify a substitution for the whole matching string. That means that you need to repeat the parts that you don't want to replace. This works:
myRe = re.compile(r"(myFunc\(.+?\,.+?\,)(.+?)(\,.+?\,.+?\,.+?\,.+?\))")
print myRe.sub(r'\1"noversion"\3', val)

If your only tool is a hammer, all problems look like nails. A regular expression is a powerfull hammer but is not the best tool for every task.
Some tasks are better handled by a parser. In this case the argument list in the string is just like a Python tuple, sou you can cheat: use the Python builtin parser:
>>> strdata = "myFunc('element','node','elementVersion','ext',12,0,0)"
>>> args = re.search(r'\(([^\)]+)\)', strdata).group(1)
>>> eval(args)
('element', 'node', 'elementVersion', 'ext', 12, 0, 0)
If you can't trust the input ast.literal_eval is safer than eval for this. Once you have the argument list in the string decontructed I think you can figure out how to manipulate and reassemble it again, if needed.

Read the documentation: re.sub returns a copy of the string where every occurrence of the entire pattern is replaced with the replacement. It cannot in any case modify the original string, because Python strings are immutable.
Try using look-ahead and look-behind assertions to construct a regex that only matches the element itself:
myRe = re.compile(r"(?<=myFunc\(.+?\,.+?\,)(.+?)(?=\,.+?\,.+?\,.+?\,.+?\))")

Have you tried using named groups? http://docs.python.org/howto/regex.html#search-and-replace
Hopefully that will let you just target the 3rd match.

If you want to do this without using regex:
>>> s = "myFunc('element','node','elementVersion','ext',12,0,0)"
>>> l = s.split(",")
>>> l[2]="'noVersion'"
>>> s = ",".join(l)
>>> s
"myFunc('element','node','noVersion','ext',12,0,0)"

Difference in regex behavior between Perl and Python?

I have a couple email addresses, 'support#company.com' and '1234567#tickets.company.com'.
In perl, I could take the To: line of a raw email and find either of the above addresses with
/\w+#(tickets\.)?company\.com/i
In python, I simply wrote the above regex as'\w+#(tickets\.)?company\.com' expecting the same result. However, support#company.com isn't found at all and a findall on the second returns a list containing only 'tickets.'. So clearly the '(tickets\.)?' is the problem area, but what exactly is the difference in regular expression rules between Perl and Python that I'm missing?

The documentation for re.findall:
findall(pattern, string, flags=0)
Return a list of all non-overlapping matches in the string.
If one or more groups are present in the pattern, return a
list of groups; this will be a list of tuples if the pattern
has more than one group.
Empty matches are included in the result.
Since (tickets\.) is a group, findall returns that instead of the whole match. If you want the whole match, put a group around the whole pattern and/or use non-grouping matches, i.e.
r'(\w+#(tickets\.)?company\.com)'
r'\w+#(?:tickets\.)?company\.com'
Note that you'll have to pick out the first element of each tuple returned by findall in the first case.

I think the problem is in your expectations of extracted values. Try using this in your current Python code:
'(\w+#(?:tickets\.)?company\.com)'

Two problems jump out at me:
You need to use a raw string to avoid having to escape "\"
You need to escape "."
So try:
r'\w+#(tickets\.)?company\.com'
EDIT
Sample output:
>>> import re
>>> exp = re.compile(r'\w+#(tickets\.)?company\.com')
>>> bool(exp.match("s#company.com"))
True
>>> bool(exp.match("1234567#tickets.company.com"))
True

There isn't a difference in the regexes, but there is a difference in what you are looking for. Your regex is capturing only "tickets." if it exists in both regexes. You probably want something like this
#!/usr/bin/python
import re
regex = re.compile("(\w+#(?:tickets\.)?company\.com)");
a = [
"foo#company.com",
"foo#tickets.company.com",
"foo#ticketsacompany.com",
"foo#compant.org"
];
for string in a:
print regex.findall(string)