Searching list of strings using regular expressions in python - python

I have a list of Strings. I want to select the strings which match a certain pattern using regular expression.
Python regular expressions dont take a list and I dont want to use loops.
Any suggestion?

Try:
def searcher(s):
if COMPILED_REGEXP_OBJECT.search(s):
return s
matching_strings = filter(searcher, YOUR_LIST_OF_STRING)
searcher() returns the string if it matches, else returns None. filter() only returns "true" objects, so will skip the Nones. It will also skip empty strings, but doubt that's a problem.
Or, better, as #JonClements pointed out:
matching_strings = filter(COMPILED_REGEXP_OBJECT.search, YOUR_LIST_OF_STRING)
Not only shorter, it only looks up the .search method once (instead of once per string).

Related

Filter words in Regular expression

So, quite recently I have been introduced to regular expressions in Python and I've come across with some code online to filter words from a string list that are contained on other substrings.
def Filter(string, substr):
return [str for str in string
if re.match(r'[^\d]+|^', str).group(0) in substr]
It seems pretty straightforward and it works pretty well for my specific problem I'm meeting, but I really can't wrap my head around the meaning of it and how it is working. It just seems very confusing. Can anyone explain to me as if I was a baby or something? My coding skills are not that great, and I'm still a rookie.
Just to be clear, the code works, and I'm happy to move on, I just don't understand this bit.
[^\d] matches any character that isn't a numeric digit; this can also be written as \D.
+ after a pattern means to match any sequence of characters that match the pattern, so [^\d]+ matches a sequence of non-digits.
| separates alternative patterns to match.
The second alternative ^ matches the beginning of the string. Every string will match this. I think they use this just to avoid the match failing, so that you can always call .group(0) on the result. They could accomplish the same thing by changing + to * in the first alternative, since this means that the matched sequence can be 0 repetitions.
re.match() looks for a match of the regexp at the beginning of the argument string. And .group(0) returns what was matched by the entire regexp. So this whole thing returns the initial sequence of non-digits in str.
Finally, the list comprehension returns any of the items in strings whose initial sequence of non-digits is in substr.
With the simplifications I mentioned above, this can be rewritten:
def Filter(string, substr):
return [item for item in string
if re.match(r'\D*', item).group(0) in substr]
Note that if any of the items begin with a digit, the result of the regexp will be an empty string, and an empty string is a substring of every string. So these items will be included in the filter result. I suspect this is not the intended result.
I will try to to explain this for you.
So basically we are creating a method named "filter" and passing two arguments i.e "string (to be searched in)" and "substring (to be searched for)". Then we are using re.match inside a python return function along with an if condition within a for loop (the for loop helps us traverse through the main string one by one).
As for: (r'[^\d]+|^': this is a regular expression pattern where, \d is regex pattern for digit and + means at least one or more and finally they are closed within () that means the group that you want to capture.
re.match:
re.match is a function that searches only from the beginning of the string and returns the matched object (if found). However, if the substring is found somewhere in the middle then it will simply return none.

Equivalent to vlookup for strings in Python?

Is there a way to search for a substring in a larger string, and then return a different substring x places left or right of the original substring?
I want to look through a string like "blahblahlink:"www.example.com"blahblah for the string "link:" and return the subsequent url.
Thanks!
Python 3, if that matters.
I think you should use regular expressions. There is module called re in python for that.
Pure Python solution would be to use index which tells you where in the string the first match occurs, then use the [start:end] slicing notation to select the string from that point:
"blahblahlink:www.example.comblahblah".index("link")
# returns 8
i = "blahblahlink:www.example.comblahblah".index("link")
"blahblahlink:www.example.comblahblah"[i+5:i+20]
# returns 'www.example.com'

Matching characters in two Python strings

I am trying to print the shared characters between 2 sets of strings in Python, I am doing this with the hopes of actually finding how to do this using nothing but python regular expressions (I don't know regex so this might be a good time to learn it).
So if first_word = "peepa" and second_word = "poopa" I want the return value to be: "pa"
since in both variables the characters that are shared are p and a. So far I am following the documentation on how to use the re module, but I can't seem to grasp the basic concepts of this.
Any ideas as to how would I solve this problem?
This sounds like a problem where you want to find the intersection of characters between the two strings. The quickest way would be to do this:
>>> set(first_word).intersection(second_word)
set(['a', 'p'])
I don't think regular expressions are the right fit for this problem.
Use sets. Casting a string to a set returns an iterable with unique letters. Then you can retrieve the intersection of the two sets.
match = set(first_word.lower()) & set(second_word.lower())
Using regular expressions
This problem is tailor made for sets. But, you ask for "how to do this using nothing but python regular expressions."
Here is a start:
>>> import re
>>> re.sub('[^peepa]', '', "poopa")
'ppa'
The above uses regular expressions to remove from "poopa" every letter that was not already in "peepa". (As you see it leaves duplicated letters which sets would not do.)
In more detail, re.sub does substitutions based on regular expressions. [peepa] is a regular expression that means any of the letters peepa. The regular expression [^peepa] means anything that is not in peepa. Anything matching this regular expression is replaced with the empty string "", that is, it is removed. What remains are only the common letters.

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)

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