Python regex to match multiple times - python

I'm trying to match a pattern against strings that could have multiple instances of the pattern. I need every instance separately. re.findall() should do it but I don't know what I'm doing wrong.
pattern = re.compile('/review: (http://url.com/(\d+)\s?)+/', re.IGNORECASE)
match = pattern.findall('this is the message. review: http://url.com/123 http://url.com/456')
I need 'http://url.com/123', http://url.com/456 and the two numbers 123 & 456 to be different elements of the match list.
I have also tried '/review: ((http://url.com/(\d+)\s?)+)/' as the pattern, but no luck.

Use this. You need to place 'review' outside the capturing group to achieve the desired result.
pattern = re.compile(r'(?:review: )?(http://url.com/(\d+))\s?', re.IGNORECASE)
This gives output
>>> match = pattern.findall('this is the message. review: http://url.com/123 http://url.com/456')
>>> match
[('http://url.com/123', '123'), ('http://url.com/456', '456')]

You've got extra /'s in the regex. In python the pattern should just be a string. e.g. instead of this:
pattern = re.compile('/review: (http://url.com/(\d+)\s?)+/', re.IGNORECASE)
It should be:
pattern = re.compile('review: (http://url.com/(\d+)\s?)+', re.IGNORECASE)
Also typically in python you'd actually use a "raw" string like this:
pattern = re.compile(r'review: (http://url.com/(\d+)\s?)+', re.IGNORECASE)
The extra r on the front of the string saves you from having to do lots of backslash escaping etc.

Use a two-step approach: First get everything from "review:" to EOL, then tokenize that.
msg = 'this is the message. review: http://url.com/123 http://url.com/456'
review_pattern = re.compile('.*review: (.*)$')
urls = review_pattern.findall(msg)[0]
url_pattern = re.compile("(http://url.com/(\d+))")
url_pattern.findall(urls)

Related

How to extract function name python regex

Hello I am trying to extract the function name in python using Regex however I am new to Python and nothing seems to be working for me. For example: if i have a string "def myFunction(s): ...." I want to just return myFunction
import re
def extractName(s):
string = []
regexp = re.compile(r"\s*(def)\s+\([^\)]*\)\s*{?\s*")
for m in regexp.finditer(s):
string += [m.group()]
return string
Assumption: You want the name myFunction from "...def myFunction(s):..."
I find something missing in your regex and the way it is structured.
\s*(def)\s+\([^\)]*\)\s*{?\s*
Lets look at it step by step:
\s*: match to zero or more white spaces.
(def): match to the word def.
\s+: match to one or more white spaces.
\([^\)]*\): match to balanced ()
\s*: match to zero or more white spaces.
After that pretty much doesn't matter if you are going for just the name of the function. You are not matching the exact thing you want out of the regex.
You can try this regex if you are interested in doing it by regex:
\s*(def)\s([a-zA-Z]*)\([a-zA-z]*\)
Now the way I have structured the regex, you will get def myFunction(s) in group0, def in group1 and myFunction in group2. So you can use the following code to get you result:
import re
def extractName(s):
string = ""
regexp = re.compile(r"(def)\s([a-zA-Z]*)\([a-zA-z]*\)")
for m in regexp.finditer(s):
string += m.group(2)
return string
You can check your regex live by going on this site.
Hope it helps!

Parse a string using regex to obtain matches beginning with a certain word

I tried to search but the information that I am getting seems to be kinda overwhelming and far from what I need. I can't seem to get it to work.
The requirement is to get the function that starts with "meta" and its parentheses.
input:
one metaOmph(uno)
one metaAsdf(dos)
one metaPoil(tres)
output:
[ metaOmph , (uno) ]
[ metaAsdf, (dos) ]
[ metaPoil, (tres)]
The one that I currently have just gets the entire line if it starts with "meta". so I have the entire "one meta<>" if it's a match, would it be possible do what I'm aiming for?
Edit: It's one input/line at a time.
I'd love to post what I did earlier but I closed repl.it due to my frustration. I'll keep it in mind on my next post. (quite new here)
import re
s = """one metaOmph(uno)
one metaAsdf(dos)
one metaPoil(tres)"""
print(re.findall(".+(meta\w+)(\(\w+\))", s))
Outputs:
[('metaOmph', '(uno)'), ('metaAsdf', '(dos)'), ('metaPoil', '(tres)')]
re.findall() approach with valid regex pattern:
import re
s = '''
one metaOmph(uno)
one metaAsdf(dos)
one metaPoil(tres)
'''
result = re.findall(r'\b(meta\w+)(\([^()]+\))', s)
print(result)
The output:
[('metaOmph', '(uno)'), ('metaAsdf', '(dos)'), ('metaPoil', '(tres)')]
If you are going to pass a multiline string, it would seem simple to use the module level re.findall function.
text = '''one metaOmph(uno)
one metaAsdf(dos)
one metaPoil(tres)'''
r = re.findall(r'\b(meta.*?)(\(.*?\))', text, re.M)
print(r)
[('metaOmph', '(uno)'), ('metaAsdf', '(dos)'), ('metaPoil', '(tres)')]
If you are going to be passing 1-line strings as input to a loop, it might make more sense to compile the pattern beforehand, using re.compile and re.search inside a function:
pat = re.compile(r'\b(meta.*?)(\(.*?\))')
def find(text):
return pat.search(text)
for text in list_of_texts: # assuming you're passing in your strings from a list, or elsewhere
m = find(text)
if m:
print(list(m.groups()))
['metaOmph', '(uno)']
['metaAsdf', '(dos)']
['metaPoil', '(tres)']
Note that m might return a match object or None depending on whether a search was found. You'll want to query the return value, otherwise you'll receive an AttributeError: 'NoneType' object has no attribute 'groups', or something along those lines.
Alternatively, if you want to append the result to a list, you might instead use:
r_list = []
for text in list_of_texts:
m = find(text)
if m:
r_list.append(list(m.groups()))
print(r_list)
[['metaOmph', '(uno)'], ['metaAsdf', '(dos)'], ['metaPoil', '(tres)']]
Regex Details
\b # word boundary (thought to add this in thanks to Roman's answer)
(
meta # literal 'meta'
.*? # non-greedy matchall
)
(
\( # literal opening brace (escaped)
.*?
\) # literal closing brace (escaped)
)

Python split before a certain character

I have following string:
BUCKET1:/dir1/dir2/BUCKET1:/dir3/dir4/BUCKET2:/dir5/dir6
I am trying to split it in a way I would get back the following dict / other data structure:
BUCKET1 -> /dir1/dir2/, BUCKET1 -> /dir3/dir4/, BUCKET2 -> /dir5/dir6/
I can somehow split it if I only have one BUCKET, not multiple, like this:
res.split(res.split(':', 1)[0].replace('.', '').upper()) -> it's not perfect
Input: ADRIAN:/dir1/dir11/DANIEL:/dir2/ADI_BUCKET:/dir3/CULEA:/dir4/ADRIAN:/dir5/ADRIAN:/dir6/
Output: [(ADRIAN, /dir1/dir11), (DANIEL, /dir2/), (CULEA, /dir3/), (ADRIAN, /dir5/), (ADRIAN, /dir6/)
As per Wiktor Stribiżew comments, the following regex does the job:
r"(BUCKET1|BUCKET2):(.*?)(?=(?:BUCKET1|BUCKET2)|$)"
If you're experienced, I'd recommend learning Regex just as the others have suggested. However, if you're looking for an alternative, here's a way of doing such without Regex. It also produces the output you're looking for.
string = input("Enter:") #Put your own input here.
tempList = string.replace("BUCKET",':').split(":")
outputList = []
for i in range(1,len(tempList)-1,2):
someTuple = ("BUCKET"+tempList[i],tempList[i+1])
outputList.append(someTuple)
print(outputList) #Put your own output here.
This will produce:
[('BUCKET1', '/dir1/dir2/'), ('BUCKET1', '/dir3/dir4/'), ('BUCKET2', '/dir5/dir6')]
This code is hopefully easier to understand and manipulate if you're unfamiliar with Regex, although I'd still personally recommend Regex to solve this if you're familiar with how to use it.
Use re.findall() function:
s = "ADRIAN:/dir1/dir11/DANIEL:/dir2/ADI_BUCKET:/dir3/CULEA:/dir4/ADRIAN:/dir5/ADRIAN:/dir6/"
result = re.findall(r'(\w+):([^:]+\/)', s)
print(result)
The output:
[('ADRIAN', '/dir1/dir11/'), ('DANIEL', '/dir2/'), ('ADI_BUCKET', '/dir3/'), ('CULEA', '/dir4/'), ('ADRIAN', '/dir5/'), ('ADRIAN', '/dir6/')]
Use regex instead?
impore re
test = 'BUCKET1:/dir1/dir2/BUCKET1:/dir3/dir4/BUCKET2:/dir5/dir6'
output = re.findall(r'(?P<bucket>[A-Z0-9]+):(?P<path>[/a-z0-9]+)', test)
print(output)
Which gives
[('BUCKET1', '/dir1/dir2/'), ('BUCKET1', '/dir3/dir4/'), ('BUCKET2', '/dir5/dir6')]
It appears you have a list of predefined "buckets" that you want to use as boundaries for the records inside the string.
That means, the easiest way to match these key-value pairs is by matching one of the buckets, then a colon and then any chars not starting a sequence of chars equal to those bucket names.
You may use
r"(BUCKET1|BUCKET2):(.*?)(?=(?:BUCKET1|BUCKET2)|$)"
Compile with re.S / re.DOTALL if your values span across multiple lines. See the regex demo.
Details:
(BUCKET1|BUCKET2) - capture group one that matches and stores in .group(1) any of the bucket names
: - a colon
(.*?) - any 0+ chars, as few as possible (as *? is a lazy quantifier), up to the first occurrence of (but not inlcuding)...
(?=(?:BUCKET1|BUCKET2)|$) - any of the bucket names or end of string.
Build it dynamically while escaping bucket names (just to play it safe in case those names contain * or + or other special chars):
import re
buckets = ['BUCKET1','BUCKET2']
rx = r"({0}):(.*?)(?=(?:{0})|$)".format("|".join([re.escape(bucket) for bucket in buckets]))
print(rx)
s = "BUCKET1:/dir1/dir2/BUCKET1:/dir3/dir4/BUCKET2:/dir5/dir6"
print(re.findall(rx, s))
# => (BUCKET1|BUCKET2):(.*?)(?=(?:BUCKET1|BUCKET2)|$)
[('BUCKET1', '/dir1/dir2/'), ('BUCKET1', '/dir3/dir4/'), ('BUCKET2', '/dir5/dir6')]
See the online Python demo.

Regex issue in python

I have a regex "value=4020a345-f646-4984-a848-3f7f5cb51f21"
if re.search( "value=\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*", x ):
x = re.search( "value=\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*", x )
m = x.group(1)
m only gives me 4020a345, not sure why it does not give me the entire "4020a345-f646-4984-a848-3f7f5cb51f21"
Can anyone tell me what i am doing wrong?
try out this regex, looks like you are trying to match a GUID
value=[0-9a-f]{8}-[0-9a-f]{4}-[0-9a-f]{4}-[0-9a-f]{4}-[0-9a-f]{12}
This should match what you want, if all the strings are of the form you've shown:
value=((\w*\d*\-?)*)
You can also use this website to validate your regular expressions:
http://regex101.com/
The below regex works as you expect.
value=([\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*\-\w*|\d*]+)
You are trying to match on some hex numbers, that is why this regex is more correct than using [\w\d]
pattern = "value=([0-9a-fA-F]{8}-([0-9a-fA-F]{4}-){3}[0-9a-fA-F]{12})"
data = "value=4020a345-f646-4984-a848-3f7f5cb51f21"
res = re.search(pattern, data)
print(res.group(1))
If you dont care about the regex safety, aka checking that it is correct hex, there is no reason not to use simple string manipulation like shown below.
>>> data = "value=4020a345-f646-4984-a848-3f7f5cb51f21"
>>> print(data[7:])
020a345-f646-4984-a848-3f7f5cb51f21
>>> # or maybe
...
>>> print(data[7:].replace('-',''))
020a345f6464984a8483f7f5cb51f21
You can get the subparts of the value as a list
txt = "value=4020a345-f646-4984-a848-3f7f5cb51f21"
parts = re.findall('\w+', txt)[1:]
parts is ['4020a345', 'f646', '4984', 'a848', '3f7f5cb51f21']
if you really want the entire string
full = "-".join(parts)
A simple way
full = re.findall("[\w-]+", txt)[-1]
full is 4020a345-f646-4984-a848-3f7f5cb51f21
value=([\w\d]*\-[\w\d]*\-[\w\d]*\-[\w\d]*\-[\w\d]*)
Try this.Grab the capture.Your regex was not giving the whole as you had used | operator.So if regex on left side of | get satisfied it will not try the latter part.
See demo.
http://regex101.com/r/hQ1rP0/45

What should I use the Non-greedy match in this case

Assume I have a string which includes some data fields that are separated by "|", like
|1|2|3|4|5|6|7|8|
My purpose is to get the 8th field. This is what I'm doing:
pattern = re.compile(r'^\s+(\|.*?\|){8}')
match = pattern.match(test_line)
if match:
print:match.group(8)
But looks like it can not match. I know in this case I need to use ? for non-greedy match, but why I can not get the 8th field?
Thanks
Regex might be complicating this problem rather than simplifying it. A simple way to get an eighth item from a | delimited string is using split():
a = '|here|is|some|data|separated|by|bars|hooray!|'
print a.split('|')[8]
RETURNS
hooray!
Using regex, one way to get it would be:
import re
a = '|here|is|some|data|separated|by|bars|hooray!|'
pattern = re.compile(r'([^\|]+)')
match = pattern.findall(a)
print match[7]
RETURNS
hooray!

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