I want to get the value of name from fstr using RegEx in Python. I tried as below, but couldn't find the intended result.
Any help will be highly appreciaaed.
fstr = "MCode=1,FCode=1,Name=XYZ,Extra=whatever" #",Extra=whatever" this portion is optional
myobj = re.search( r'(.*?),Name(.*?),*(.*)', fstr, re.M|re.I)
print(myobj.group(2))
You may not believe, but the actual problem was ,*, in your regular expression. It makes matching , optional. So, the second capturing group in your regex matches nothing (.*? means match between zero to unlimited and match lazily) and it checks the next item ,*, it also means match , zero or more times. So it matches zero times and the last capturing groups matches the rest of the string.
If you want to fix your RegEx, you can simply remove the * after the comma, like this
myobj = re.search( r'(.*?),Name(.*?),(.*)', fstr, re.I)
print(myobj.group(2))
# =XYZ
Online RegEx demo (with the mistake)
Online RegEx demo (after fixing it)
Debuggex Demo
But as the other answer shows, you don't have to create additional capture groups.
BTW, I like to use RegEx only when it is particularly needed. In this case, I would have solved it, without RegEx, like this
fstr = "MCode=1,FCode=1,Name=XYZ,Extra=whatever"
d = dict(item.split("=") for item in fstr.split(","))
# {'FCode': '1', 'Extra': 'whatever', 'Name': 'XYZ', 'MCode': '1'}
Now that I have all the information, I can access them like this
print d["Name"]
# XYZ
Simple, huh? :-)
Edit: If you want to use the same regex for one million records, we can slightly improve the performance by precompiling the RegEx, like this
import re
pattern = re.compile(r"Name=([^,]+)", re.I)
match = re.search(pattern, data)
if match:
match.group(1)
You can do it as follows:
import re
fstr = "MCode=1,FCode=1,Name=XYZ,Extra=whatever"
myobj = re.search( r'Name=([^,]+)', fstr, re.M|re.I)
>>> print myobj.group(1)
XYZ
try it
rule = re.compile(r"Name=(?P<Name>\w*),")
res = rule.search(fstr)
res.group("Name")
Related
I currently investigate a problem that I want to replace something in a string.
For example. I have the following string:
'123.49, 19.30, 02\n'
I only want the first two numbers like '123.49, 19.30'. The split function is not possible, because a I have a lot of data and some with and some without the last number.
I tried something like this:
import re as regex
#result = regex.match(', (.*)\n', string)
result = re.search(', (.*)\\n', string)
print(result.group(1))
This is not working finde. Can someone help me?
Thanks in advance
You could do something like this:
reg=r'(\d+\.\d+), (\d+\.\d+).*'
if(re.search(reg, your_text)):
match = re.search(reg, your_text)
first_num = match.group(1)
second_num = match.group(2)
Alternatively, also adding the ^ sign at the beginning, making sure to always only take the first two.
import re
string = '123.49, 19.30, 02\n'
pattern = re.compile('^(\d*.?\d*), (\d*.?\d*)')
result = re.findall(pattern, string)
result
Output:
[('123.49', '19.30')]
In the code you are using import re as regex. If you do that, you would have to use regex.search instead or re.search.
But in this case you can just use re.
If you use , (.*) you would capture all after the first occurrence of , and you are not taking digits into account.
If you want the first 2 numbers as stated in the question '123.49, 19.30' separated by comma's you can match them without using capture groups:
\b\d+\.\d+,\s*\d+\.\d+\b
Or matching 1 or more repetitions preceded by a comma:
\b\d+\.\d+(?:,\s*\d+\.\d+)+\b
regex demo | Python demo
As re.search can also return None, you can first check if there is a result (no need to run re.search twice)
import re
regex = r"\b\d+\.\d+(?:,\s*\d+\.\d+)+\b"
s = "123.49, 19.30, 02"
match = re.search(regex, s)
if match:
print(match.group())
Output
123.49, 19.30
Given:
lst = ['(abc): my name is ?123']
I'm trying to return everything from ': ' till the end of lst[0], for that I tried a regex expression:
result = re.search(r': (.*?)', lst[0]).group(1)
It returns an empty string.
How can this be done using regex correctly?
Expected output :
'my name is ?123'
Resources used : Regex wiki
The issue is that you made your .* lazy by placing ? at the end. Lazy means match as little as possible, for a valid match. In this case, since your pattern does not have anything to match beyond the (.*?), the regex engine is matching empty string. Just use (.*), the non lazy version, and it will work.
lst = ['(abc): my name is ?123']
result = re.search(r': (.*)', lst[0]).group(1)
print(result)
This prints:
my name is ?123
I am using Python and would like to match all the words after test till a period (full-stop) or space is encountered.
text = "test : match this."
At the moment, I am using :
import re
re.match('(?<=test :).*',text)
The above code doesn't match anything. I need match this as my output.
Everything after test, including test
test.*
Everything after test, without test
(?<=test).*
Example here on regexr.com
You need to use re.search since re.match tries to match from the beging of the string. To match until a space or period is encountered.
re.search(r'(?<=test :)[^.\s]*',text)
To match all the chars until a period is encountered,
re.search(r'(?<=test :)[^.]*',text)
In a general case, as the title mentions, you may capture with (.*) pattern any 0 or more chars other than newline after any pattern(s) you want:
import re
p = re.compile(r'test\s*:\s*(.*)')
s = "test : match this."
m = p.search(s) # Run a regex search anywhere inside a string
if m: # If there is a match
print(m.group(1)) # Print Group 1 value
If you want . to match across multiple lines, compile the regex with re.DOTALL or re.S flag (or add (?s) before the pattern):
p = re.compile(r'test\s*:\s*(.*)', re.DOTALL)
p = re.compile(r'(?s)test\s*:\s*(.*)')
However, it will retrun match this.. See also a regex demo.
You can add \. pattern after (.*) to make the regex engine stop before the last . on that line:
test\s*:\s*(.*)\.
Watch out for re.match() since it will only look for a match at the beginning of the string (Avinash aleady pointed that out, but it is a very important note!)
See the regex demo and a sample Python code snippet:
import re
p = re.compile(r'test\s*:\s*(.*)\.')
s = "test : match this."
m = p.search(s) # Run a regex search anywhere inside a string
if m: # If there is a match
print(m.group(1)) # Print Group 1 value
If you want to make sure test is matched as a whole word, add \b before it (do not remove the r prefix from the string literal, or '\b' will match a BACKSPACE char!) - r'\btest\s*:\s*(.*)\.'.
I don't see why you want to use regex if you're just getting a subset from a string.
This works the same way:
if line.startswith('test:'):
print(line[5:line.find('.')])
example:
>>> line = "test: match this."
>>> print(line[5:line.find('.')])
match this
Regex is slow, it is awkward to design, and difficult to debug. There are definitely occassions to use it, but if you just want to extract the text between test: and ., then I don't think is one of those occasions.
See: https://softwareengineering.stackexchange.com/questions/113237/when-you-should-not-use-regular-expressions
For more flexibility (for example if you are looping through a list of strings you want to find at the beginning of a string and then index out) replace 5 (the length of 'test:') in the index with len(str_you_looked_for).
I'm working on a Flask API, which takes the following regex as an endpoint:
([0-9]*)((OK)|(BACK)|(X))*
That means I'm expecting a series of numbers, and the OK, BACK, X keywords multiple times in succession after the numbers.
I want to split this regex and do different stuff depending which capture groups were present.
My approach was the following:
endp = endp.encode('ASCII', 'ignore')
match = re.search(r"([0-9]*)", str(endp), re.I)
if match:
n = match.groups()
logging.info('nums: ' + str(n[0]))
match = re.search(r"((OK)|(BACK)|(X))*", str(endp), re.I)
if match:
s1 = match.groups()
for i in s1:
logging.info('str: ' + str(i[0]))
Using the /12OK endpoint, getting the numbers works just fine, but for some reason capturing the rest of the keywords are unsuccessful. I tried reducing the second capture group to only
match = re.search(r"(OK)*", str(endp), re.I)
I constantly find the following in s1 (using the reduced regex):
(None,)
originally (with the rest of the keywords):
(None, None, None, None)
Which I suppose means the regex pattern does not match anything in my endp string (why does it have 4 Nones? 1 for each keyword, but what the 4th is there for?). I validated my endpoint (the regex against the same string too) with a regex validator, it seems fine to me. I understand that re.match is supposed to get matches from the beginning, therefore I used the re.search method, as the documentation points out it's supposed to match anywhere in the string.
What am I missing here? Please advise, I'm a beginner in the python world.
Indeed it is a bit surprising that searching with * returns `None:
>>> re.search("(OK|BACK|X)*", u'/12OK').groups()
(None,)
But it's "correct", since * matches zero or more, and any pattern matches zero times in any string, that's why you see None. Searching with + somewhat solves it:
>>> re.search("(OK|BACK|X)+", u'/12OK').groups()
('OK',)
But now, searching with this pattern in /12OKOK still only finds one match because + means one or more, and it matched one time at the first OK. To find all occurrences you need to use re.findall:
>>> re.findall("(OK|BACK|X)", u'/12OKOK')
['OK', 'OK']
With those findings, your code would look as follows: (note that you don't need to write i[0] since i is already a string, unless you want to log only the first char of the string):
import re
endp = endp.encode('ASCII', 'ignore')
match = re.search(r"([0-9]+)", str(endp))
if match:
n = match.groups()
logging.info('nums: ' + str(n))
match = re.findall(r"(OK|BACK|X)", str(endp), re.I)
for i in match:
logging.info('str: ' + str(i))
If you want to match at least ONE of the groups, use + instead of *.
>>> endp = '/12OK'
>>> match = re.search(r"((OK)|(BACK)|(X))+", str(endp), re.I)
>>> if match:
... s1 = match.groups()
... for i in s1:
... print s1
...
('OK', 'OK', None, None)
>>> endp = '/12X'
>>> match = re.search(r"((OK)|(BACK)|(X))+", str(endp), re.I)
>>> match.groups()
('X', None, None, 'X')
Notice that you have 4 matching groups in your expression, one for each pair of parentheses. The first match is the outer parenthesis and the second one is the first of the nested groups. In the second example, you still get the first match for the outer parenthesis and then the last one is the third of the nested ones.
"((OK)|(BACK)|(X))*" will search for OK or BACK or X, 0 or more times. Note that the * means 0 or more, not more than 0. The above expression should have a + at the end not * as + means 1 or more.
I think you're having two different issues, and their intersection is causing more confusion than either of them would cause on their own.
The first issue is that you're using repeated groups. Python's re library is not able to capture multiple matches when a group is repeated. Matching with a pattern like (X)+ against 'XXXX' will only capture a single 'X' in the first group even though the whole string will be matched. The regex library (which is not part of the standard library) can do multiple captures, though I'm not sure of the exact commands required.
The second issue is using the * repetition operator in your pattern. The pattern you show at the top of the question will match on an empty string. Obviously, none of the gropus will capture anything in that situation (which may be why you're seeing a lot of None entries in your results). You probably need to modify your pattern so that it requires some minimal amount of valid text to count as a match. Using + instead of * might be one solution, but it's not clear to me exactly what you want to match against so I can't suggest a specific pattern.
I have some strings (in fact they are lines read from a file). The lines are just copied to some other file, but some of them are "special" and need a different treatment.
These lines have the following syntax:
someText[SUBSTRING1=SUBSTRING2]someMoreText
So, what I want is: When I have a line on which this "mask" can be applied, I want to store SUBSTRING1 and SUBSTRING2 into variables. The braces and the = shall be stripped.
I guess this consists of several tasks:
Decide if a line contains this mask
If yes, get the positions of the substrings
Extract the substrings
I'm sure this is a easy task with regex, however, I'm not used to it. I can write a huge monster function using string manipulation, but I guess this is not the "Python Way" to do this.
Any suggestions on this?
re.search() returns None if it doesn't find a match. \w matches an alphanumeric, + means 1 or more. Parenthesis indicate the capturing groups.
s = """
bla bla
someText[SUBSTRING1=SUBSTRING2]someMoreText"""
results = {}
for line_num, line in enumerate(s.split('\n')):
m = re.search(r'\[(\w+)=(\w+)\]', line)
if m:
results.update({line_num: {'first': m.group(0), 'second': m.group(1)}})
print(results)
^[^\[\]]*\[([^\]\[=]*)=([^\]\[=]*)\][^\]\[]*$
You can try this.Group 1and Group 2 has the two string you want.See demo.
https://regex101.com/r/pT4tM5/26
import re
p = re.compile(r'^[^\[\]]*\[([^\]\[=]*)=([^\]\[=]*)\][^\]\[]*$', re.MULTILINE)
test_str = "someText[SUBSTRING1=SUBSTRING2]someMoreText\nsomeText[SUBSTRING1=SUBSTRING2someMoreText\nsomeText[SUBSTRING1=SUBSTRING2]someMoreText"
re.findall(p, test_str)