I have a string in python and I want to replace multiple consecutive repeating character into 1.
For example:
st = "UUUURRGGGEENNTTT"
print(st.replace(r'(\w){2,}',r'\1'))
But this command doesn't seems to be working, please can anybody help in finding what's wrong with this command?
There is one more way to solve this but wanted to understand why the above command fails and is there any way to correct it:
print(re.sub(r"([a-z])\1+",r"\1",st)) -- print URGENT
you need to use regex.
so you can do this:
import re
re.sub(r'[^\w\s]|(.)(?=\1)', '', 'UUURRRUU')
the result is UR.
this is a snapshot of what I have got:
for this regex: (.)(?=.*\1)
(.) means: match any char except new lines (line breaks)
?=. means: lookahead every char except new line (.)
* means: match a preceding token
\1 means: to mach the result of captured group, which is the U or R ...
then replace all matches with ''
also you can check this:
lookahead
also check this tool I solve my regex using it,
it describe everything and you can learn a lot from it:
regexer
The reason for why your code does not work is because str.replace does not support regex, you can only replace a substring with another string. You will need to use the re module if you want to replace by matching a regex pattern.
Secondly, your regex pattern is also incorrect, (\w){2,} will match any characters that occurs 2 or more times (doesn’t have to be the same character though), so it will not work. You will need to do something like this:
import re
st = "UUUURRGGGEENNTTT"
print(re.sub(r'(\w)\1+',r'\1', st)))
# URGENT
Now this will only match the same character 2 or more times.
An alternative, “unique” solution to this is that you can use the unique_justseen recipe that itertools provides:
from itertools import groupby
from operator import itemgetter
st = "UUUURRGGGEENNTTT"
new ="".join(map(next, map(itemgetter(1), groupby(st))))
print(new)
# URGENT
string.replace(s, old, new[, maxreplace]) only does substring replacement:
>>> '(\w){2,}'.replace(r'(\w){2,}',r'\1')
'\\1'
That's why it fails and it can't work with regex expression so no way to correct the first command.
Related
I have the following path stored as a python string 'C:\ABC\DEF\GHI\App\Module\feature\src' and I would like to extract the word Module that is located between words \App\ and \feature\ in the path name. Note that there are file separators '\' in between which ought not to be extracted, but only the string Module has to be extracted.
I had the few ideas on how to do it:
Write a RegEx that matches a string between \App\ and \feature\
Write a RegEx that matches a string after \App\ --> App\\[A-Za-z0-9]*\\, and then split that matched string in order to find the Module.
I think the 1st solution is better, but that unfortunately it goes over my RegEx knowledge and I am not sure how to do it.
I would much appreciate any help.
Thank you in advance!
The regex you want is:
(?<=\\App\\).*?(?=\\feature\\)
Explanation of the regex:
(?<=behind)rest matches all instances of rest if there is behind immediately before it. It's called a positive lookbehind
rest(?=ahead) matches all instances of rest where there is ahead immediately after it. This is a positive lookahead.
\ is a reserved character in regex patterns, so to use them as part of the pattern itself, we have to escape it; hence, \\
.* matches any character, zero or more times.
? specifies that the match is not greedy (so we are implicitly assuming here that \feature\ only shows up once after \App\).
The pattern in general also assumes that there are no \ characters between \App\ and \feature\.
The full code would be something like:
str = 'C:\\ABC\\DEF\\GHI\\App\\Module\\feature\\src'
start = '\\App\\'
end = '\\feature\\'
pattern = rf"(?<=\{start}\).*?(?=\{end}\)"
print(pattern) # (?<=\\App\\).*?(?=\\feature\\)
print(re.search(pattern, str)[0]) # Module
A link on regex lookarounds that may be helpful: https://www.regular-expressions.info/lookaround.html
We can do that by str.find somethings like
str = 'C:\\ABC\\DEF\\GHI\\App\\Module\\feature\\src'
import re
start = '\\App\\'
end = '\\feature\\'
print( (str[str.find(start)+len(start):str.rfind(end)]))
print("\n")
output
Module
Your are looking for groups. With some small modificatians you can extract only the part between App and Feature.
(?:App\\\\)([A-Za-z0-9]*)(?:\\\\feature)
The brackets ( ) define a Match group which you can get by match.group(1). Using (?:foo) defines a non-matching group, e.g. one that is not included in your result. Try the expression here: https://regex101.com/r/24mkLO/1
I am new to regexes.
I have the following string : \n(941)\n364\nShackle\n(941)\nRivet\n105\nTop
Out of this string, I want to extract Rivet and I already have (941) as a string in a variable.
My thought process was like this:
Find all the (941)s
filter the results by checking if the string after (941) is followed by \n, followed by a word, and ending with \n
I made a regex for the 2nd part: \n[\w\s\'\d\-\/\.]+$\n.
The problem I am facing is that because of the parenthesis in (941) the regex is taking 941 as a group. In the 3rd step the regex may be wrong, which I can fix later, but 1st I needed help in finding the 2nd (941) so then I can apply the 3rd step on that.
PS.
I know I can use python string methods like find and then loop over the searches, but I wanted to see if this can be done directly using regex only.
I have tried the following regex: (?:...), (941){1} and the make regex literal character \ like this \(941\) with no useful results. Maybe I am using them wrong.
Just wanted to know if it is possible to be done using regex. Though it might be useful for others too or a good share for future viewers.
Thanks!
Assuming:
You want to avoid matching only digits;
Want to match a substring made of word-characters (thus including possible digits);
Try to escape the variable and use it in the regular expression through f-string:
import re
s = '\n(941)\n364\nShackle\n(941)\nRivet\n105\nTop'
var1 = '(941)'
var2 = re.escape(var1)
m = re.findall(fr'{var2}\n(?!\d+\n)(\w+)', s)[0]
print(m)
Prints:
Rivet
If you have text in a variable that should be matched exactly, use re.escape() to escape it when substituting into the regexp.
s = '\n(941)\n364\nShackle\n(941)\nRivet\n105\nTop'
num = '(941)'
re.findall(rf'(?<=\n{re.escape(num)}\n)[\w\s\'\d\-\/\.]+(?=\n)', s)
This puts (941)\n in a lookbehind, so it's not included in the match. This avoids a problem with the \n at the end of one match overlapping with the \n at the beginning of the next.
Lets say I have an arrays of strings.I want to find all the string which contain the following
substring , charachter digit digit digit charachter (CDDDC will be the pattern). For instance the format would be as following:
H554L
K007K
Is there any fast string expression matching to find such occurrences ?
Things like this are the field of "regex". Regex is made for pattern matching. It of itself is a broad ttopic too much to explain here (check regexbuddy or another site).
python has a regex compiler build in, under the re (as well as regex module). A simple solution would hence be:
word for word in somelist if re.search(r"[a-zA-Z]\d{3}[a-zA-Z]", word)
Which iterates over somelist, and selects anything that matches (completely) a character in one of the two "ranges", followed by 3 digits, followed by a character in the range.
A noted as in the comments: re.search will match (find) any item which has a "part" of that item matching the "pattern". So it will match a123b as well as abc b123cd. If you wish to make sure that the full "word" in the array matches the substring use re.fullmatch instead.
Fullmatch will match a123b but not abc b123cd and not ab123cd
Try this example with this regex:
regex: (?i)[A-Z]\d\d\d[A-Z]
import re
xx = ['aeeea','5eeae','H554L','juan','K007K']
for i in xx:
r1 = re.findall(r"(?i)[A-Z]\d\d\d[A-Z]", i)
print (', '.join(r1)
)
Run the example online
I am using Python 2.7 and have a question with regards to regular expressions. My string would be something like this...
"SecurityGroup:Pub HDP SG"
"SecurityGroup:Group-Name"
"SecurityGroup:TestName"
My regular expression looks something like below
[^S^e^c^r^i^t^y^G^r^o^u^p^:].*
The above seems to work but I have the feeling it is not very efficient and also if the string has the word "group" in it, that will fail as well...
What I am looking for is the output should find anything after the colon (:). I also thought I can do something like using group 2 as my match... but the problem with that is, if there are spaces in the name then I won't be able to get the correct name.
(SecurityGroup):(\w{1,})
Why not just do
security_string.split(':')[1]
To grab the second part of the String after the colon?
You could use lookbehind:
pattern = re.compile(r"(?<=SecurityGroup:)(.*)")
matches = re.findall(pattern, your_string)
Breaking it down:
(?<= # positive lookbehind. Matches things preceded by the following group
SecurityGroup: # pattern you want your matches preceded by
) # end positive lookbehind
( # start matching group
.* # any number of characters
) # end matching group
When tested on the string "something something SecurityGroup:stuff and stuff" it returns matches = ['stuff and stuff'].
Edit:
As mentioned in a comment, pattern = re.compile(r"SecurityGroup:(.*)") accomplishes the same thing. In this case you are matching the string "SecurityGroup:" followed by anything, but only returning the stuff that follows. This is probably more clear than my original example using lookbehind.
Maybe this:
([^:"]+[^\s](?="))
Regex live here.
I have this string:
'Is?"they'
I want to find the question mark (?) in the string, and put it at the end of the string. The output should look like this:
'Is"they?'
I am using the following regular expression in python 2.7. I don't know why my regex is not working.
import re
regs = re.sub('(\w*)(\?)(\w*)', '\\1\\3\\2', 'Is?"they')
print regs
Is?"they # this is the output of my regex.
Your regex doesn't match because " is not in the \w character class. You would need to change it to something like:
regs = re.sub('(\w*)(\?)([^"\w]*)', '\\1\\3\\2', 'Is?"they')
As shown here, " is not captured by \w. Hence, it would probably be best to just use a .:
>>> import re
>>> re.sub("(.*)(\?)(.*)", r'\1\3\2', 'Is?"they')
'Is"they?'
>>>
. captures anything/everything in Regex (except newlines).
Also, you'll notice that I used a raw-string for the second argument of re.sub. Doing so is cleaner than having all those backslashes.