Regex to exclude specific special characters, spaces and alphabets [duplicate] - python

This question already has answers here:
What special characters must be escaped in regular expressions?
(13 answers)
Closed 4 years ago.
I want a regular expression which converts this:
91009-01-28-00 Maximum (c/s)................ 1543.5
to this:
91009-01-28-00 1543.5
So basically, a regular expression that escapes alphabets, spaces, forward slashes and brackets.
I have written the following python code so far:
with open('lcstats.txt', 'r') as lcstats_file:
with open (lcstats_full_path + '_lcstats_full.txt', "a+") as lcstats_full_file:
lcstats_full_file.write(obsid )
for line in lcstats_file.readlines():
if not re.search(r'Maximum [(c/s)]', line):
continue
line = (re.sub(**REGEX**,'',line))
lcstats_full_file.write(line)


It appears you want to have first and last part of the string. If that is the case for every line than spliting it accordingly can be helpful, as in the following code
import re
line = "91009-01-28-00 Maximum (c/s) ................ 1543.5"
line=line.split(' ')
line=line[0]+' '+ line[-1]
print(line)
Output:
91009-01-28-00 1543.5


In your code you are using search to check if you can match Maximum (c/s) and then you want to use a regex to remove that.
I think with your regex Maximum [(c/s)] you mean Maximum \(c/s\). The square brackets make it a character class and (c/s) captures c/s in a capturing group which is not required if you only want to match it.
Wat you could do is match Maximum (c/s) and match one or more times a whitespace or a comma using a character class [ .]+ and replace with an empty string.
Maximum \(c/s\)[ .]+
import re
s = "91009-01-28-00 Maximum (c/s)................ 1543.5"
print( re.sub(r"Maximum \(c/s\)[ .]+", "", s))
Demo


Try using this regex /\s[^0-9]+/ This will match from the first space followed by 1 or more not digit characters. You will need to add a space in the replacement string to keep the two bits of remaining data separate.


Regex:
((?<!\d)\D)
Match all non digits\D which is not followed by a digit \d

Related

Split a split (regex) in python

I do have got the below string and I am looking for a way to split it in order to consistently end up with the following output
'1GB 02060250396L7.067,702BE 129517720L6.633,403NL 134187650L3.824,234DE 165893440L3.111,005PL 65775644897L1.010,006DE 811506926L3.547,407AT U16235008L-830,008SE U57469158L3.001,30'
['1GB 02060250396L1.060,70',
'2BE 129517720L2.639,40',
'3NL 134187650L4.024,23',
'4DE 165893440L8.111,00',
'5PL 65775644897L3.010,00',
'6DE 811506926L3.547,40',
'7AT U16235008L-830,00',
'8SE U57469158L8.0221,30']
My current approach
re.split("([0-9][0-9][0-9][A-Z][A-Z])", input) however is also splitting my delimiter which gives and there is no other split possible than the one I am currently using in order to remain consistent. Is it possible to split my delimiter as well and assign a part of it "70" to the string in front and a part "2BE" to the following string?

Use re.findall() instead of re.split().
You want to match
a number \d, followed by
two letters [A-Z]{2}, followed by
a space \s, followed by
a bunch of characters until you encounter a comma [^,]+, followed by
two digits \d{2}
Try it at regex101
So do:
input_str = '1GB 02060250396L7.067,702BE 129517720L6.633,403NL 134187650L3.824,234DE 165893440L3.111,005PL 65775644897L1.010,006DE 811506926L3.547,407AT U16235008L-830,008SE U57469158L3.001,30'
re.findall(r"\d[A-Z]{2}\s[^,]+,\d{2}", input_str)
Which gives
['1GB 02060250396L7.067,70',
'2BE 129517720L6.633,40',
'3NL 134187650L3.824,23',
'4DE 165893440L3.111,00',
'5PL 65775644897L1.010,00',
'6DE 811506926L3.547,40',
'7AT U16235008L-830,00',
'8SE U57469158L3.001,30']
Alternatively, if you don't want to be so specific with your pattern, you could simply use the regex
[^,]+,\d{2} Try it at regex101
This will match as many of any character except a comma, then a single comma, then two digits.
re.findall(r"[^,]+,\d{2}", input_str)
# Output:
['1GB 02060250396L7.067,70',
'2BE 129517720L6.633,40',
'3NL 134187650L3.824,23',
'4DE 165893440L3.111,00',
'5PL 65775644897L1.010,00',
'6DE 811506926L3.547,40',
'7AT U16235008L-830,00',
'8SE U57469158L3.001,30']

Is it possible to split my delimiter as well and assign a part of it "70" to the string in front and a part "2BE" to the following string?
If you must use re.split AT ANY PRICE then you might exploit zero-length assertion for this task following way
import re
text = '1GB 02060250396L7.067,702BE 129517720L6.633,403NL 134187650L3.824,234DE 165893440L3.111,005PL 65775644897L1.010,006DE 811506926L3.547,407AT U16235008L-830,008SE U57469158L3.001,30'
parts = re.split(r'(?<=,[0-9][0-9])', text)
print(parts)
output
['1GB 02060250396L7.067,70', '2BE 129517720L6.633,40', '3NL 134187650L3.824,23', '4DE 165893440L3.111,00', '5PL 65775644897L1.010,00', '6DE 811506926L3.547,40', '7AT U16235008L-830,00', '8SE U57469158L3.001,30', '']
Explanation: This particular one is positive lookbehind, it does find zero-length substring preceded by , digit digit. Note that parts has superfluous empty str at end.

how to write a regular expression to match a small part of a repeating pattern?

I have the following pattern to match :
(10,'more random stuff 21325','random stuff','2014-10-26 04:50:23','','uca-default-u-kn','page')
For some context, it's part of a larger file , which contains many similar patterns separated by commas :
(10,'more random stuff 21325','random stuff','2014-10-26 04:50:23','','uca-default-u-kn','page'),
(11,'more random stuff 1nyny5','random stuff','2014-10-26 04:50:23','','uca-default-u-kn','subcat'),
(14,'more random stuff 21dd5','random stuff','2014-10-26 04:50:23','','uca-default-u-kn','page')
my goal is to ditch all patterns ending with 'page' and to keep the rest. For that, I'm trying to use
regular expressions to identify those patterns. Here is the one I come out with for now :
"\(.*?,\'page\'\)"
However, it's not working as expected.
In the following python code, I use this regex, and replace every match with an empty string :
import re
txt = "(10,'Redirects_from_moves','*..2NN:,#2.FBHRP:D6ܽ�','2014-10-26 04:50:23','','uca-default-u-kn','page'),"
txt += "(11,'Redirects_with_old_history','*..2NN:,#2.FBHRP:D6ܽ�','2010-08-26 22:38:36','','uca-default-u-kn','page'),"
txt += "(12,'Unprintworthy_redirects','*..2NN:,#2.FBHRP:D6ܽ�','2010-08-26 22:38:36','','uca-default-u-kn','subcat'),"
txt += "(13,'Anarchism','random_stuff','2020-01-23 13:27:44',' ','uca-default-u-kn','page'),"
txt += "(14,'Anti-capitalism','random_stuff','2020-01-23 13:27:44','','uca-default-u-kn','subcat'),"
txt += "(15,'Anti-fascism','*D*L.8:NB\r�','2020-01-23 13:27:44','','uca-default-u-kn','subcat'),"
txt += "(16,'Articles_containing_French-language_text','*D*L.8:NB\r�','2020-01-23 13:27:44','','uca-default-u-kn','page'),"
txt += "(17,'Articles_containing_French-language_text','*D*L.8:NB\r�','2020-01-23 13:27:44','','uca-default-u-kn','page')"
new_txt = re.sub("\(.*?,\'page\'\)", "",txt)
I was expecting that new_text would contains all patterns ending with 'subcat', and remove all
patterns ending with 'page', however, I obtain :
new_txt = ,,,,
What's happening here ? How can I change my regex to obtain the desired result ?

We might be tempted to do a regex replacement here, but that would basically always leave open edge cases, as #Wiktor has correctly pointed out in a comment below. Instead, a more foolproof approach is to use re.findall and simply extract every tuple with does not end in 'page'. Here is an example:
parts = re.findall(r"\(\d+,'[^']*?'(?:,'[^']*?'){4},'(?!page')[^']*?'\),?", txt)
print(''.join(parts))
This prints:
(12,'Unprintworthy_redirects','*..2NN:,#2.FBHRP:D6ܽ�','2010-08-26 22:38:36','','uca-default-u-kn','subcat'),(14,'Anti-capitalism','random_stuff','2020-01-23 13:27:44','','uca-default-u-kn','subcat'),(15,'Anti-fascism','DL.8:NB�','2020-01-23 13:27:44','','uca-default-u-kn','subcat'),
The regex pattern used above just matches a leading number, followed by 5 singly quoted terms, and then a sixth singly quoted term which is not 'page'. Then, we string join the tuples in the list output to form a string.

What happens is that you concatenate the string, then then remove all until the first occurrence of ,'page') leaving only the trailing comma's.
Another workaround might be using a list of the strings, and join them with a newline instead of concatenating them.
Then use your pattern matching an optional comma and newline at the end to remove the line, leaving the ones that end with subcat
import re
lines = [
"(10,'Redirects_from_moves','*..2NN:,#2.FBHRP:D6ܽ�','2014-10-26 04:50:23','','uca-default-u-kn','page'),",
"(11,'Redirects_with_old_history','*..2NN:,#2.FBHRP:D6ܽ�','2010-08-26 22:38:36','','uca-default-u-kn','page'),",
"(12,'Unprintworthy_redirects','*..2NN:,#2.FBHRP:D6ܽ�','2010-08-26 22:38:36','','uca-default-u-kn','subcat'),",
"(13,'Anarchism','random_stuff','2020-01-23 13:27:44',' ','uca-default-u-kn','page'),",
"(14,'Anti-capitalism','random_stuff','2020-01-23 13:27:44','','uca-default-u-kn','subcat'),",
"(15,'Anti-fascism','*D*L.8:NB\r�','2020-01-23 13:27:44','','uca-default-u-kn','subcat'),",
"(16,'Articles_containing_French-language_text','*D*L.8:NB\r�','2020-01-23 13:27:44','','uca-default-u-kn','page'),",
"(17,'Articles_containing_French-language_text','*D*L.8:NB\r�','2020-01-23 13:27:44','','uca-default-u-kn','page')"
]
new_txt = re.sub("\(.*,'page'\)(?:,\n)?", "", '\n'.join(lines))
print(new_txt)
Output
(12,'Unprintworthy_redirects','*..2NN:,#2.FBHRP:D6ܽ�','2010-08-26 22:38:36','','uca-default-u-kn','subcat'),
(14,'Anti-capitalism','random_stuff','2020-01-23 13:27:44','','uca-default-u-kn','subcat'),
�','2020-01-23 13:27:44','','uca-default-u-kn','subcat'),
Or you can use a list comprehension to keep the lines that do not match the pattern.
result = [line for line in lines if not re.match(r"\(.*,'page'\),?$", line)]
print('\n'.join(result))
Output
(12,'Unprintworthy_redirects','*..2NN:,#2.FBHRP:D6ܽ�','2010-08-26 22:38:36','','uca-default-u-kn','subcat'),
(14,'Anti-capitalism','random_stuff','2020-01-23 13:27:44','','uca-default-u-kn','subcat'),
�','2020-01-23 13:27:44','','uca-default-u-kn','subcat'),
Another option to match the parts that end with 'page') for the example data:
\(\d+,[^)]*(?:\)(?!,\s*\(\d+,)[^)]*)*,'page'\),?
The pattern matches:
\(\d+, Match ( followed by 1+ digits and a comma
[^)]* Optionally match any char except )
(?: Non capture group
\)(?!,\s*\(\d+,)[^)]* Only match a ) when not directly followed by the pattern ,\s*\(\d+, which matches the start of the parts in the example data
)* Close group and optionally repeat
,'page'\),? Match ,'page') with an optional comma
Regex demo

only keep digits after ":" in regular expression in python [duplicate]

This question already has answers here:
How to grab number after word in python
(4 answers)
Closed 2 years ago.
I want to extract the numbers for each parameter below:
import re
parameters = '''
NO2: 42602
SO2: 42401
CO: 42101
'''
The desired output should be:['42602','42401','42101']
I first tried re.findall(r'\d+',parameters), but it also returns the "2" from "NO2" and "SO2".
Then I tried re.findall(':.*',parameters), but it returns [': 42602', ': 42401', ': 42101']
If I can not rename the "NO2" to "Nitrogen dioxide", is there a way just to collect numbers on the right (after ":")?
Many thanks.

If you do not want to use capturing groups, you could use look behind.
(?<=:\s)\d+
Details:
(?<=:\s): gets string after :\s
\d+: gets digits
I also tried result on python.
import re
parameters = '''
NO2: 42602
SO2: 42401
CO: 42101
'''
result = re.findall(r'(?<=:\s)\d+',parameters)
print (result)
Result
['42602', '42401', '42101']

You can use the following regex to capture the numbers
^\s*\w+:\s(\d+)$
Hereby, ^ in the beginning asserts the position at the start of the line. \s* means that there may be 0 or more whitespaces before the content. \w+:\s matches a word character followed by ":" and space, that is "NO2: ".
Finally, (\d+) matches the following digits you want as a group. $ matches the end of the line.
To get all the matches as a list you can use
matches = re.findall(r'^\s*\w+:\s(\d+)$', parameters, re.MULTILINE)
As re.MULTILINE is specified,
the pattern character '^' matches at the beginning of the string and
at the beginning of each line.
as stated in the docs.
The result is as follows
>> print(matches)
['42602', '42401', '42101']

To put my two cents in, you could simpley use
re.findall(r'(\b\d+\b)', parameters)
See a demo on regex101.com.
If you happen to have other digits floating around somewhere in your string, be more precise with
\w+:\s*(\d+)
See another demo on regex101.com.

re.findall(r'(?<=:\s)\d+', parameters)
Should work. You can learn more about look-behind from here.

You just need to specify where in your string do you want to search for digits, you can use:
re.findall(r': (\d+)', parameters)
This tells Python to look for digits in the part of the string after ":" and the "space".

How to find values in specific format including parenthesis using regular expression in python [duplicate]

This question already has answers here:
Regular expression to return text between parenthesis
(11 answers)
Closed 2 years ago.
I have long string S, and I want to find value (numeric) in the following format "Value(**)", where ** is values I want to extract.
For example, S is "abcdef Value(34) Value(56) Value(13)", then I want to extract values 34, 56, 13 from S.
I tried to use regex as follows.
import re
regex = re.compile('\Value(.*'))
re.findall(regex, S)
But the code yields the result I did not expect.
Edit. I edited some mistakes.

You should escape the parentheses, correct the typo of Value (as opposed to Values), use a lazy repeater *? instead of *, add the missing right parenthesis, and capture what's enclosed in the escaped parentheses with a pair of parentheses:
regex = re.compile(r'Value\((.*?)\)')

Only one of your numbers follows the word 'Value', so you can extract anything inside parentheses. You also need to escape the parentheses which are special characters.
regex = re.compile('\(.*?\)')
re.findall(regex, S)
Output:
['(34)', '(56)', '(13)']

I think what you're looking for is a capturing group that can return multiple matches. This string is: (\(\d{2}\))?. \d matches an digit and {2} matches exactly 2 digits. {1,2} will match 1 or 2 digits ect. ? matches 0 to unlimited number of times. Each group can be indexed and combined into a list. This is a simple implementation and will return the numbers within parentheses.
eg. 'asdasd Value(102222), fgdf(20), he(77)' will match 20 and 77 but not 102222.

My regular expression is not getting matched exactly in python [duplicate]

This question already has answers here:
Checking whole string with a regex
(5 answers)
Closed 6 years ago.
Here's my code...
import re
l=["chap","chap11","chapa","chapb","chapc","chap3","chap2","chapf","chap4","chap55","chapf","chap33","chap54","chapgk"]
for i in l:
matchobj=re.match(r'chap[0-9]',i,re.M|re.I)
if matchobj:
print(i)
as I have mentioned chap[0-9].. so it should only those strings which follow only one integer after chap
so I should get the following output..
chap3
chap2
chap4
but I am getting the following output...
chap11
chap3
chap2
chap4
chap55
chap33
chap54

match matches your pattern at the beginning of the string. Append e.g. end of string '$' or word boundary '\b' to your pattern:
matchobj=re.match(r'chap\d$',i,re.M|re.I)
# \d (digit) is shortcut for [0-9]
From the docs on re.match:
If zero or more characters at the beginning of string match the regular expression pattern, return a corresponding MatchObject instance.

You should add a dollar sign to the end of your regex expression. The dollar ($) means the end of the string, and for future reference, the carat (^) signifies the beginning.
import re
l=["chap","chap11","chapa","chapb","chapc","chap3","chap2","chapf","chap4","chap55","chapf","chap33","chap54","chapgk"]
for i in l:
matchobj=re.match(r'chap[0-9]$',i,re.M|re.I)
if matchobj:
print(i)
Output
chap3
chap2
chap4

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