Develop Reference

How I can use regex to remove repeated characters from string

I have a string as follows where I tried to remove similar consecutive characters.
import re
input = "abccbcbbb";
for i in input :
input = re.sub("(.)\\1+", "",input);
print(input)
Now I need to let the user specify the value of k.
I am using the following python code to do it, but I got the error message TypeError: can only concatenate str (not "int") to str
import re
input = "abccbcbbb";
k=3
for i in input :
input= re.sub("(.)\\1+{"+(k-1)+"}", "",input)
print(input)

The for i in input : does not do what you need. i is each character in the input string, and your re.sub is supposed to take the whole input as a char sequence.
If you plan to match a specific amount of chars you should get rid of the + quantifier after \1. The limiting {min,} / {min,max} quantifier should be placed right after the pattern it modifies.
Also, it is more convenient to use raw string literals when defining regexps.
You can use
import re
input_text = "abccbcbbb";
k=3
input_text = re.sub(fr"(.)\1{{{k-1}}}", "", input_text)
print(input_text)
# => abccbc
See this Python demo.
The fr"(.)\1{{{k-1}}}" raw f-string literal will translate into (.)\1{2} pattern. In f-strings, you need to double curly braces to denote a literal curly brace and you needn't escape \1 again since it is a raw string literal.

If I were you, I would prefer to do it like suggested before. But since I've already spend time on answering this question here is my handmade solution.
The pattern described below creates a named group named "letter". This group updates iterative, so firstly it is a, then b, etc. Then it looks ahead for all the repetitions of the group "letter" (which updates for each letter).
So it finds all groups of repeated letters and replaces them with empty string.
import re
input = 'abccbcbbb'
result = 'abcbcb'
pattern = r'(?P<letter>[a-z])(?=(?P=letter)+)'
substituted = re.sub(pattern, '', input)
assert substituted == result

Just to make sure I have the question correct you mean to turn "abccbcbbb" into "abcbcb" only removing sequential duplicate characters. Is there a reason you need to use regex? you could likely do a simple list comprehension. I mean this is a really cut and dirty way to do it but you could just put
input = "abccbcbbb"
input = list(input)
previous = input.pop(0)
result = [previous]
for letter in input:
if letter != previous : result += letter
previous = letter
result = "".join(result)
and with a method like this, you could make it easier to read and faster with a bit of modification id assume.

Regular expression to retrieve string parts within parentheses separated by commas

I have a String from which I want to take the values within the parenthesis. Then, get the values that are separated from a comma.
Example: x(142,1,23ERWA31)
I would like to get:
142
1
23ERWA31
Is it possible to get everything with one regex?
I have found a method to do so, but it is ugly.
This is how I did it in python:
import re
string = "x(142,1,23ERWA31)"
firstResult = re.search("\((.*?)\)", string)
secondResult = re.search("(?<=\()(.*?)(?=\))", firstResult.group(0))
finalResult = [x.strip() for x in secondResult.group(0).split(',')]
for i in finalResult:
print(i)
142
1
23ERWA31

This works for your example string:
import re
string = "x(142,1,23ERWA31)"
l = re.findall (r'([^(,)]+)(?!.*\()', string)
print (l)
Result: a plain list
['142', '1', '23ERWA31']
The expression matches a sequence of characters not in (,,,) and – to prevent the first x being picked up – may not be followed by a ( anywhere further in the string. This makes it also work if your preamble x consists of more than a single character.
findall rather than search makes sure all items are found, and as a bonus it returns a plain list of the results.

You can make this a lot simpler. You are running your first Regex but then not taking the result. You want .group(1) (inside the brackets), not .group(0) (the whole match). Once you have that you can just split it on ,:
import re
string = "x(142,1,23ERWA31)"
firstResult = re.search("\((.*?)\)", string)
for e in firstResult.group(1).split(','):
print(e)

A little wonky looking, and also assuming there's always going to be a grouping of 3 values in the parenthesis - but try this regex
\((.*?),(.*?),(.*?)\)
To extract all the group matches to a single object - your code would then look like
import re
string = "x(142,1,23ERWA31)"
firstResult = re.search("\((.*?),(.*?),(.*?)\)", string).groups()
You can then call the firstResult object like a list
>> print(firstResult[2])
23ERWA31

Parsing Korean text into a list using regex

I have some data stored as pandas data frame and one of the columns contains text strings in Korean. I would like to process each of these text strings as follows:
my_string = '모질상태불량(피부상태불량, 심하게 야윔), 치석심함, 양측 수정체 백탁, 좌측 화농성 눈곱심함(7/22), 코로나음성(활력저하)'
Into a list like this:
parsed_text = '모질상태불량, 피부상태불량, 심하게 야윔, 치석심함, 양측 수정체 백탁, 좌측 화농성 눈곱심함(7/22), 코로나음성, 활력저하'
So the problem is to identify cases where a word (or several words) are followed by parentheses with text only (can be one words or several words separated by commas) and replace them by all the words (before and inside parentheses) separated by comma (for later processing). If a word is followed by parentheses containing numbers (as in this case 7/22), it should be kept as it is. If a word is not followed by any parentheses, it should also be kept as it is. Furthermore, I would like to preserve the order of words (as they appeared in the original string).
I can extract text in parentheses by using regex as follows:
corrected_string = re.findall(r'(\w+)\((\D.*?)\)', my_string)
which yields this:
[('모질상태불량', '피부상태불량, 심하게 야윔'), ('코로나음성', '활력저하')]
But I'm having difficulty creating my resulting string, i.e. replacing my original text with the pattern I've matched. Any suggestions? Thank you.

You can use re.findall with a pattern that optionally matches a number enclosed in parentheses:
corrected_string = re.findall(r'[^,()]+(?:\([^)]*\d[^)]*\))?', my_string)

It's little bit clumsy but you can try:
my_string_list = [x.strip() for x in re.split(r"\((?!\d)|(?<!\d)\)|,", my_string) if x]
# you can make string out of list then.

Search for string in file while ignoring id and replacing only a substring

I’ve got a master .xml file generated by an external application and want to create several new .xmls by adapting and deleting some rows with python. The search strings and replace strings for these adaptions are stored within an array, e.g.:
replaceArray = [
[u'ref_layerid_mapping="x4049" lyvis="off" toc_visible="off"',
u'ref_layerid_mapping="x4049" lyvis="on" toc_visible="on"'],
[u'<TOOL_BUFFER RowID="106874" id_tool_base="3651" use="false"/>',
u'<TOOL_BUFFER RowID="106874" id_tool_base="3651" use="true"/>'],
[u'<TOOL_SELECT_LINE RowID="106871" id_tool_base="3658" use="false"/>',
u'<TOOL_SELECT_LINE RowID="106871" id_tool_base="3658" use="true"/>']]
So I'd like to iterate through my file and replace all occurences of 'ref_layerid_mapping="x4049" lyvis="off" toc_visible="off"' with 'ref_layerid_mapping="x4049" lyvis="on" toc_visible="on"' and so on.
Unfortunately the ID values of "RowID", “id_tool_base” and “ref_layerid_mapping” might change occassionally. So what I need is to search for matches of the whole string in the master file regardless which id value is inbetween the quotation mark and only to replace the substring that is different in both strings of the replaceArray (e.g. use=”true” instead of use=”false”). I’m not very familiar with regular expressions, but I think I need something like that for my search?
re.sub(r'<TOOL_SELECT_LINE RowID="\d+" id_tool_base="\d+" use="false"/>', "", sentence)
I'm happy about any hint that points me in the right direction! If you need any further information or if something is not clear in my question, please let me know.

One way to do this is to have a function for replacing text. The function would get the match object from re.sub and insert id captured from the string being replaced.
import re
s = 'ref_layerid_mapping="x4049" lyvis="off" toc_visible="off"'
pat = re.compile(r'ref_layerid_mapping=(.+) lyvis="off" toc_visible="off"')
def replacer(m):
return "ref_layerid_mapping=" + m.group(1) + 'lyvis="on" toc_visible="on"';
re.sub(pat, replacer, s)
Output:
'ref_layerid_mapping="x4049"lyvis="on" toc_visible="on"'
Another way is to use back-references in replacement pattern. (see http://www.regular-expressions.info/replacebackref.html)
For example:
import re
s = "Ab ab"
re.sub(r"(\w)b (\w)b", r"\1d \2d", s)
Output:
'Ad ad'

Parsing and reformatting CSV/text data using Python

sorry if this a bit of a beginner's question, but I haven't had much experience with python, and could really use some help in figuring this out. If there is a better programming language for tackling this, I'd be more than open to hearing it
I'm working on a small project, and I have two blocks of data, formatted differently from each other. They're all spreadsheets saved as CSV files, and I'd really like to make one group match the other without having to manually edit all the data.
What I need to do is go through a CSV, and format any data saved like this:
10W
20E
15-16N
17-18S
To a format like this (respective line to respective format):
10,W
20,E
,,15,16,N
,,17,18,S
So that they can just be copied over when opened as spreadsheets
I'm able to get the files into a string in python, but I'm unsure of how to properly write something to search for a number-hyphen-number-letter format.
I'd be immensely grateful for any help I can get. Thanks

This sounds like a good use-case for regular expressions. Once you've split the lines up into individual strings and stripped the whitespace (using s.strip()) these should work (I'm assuming those are cardinal directions; you'll need to change [NESW] to something else if that assumption is incorrect.):
>>> import re
>>> re.findall('\A(\d+)([NESW])', '16N')
[('16', 'N')]
>>> re.findall('\A(\d+)([NESW])', '15-16N')
[]
>>> re.findall('\A(\d+)-(\d+)([NESW])', '15-16N')
[('15', '16', 'N')]
>>> re.findall('\A(\d+)-(\d+)([NESW])', '16N')
[]
The first regex '\A(\d+)([NESW])' matches only a string that begins with a sequence of digits followed by a capital letter N, E, S, or W. The second matches only a string that begins with a sequence of digits followed by a hyphen, followed by another sequence of digits, followed by a capital letter N, E, S, or W. Forcing it to match at the beginning ensures that these regexes don't match a suffix of a longer string.
Then you can do something like this:
>>> vals = re.findall('\A(\d+)([NESW])', '16N')[0]
>>> ','.join(vals)
'16,N'
>>> vals = re.findall('(\d+)-(\d+)([NESW])', '15-16N')[0]
>>> ',,' + ','.join(vals)
',,15,16,N'

This is a whole solution that uses regexs. #senderle has beat me to the answer, so feel free to tick his response. This is just added here as I know how difficult it was to wrap my head around re in my code at first.
import re
dash = re.compile('(\d{2})-(\d{2})([WENS])')
no_dash = re.compile( '(\d{2})([WENS])' )
raw = '''10W
20E
15-16N
17-18S'''
lines = raw.split('\n')
data = []
for l in lines:
if '-' in l:
match = re.search(dash, l).groups()
data.append( ',,%s,%s,%s' % (match[0], match[1], match[2] ) )
else:
match = re.search(no_dash, l).groups()
data.append( '%s,%s' % (match[0], match[1] ) )
print '\n'.join(data)

In your case, I think the quick solution would involve regexps
You can either use the match method to extract your different tokens when they match a given regular expression, or the split method to split your string into tokens given a separator.
However, in your case, the separator would be a single character, so you can use the split method from the str class.