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Split string if separator is not in-between two characters

I want to write a script that reads from a csv file and splits each line by comma except any commas in-between two specific characters.
In the below code snippet I would like to split line by commas except the commas in-between two $s.
line = "$abc,def$,$ghi$,$jkl,mno$"
output = line.split(',')
for o in output:
print(o)
How do I write output = line.split(',') so that I get the following terminal output?
~$ python script.py
$abc,def$
$ghi$
$jkl,mno$

You can do this with a regular expression:
In re, the (?<!\$) will match a character not immediately following a $.
Similarly, a (?!\$) will match a character not immediately before a dollar.
The | character cam match multiple options. So to match a character where either side is not a $ you can use:
expression = r"(?<!\$),|,(?!\$)"
Full program:
import re
expression = r"(?<!\$),|,(?!\$)"
print(re.split(expression, "$abc,def$,$ghi$,$jkl,mno$"))

One solution (maybe not the most elegant but it will work) is to replace the string $,$ with something like $,,$ and then split ,,. So something like this
output = line.replace('$,$','$,,$').split(',,')
Using regex like mousetail suggested is the more elegant and robust solution but requires knowing regex (not that anyone KNOWS regex)

Try regular expressions:
import re
line = "$abc,def$,$ghi$,$jkl,mno$"
output = re.findall(r"\$(.*?)\$", line)
for o in output:
print('$'+o+'$')
$abc,def$
$ghi$
$jkl,mno$

First, you can identify a character that is not used in that line:
c = chr(max(map(ord, line)) + 1)
Then, you can proceed as follows:
line.replace('$,$', f'${c}$').split(c)
Here is your example:
>>> line = '$abc,def$,$ghi$,$jkl,mno$'
>>> c = chr(max(map(ord, line)) + 1)
>>> result = line.replace('$,$', f'${c}$').split(c)
>>> print(*result, sep='\n')
$abc,def$
$ghi$
$jkl,mno$

Implement regular expression in Python to replace every occurence of "meshname = x" in a text file

I want to replace every line in a textfile with " " which starts with "meshname = " and ends with any letter/number and underscore combination. I used regex's in CS but I never really understood the different notations in Python. Can you help me with that?
Is this the right regex for my problem and how would i transform that into a Python regex?
m.e.s.h.n.a.m.e.' '.=.' '.{{_}*,{0,...,9}*,{a,...,z}*,{A,...,Z}*}*
x.y = Concatenation of x and y
' ' = whitespace
{x} = set containing x
x* = x.x.x. ... .x or empty word
What would the script look like in order to replace every string/line in a file containing meshname = ... with the Python regex? Something like this?
fin = open("test.txt", 'r')
data = fin.read()
data = data.replace("^meshname = [[a-z]*[A-Z]*[0-9]*[_]*]+", "")
fin.close()
fin = open("test.txt", 'w')
fin.write(data)
fin.close()
or is this completely wrong? I've tried to get it working with this approach, but somehow it never matched the right string: How to input a regex in string.replace?

Following the current code logic, you can use
data = re.sub(r'^meshname = .*\w$', ' ', data, flags=re.M)
The re.sub will replace with a space any line that matches
^ - line start (note the flags=re.M argument that makes sure the multiline mode is on)
meshname - a meshname word
= - a = string
.* - any zero or more chars other than line break chars as many as possible
\w - a letter/digit/_
$ - line end.

How to remove words after certain character in a line in python [duplicate]

I have a string. How do I remove all text after a certain character? (In this case ...)
The text after will ... change so I that's why I want to remove all characters after a certain one.

Split on your separator at most once, and take the first piece:
sep = '...'
stripped = text.split(sep, 1)[0]
You didn't say what should happen if the separator isn't present. Both this and Alex's solution will return the entire string in that case.

Assuming your separator is '...', but it can be any string.
text = 'some string... this part will be removed.'
head, sep, tail = text.partition('...')
>>> print head
some string
If the separator is not found, head will contain all of the original string.
The partition function was added in Python 2.5.
S.partition(sep) -> (head, sep, tail)
Searches for the separator sep in S, and returns the part before it,
the separator itself, and the part after it. If the separator is not
found, returns S and two empty strings.

If you want to remove everything after the last occurrence of separator in a string I find this works well:
<separator>.join(string_to_split.split(<separator>)[:-1])
For example, if string_to_split is a path like root/location/child/too_far.exe and you only want the folder path, you can split by "/".join(string_to_split.split("/")[:-1]) and you'll get
root/location/child

Without a regular expression (which I assume is what you want):
def remafterellipsis(text):
where_ellipsis = text.find('...')
if where_ellipsis == -1:
return text
return text[:where_ellipsis + 3]
or, with a regular expression:
import re
def remwithre(text, there=re.compile(re.escape('...')+'.*')):
return there.sub('', text)

import re
test = "This is a test...we should not be able to see this"
res = re.sub(r'\.\.\..*',"",test)
print(res)
Output: "This is a test"

The method find will return the character position in a string. Then, if you want remove every thing from the character, do this:
mystring = "123⋯567"
mystring[ 0 : mystring.index("⋯")]
>> '123'
If you want to keep the character, add 1 to the character position.

From a file:
import re
sep = '...'
with open("requirements.txt") as file_in:
lines = []
for line in file_in:
res = line.split(sep, 1)[0]
print(res)

This is in python 3.7 working to me
In my case I need to remove after dot in my string variable fees
fees = 45.05
split_string = fees.split(".", 1)
substring = split_string[0]
print(substring)

Yet another way to remove all characters after the last occurrence of a character in a string (assume that you want to remove all characters after the final '/').
path = 'I/only/want/the/containing/directory/not/the/file.txt'
while path[-1] != '/':
path = path[:-1]

another easy way using re will be
import re, clr
text = 'some string... this part will be removed.'
text= re.search(r'(\A.*)\.\.\..+',url,re.DOTALL|re.IGNORECASE).group(1)
// text = some string

regex replace '...' at the end of the string

I have a string like:
text1 = 'python...is...fun...'
I want to replace the multiple '.'s to one '.' only when they are at the end of the string, i want the output to be:
python...is...fun.
So when there is only one '.' at the end of the string, then it won't be replaced
text2 = 'python...is...fun.'
and the output is just the same as text2
My regex is like this:
text = re.sub(r'(.*)\.{2,}$', r'\1.', text)
which i want to match any string then {2 to n} of '.' at the end of the string, but the output is:
python...is...fun..
any ideas how to do this?
Thx in advance!

You are making it a bit complex, you can easily do it by using regex as \.+$ and replace the regex pattern with single . character.
>>> text1 = 'python...is...fun...'
>>> new_text = re.sub(r"\.+$", ".", text1)
>>> 'python...is...fun.'
You may extend this regex further to handle the cases with input such as ... only, etc but the main concept was that there is no need to counting the number of ., as you have done in your answer.

Just look for the string ending with three periods, and replace them with a single one.
import re
x = "foo...bar...quux..."
print(re.sub('\.{2,}$', '.', x))
// foo...bar...quux.

import re
print(re.sub(r'\.{2,}$', '.', 'I...love...python...'))
As simple as that. Note that you need to escape the . because otherwise, it means whichever char
except \n.

I want to replace the multiple '.'s to one '.' only when they are at
the end of the string
For such simple case it's easier to substitute without importing re module, checking the value of the last 3 characters:
text1 = 'python...is...fun...'
text1 = text1[:-2] if text1[-3:] == '...' else text1
print(text1)
The output:
python...is...fun.

python regex for repeating string

I am wanting to verify and then parse this string (in quotes):
string = "start: c12354, c3456, 34526; other stuff that I don't care about"
//Note that some codes begin with 'c'
I would like to verify that the string starts with 'start:' and ends with ';'
Afterward, I would like to have a regex parse out the strings. I tried the following python re code:
regx = r"start: (c?[0-9]+,?)+;"
reg = re.compile(regx)
matched = reg.search(string)
print ' matched.groups()', matched.groups()
I have tried different variations but I can either get the first or the last code but not a list of all three.
Or should I abandon using a regex?
EDIT: updated to reflect part of the problem space I neglected and fixed string difference.
Thanks for all the suggestions - in such a short time.

In Python, this isn’t possible with a single regular expression: each capture of a group overrides the last capture of that same group (in .NET, this would actually be possible since the engine distinguishes between captures and groups).
Your easiest solution is to first extract the part between start: and ; and then using a regular expression to return all matches, not just a single match, using re.findall('c?[0-9]+', text).

You could use the standard string tools, which are pretty much always more readable.
s = "start: c12354, c3456, 34526;"
s.startswith("start:") # returns a boolean if it starts with this string
s.endswith(";") # returns a boolean if it ends with this string
s[6:-1].split(', ') # will give you a list of tokens separated by the string ", "

This can be done (pretty elegantly) with a tool like Pyparsing:
from pyparsing import Group, Literal, Optional, Word
import string
code = Group(Optional(Literal("c"), default='') + Word(string.digits) + Optional(Literal(","), default=''))
parser = Literal("start:") + OneOrMore(code) + Literal(";")
# Read lines from file:
with open('lines.txt', 'r') as f:
for line in f:
try:
result = parser.parseString(line)
codes = [c[1] for c in result[1:-1]]
# Do something with teh codez...
except ParseException exc:
# Oh noes: string doesn't match!
continue
Cleaner than a regular expression, returns a list of codes (no need to string.split), and ignores any extra characters in the line, just like your example.

import re
sstr = re.compile(r'start:([^;]*);')
slst = re.compile(r'(?:c?)(\d+)')
mystr = "start: c12354, c3456, 34526; other stuff that I don't care about"
match = re.match(sstr, mystr)
if match:
res = re.findall(slst, match.group(0))
results in
['12354', '3456', '34526']