I'm trying to remove specific characters from a string using Python. This is the code I'm using right now. Unfortunately it appears to do nothing to the string.
for char in line:
if char in " ?.!/;:":
line.replace(char,'')
How do I do this properly?
Strings in Python are immutable (can't be changed). Because of this, the effect of line.replace(...) is just to create a new string, rather than changing the old one. You need to rebind (assign) it to line in order to have that variable take the new value, with those characters removed.
Also, the way you are doing it is going to be kind of slow, relatively. It's also likely to be a bit confusing to experienced pythonators, who will see a doubly-nested structure and think for a moment that something more complicated is going on.
Starting in Python 2.6 and newer Python 2.x versions *, you can instead use str.translate, (see Python 3 answer below):
line = line.translate(None, '!##$')
or regular expression replacement with re.sub
import re
line = re.sub('[!##$]', '', line)
The characters enclosed in brackets constitute a character class. Any characters in line which are in that class are replaced with the second parameter to sub: an empty string.
Python 3 answer
In Python 3, strings are Unicode. You'll have to translate a little differently. kevpie mentions this in a comment on one of the answers, and it's noted in the documentation for str.translate.
When calling the translate method of a Unicode string, you cannot pass the second parameter that we used above. You also can't pass None as the first parameter. Instead, you pass a translation table (usually a dictionary) as the only parameter. This table maps the ordinal values of characters (i.e. the result of calling ord on them) to the ordinal values of the characters which should replace them, or—usefully to us—None to indicate that they should be deleted.
So to do the above dance with a Unicode string you would call something like
translation_table = dict.fromkeys(map(ord, '!##$'), None)
unicode_line = unicode_line.translate(translation_table)
Here dict.fromkeys and map are used to succinctly generate a dictionary containing
{ord('!'): None, ord('#'): None, ...}
Even simpler, as another answer puts it, create the translation table in place:
unicode_line = unicode_line.translate({ord(c): None for c in '!##$'})
Or, as brought up by Joseph Lee, create the same translation table with str.maketrans:
unicode_line = unicode_line.translate(str.maketrans('', '', '!##$'))
* for compatibility with earlier Pythons, you can create a "null" translation table to pass in place of None:
import string
line = line.translate(string.maketrans('', ''), '!##$')
Here string.maketrans is used to create a translation table, which is just a string containing the characters with ordinal values 0 to 255.
Am I missing the point here, or is it just the following:
string = "ab1cd1ef"
string = string.replace("1", "")
print(string)
# result: "abcdef"
Put it in a loop:
a = "a!b#c#d$"
b = "!##$"
for char in b:
a = a.replace(char, "")
print(a)
# result: "abcd"
>>> line = "abc##!?efg12;:?"
>>> ''.join( c for c in line if c not in '?:!/;' )
'abc##efg12'
With re.sub regular expression
Since Python 3.5, substitution using regular expressions re.sub became available:
import re
re.sub('\ |\?|\.|\!|\/|\;|\:', '', line)
Example
import re
line = 'Q: Do I write ;/.??? No!!!'
re.sub('\ |\?|\.|\!|\/|\;|\:', '', line)
'QDoIwriteNo'
Explanation
In regular expressions (regex), | is a logical OR and \ escapes spaces and special characters that might be actual regex commands. Whereas sub stands for substitution, in this case with the empty string ''.
The asker almost had it. Like most things in Python, the answer is simpler than you think.
>>> line = "H E?.LL!/;O:: "
>>> for char in ' ?.!/;:':
... line = line.replace(char,'')
...
>>> print line
HELLO
You don't have to do the nested if/for loop thing, but you DO need to check each character individually.
For the inverse requirement of only allowing certain characters in a string, you can use regular expressions with a set complement operator [^ABCabc]. For example, to remove everything except ascii letters, digits, and the hyphen:
>>> import string
>>> import re
>>>
>>> phrase = ' There were "nine" (9) chick-peas in my pocket!!! '
>>> allow = string.letters + string.digits + '-'
>>> re.sub('[^%s]' % allow, '', phrase)
'Therewerenine9chick-peasinmypocket'
From the python regular expression documentation:
Characters that are not within a range can be matched by complementing
the set. If the first character of the set is '^', all the characters
that are not in the set will be matched. For example, [^5] will match
any character except '5', and [^^] will match any character except
'^'. ^ has no special meaning if it’s not the first character in the
set.
line = line.translate(None, " ?.!/;:")
>>> s = 'a1b2c3'
>>> ''.join(c for c in s if c not in '123')
'abc'
Strings are immutable in Python. The replace method returns a new string after the replacement. Try:
for char in line:
if char in " ?.!/;:":
line = line.replace(char,'')
This is identical to your original code, with the addition of an assignment to line inside the loop.
Note that the string replace() method replaces all of the occurrences of the character in the string, so you can do better by using replace() for each character you want to remove, instead of looping over each character in your string.
I was surprised that no one had yet recommended using the builtin filter function.
import operator
import string # only for the example you could use a custom string
s = "1212edjaq"
Say we want to filter out everything that isn't a number. Using the filter builtin method "...is equivalent to the generator expression (item for item in iterable if function(item))" [Python 3 Builtins: Filter]
sList = list(s)
intsList = list(string.digits)
obj = filter(lambda x: operator.contains(intsList, x), sList)))
In Python 3 this returns
>> <filter object # hex>
To get a printed string,
nums = "".join(list(obj))
print(nums)
>> "1212"
I am not sure how filter ranks in terms of efficiency but it is a good thing to know how to use when doing list comprehensions and such.
UPDATE
Logically, since filter works you could also use list comprehension and from what I have read it is supposed to be more efficient because lambdas are the wall street hedge fund managers of the programming function world. Another plus is that it is a one-liner that doesnt require any imports. For example, using the same string 's' defined above,
num = "".join([i for i in s if i.isdigit()])
That's it. The return will be a string of all the characters that are digits in the original string.
If you have a specific list of acceptable/unacceptable characters you need only adjust the 'if' part of the list comprehension.
target_chars = "".join([i for i in s if i in some_list])
or alternatively,
target_chars = "".join([i for i in s if i not in some_list])
Using filter, you'd just need one line
line = filter(lambda char: char not in " ?.!/;:", line)
This treats the string as an iterable and checks every character if the lambda returns True:
>>> help(filter)
Help on built-in function filter in module __builtin__:
filter(...)
filter(function or None, sequence) -> list, tuple, or string
Return those items of sequence for which function(item) is true. If
function is None, return the items that are true. If sequence is a tuple
or string, return the same type, else return a list.
Try this one:
def rm_char(original_str, need2rm):
''' Remove charecters in "need2rm" from "original_str" '''
return original_str.translate(str.maketrans('','',need2rm))
This method works well in Python 3
Here's some possible ways to achieve this task:
def attempt1(string):
return "".join([v for v in string if v not in ("a", "e", "i", "o", "u")])
def attempt2(string):
for v in ("a", "e", "i", "o", "u"):
string = string.replace(v, "")
return string
def attempt3(string):
import re
for v in ("a", "e", "i", "o", "u"):
string = re.sub(v, "", string)
return string
def attempt4(string):
return string.replace("a", "").replace("e", "").replace("i", "").replace("o", "").replace("u", "")
for attempt in [attempt1, attempt2, attempt3, attempt4]:
print(attempt("murcielago"))
PS: Instead using " ?.!/;:" the examples use the vowels... and yeah, "murcielago" is the Spanish word to say bat... funny word as it contains all the vowels :)
PS2: If you're interested on performance you could measure these attempts with a simple code like:
import timeit
K = 1000000
for i in range(1,5):
t = timeit.Timer(
f"attempt{i}('murcielago')",
setup=f"from __main__ import attempt{i}"
).repeat(1, K)
print(f"attempt{i}",min(t))
In my box you'd get:
attempt1 2.2334518376057244
attempt2 1.8806643818474513
attempt3 7.214925774955572
attempt4 1.7271184513757465
So it seems attempt4 is the fastest one for this particular input.
Here's my Python 2/3 compatible version. Since the translate api has changed.
def remove(str_, chars):
"""Removes each char in `chars` from `str_`.
Args:
str_: String to remove characters from
chars: String of to-be removed characters
Returns:
A copy of str_ with `chars` removed
Example:
remove("What?!?: darn;", " ?.!:;") => 'Whatdarn'
"""
try:
# Python2.x
return str_.translate(None, chars)
except TypeError:
# Python 3.x
table = {ord(char): None for char in chars}
return str_.translate(table)
#!/usr/bin/python
import re
strs = "how^ much for{} the maple syrup? $20.99? That's[] ricidulous!!!"
print strs
nstr = re.sub(r'[?|$|.|!|a|b]',r' ',strs)#i have taken special character to remove but any #character can be added here
print nstr
nestr = re.sub(r'[^a-zA-Z0-9 ]',r'',nstr)#for removing special character
print nestr
You can also use a function in order to substitute different kind of regular expression or other pattern with the use of a list. With that, you can mixed regular expression, character class, and really basic text pattern. It's really useful when you need to substitute a lot of elements like HTML ones.
*NB: works with Python 3.x
import re # Regular expression library
def string_cleanup(x, notwanted):
for item in notwanted:
x = re.sub(item, '', x)
return x
line = "<title>My example: <strong>A text %very% $clean!!</strong></title>"
print("Uncleaned: ", line)
# Get rid of html elements
html_elements = ["<title>", "</title>", "<strong>", "</strong>"]
line = string_cleanup(line, html_elements)
print("1st clean: ", line)
# Get rid of special characters
special_chars = ["[!##$]", "%"]
line = string_cleanup(line, special_chars)
print("2nd clean: ", line)
In the function string_cleanup, it takes your string x and your list notwanted as arguments. For each item in that list of elements or pattern, if a substitute is needed it will be done.
The output:
Uncleaned: <title>My example: <strong>A text %very% $clean!!</strong></title>
1st clean: My example: A text %very% $clean!!
2nd clean: My example: A text very clean
My method I'd use probably wouldn't work as efficiently, but it is massively simple. I can remove multiple characters at different positions all at once, using slicing and formatting.
Here's an example:
words = "things"
removed = "%s%s" % (words[:3], words[-1:])
This will result in 'removed' holding the word 'this'.
Formatting can be very helpful for printing variables midway through a print string. It can insert any data type using a % followed by the variable's data type; all data types can use %s, and floats (aka decimals) and integers can use %d.
Slicing can be used for intricate control over strings. When I put words[:3], it allows me to select all the characters in the string from the beginning (the colon is before the number, this will mean 'from the beginning to') to the 4th character (it includes the 4th character). The reason 3 equals till the 4th position is because Python starts at 0. Then, when I put word[-1:], it means the 2nd last character to the end (the colon is behind the number). Putting -1 will make Python count from the last character, rather than the first. Again, Python will start at 0. So, word[-1:] basically means 'from the second last character to the end of the string.
So, by cutting off the characters before the character I want to remove and the characters after and sandwiching them together, I can remove the unwanted character. Think of it like a sausage. In the middle it's dirty, so I want to get rid of it. I simply cut off the two ends I want then put them together without the unwanted part in the middle.
If I want to remove multiple consecutive characters, I simply shift the numbers around in the [] (slicing part). Or if I want to remove multiple characters from different positions, I can simply sandwich together multiple slices at once.
Examples:
words = "control"
removed = "%s%s" % (words[:2], words[-2:])
removed equals 'cool'.
words = "impacts"
removed = "%s%s%s" % (words[1], words[3:5], words[-1])
removed equals 'macs'.
In this case, [3:5] means character at position 3 through character at position 5 (excluding the character at the final position).
Remember, Python starts counting at 0, so you will need to as well.
In Python 3.5
e.g.,
os.rename(file_name, file_name.translate({ord(c): None for c in '0123456789'}))
To remove all the number from the string
How about this:
def text_cleanup(text):
new = ""
for i in text:
if i not in " ?.!/;:":
new += i
return new
Below one.. with out using regular expression concept..
ipstring ="text with symbols!##$^&*( ends here"
opstring=''
for i in ipstring:
if i.isalnum()==1 or i==' ':
opstring+=i
pass
print opstring
Recursive split:
s=string ; chars=chars to remove
def strip(s,chars):
if len(s)==1:
return "" if s in chars else s
return strip(s[0:int(len(s)/2)],chars) + strip(s[int(len(s)/2):len(s)],chars)
example:
print(strip("Hello!","lo")) #He!
You could use the re module's regular expression replacement. Using the ^ expression allows you to pick exactly what you want from your string.
import re
text = "This is absurd!"
text = re.sub("[^a-zA-Z]","",text) # Keeps only Alphabets
print(text)
Output to this would be "Thisisabsurd". Only things specified after the ^ symbol will appear.
# for each file on a directory, rename filename
file_list = os.listdir (r"D:\Dev\Python")
for file_name in file_list:
os.rename(file_name, re.sub(r'\d+','',file_name))
Even the below approach works
line = "a,b,c,d,e"
alpha = list(line)
while ',' in alpha:
alpha.remove(',')
finalString = ''.join(alpha)
print(finalString)
output: abcde
The string method replace does not modify the original string. It leaves the original alone and returns a modified copy.
What you want is something like: line = line.replace(char,'')
def replace_all(line, )for char in line:
if char in " ?.!/;:":
line = line.replace(char,'')
return line
However, creating a new string each and every time that a character is removed is very inefficient. I recommend the following instead:
def replace_all(line, baddies, *):
"""
The following is documentation on how to use the class,
without reference to the implementation details:
For implementation notes, please see comments begining with `#`
in the source file.
[*crickets chirp*]
"""
is_bad = lambda ch, baddies=baddies: return ch in baddies
filter_baddies = lambda ch, *, is_bad=is_bad: "" if is_bad(ch) else ch
mahp = replace_all.map(filter_baddies, line)
return replace_all.join('', join(mahp))
# -------------------------------------------------
# WHY `baddies=baddies`?!?
# `is_bad=is_bad`
# -------------------------------------------------
# Default arguments to a lambda function are evaluated
# at the same time as when a lambda function is
# **defined**.
#
# global variables of a lambda function
# are evaluated when the lambda function is
# **called**
#
# The following prints "as yellow as snow"
#
# fleece_color = "white"
# little_lamb = lambda end: return "as " + fleece_color + end
#
# # sometime later...
#
# fleece_color = "yellow"
# print(little_lamb(" as snow"))
# --------------------------------------------------
replace_all.map = map
replace_all.join = str.join
If you want your string to be just allowed characters by using ASCII codes, you can use this piece of code:
for char in s:
if ord(char) < 96 or ord(char) > 123:
s = s.replace(char, "")
It will remove all the characters beyond a....z even upper cases.
I'm looking to find and modify some sql syntax around the convert function. I want basically any convert(A,B) or CONVERT(A,B) in all my files to be selected and converted to B::A.
So far I tried selecting them with re.findall(r"\bconvert\b\(.*?,.*\)", l, re.IGNORECASE) But it's only returning a small selection out of what I want and I also have trouble actually manipulating the A/B I mentioned.
For example, a sample line (note the nested structure here is irrelevant, I'm only getting the outer layer working if possible)
convert(varchar, '/' || convert(nvarchar, es.Item_ID) || ':' || convert(nvarchar, o.Option_Number) || '/') as LocPath
...should become...
'/' || es.Item_ID::nvarchar || ':' || o.Option_Number::nvarchar || '/' :: varchar as LocPath
Example2:
SELECT LocationID AS ItemId, convert(bigint, -1),
...should become...
SELECT LocationID AS ItemId, -1::bigint,
I think this should be possible with some kind of re.sub with groups and currently have a code structure inside a for each loop where line is the each line in the file:
matchConvert = ["convert(", "CONVERT("]
a = next((a for a in matchConvert if a in line), False)
if a:
print("convert() line")
#line = re.sub(re.escape(a) + r'', '', line)
Edit: In the end I went with a non re solution and handled each line by identifying each block and manipulate them accordingly.
This may be an X/Y problem, meaning you’re asking how to do something with Regex that may be better solved with parsing (meaning using/modifying/writing a SQL parser). An indication that this is the case is the fact that “convert” calls can be nested. I’m guessing Regex is going to be more of a headache than it’s worth here in the long run if you’re working with a lot of files and they’re at all complicated.
The task:
Swap the parameters of all the 'convert' functions in this given. Parameters can contain any character, including nested 'convert' functions.
A solution:
def convert_py(s):
#capturing start:
left=s.index('convert')
start=s[:left]
#capturing part_1:
c=0
line=''
for n1,i in enumerate(s[left+8:],start=len(start)+8):
if i==',' and c==0:
part_1=line
break
if i==')':
c-=1
if i=='(':
c+=1
line+=i
#capturing part_2:
c=0
line=''
for n2,i in enumerate(s[n1+1:],start=n1+1):
if i==')':
c-=1
if i=='(':
c+=1
if c<0:
part_2=line
break
line+=i
#capturing end:
end=s[n2+1:]
#capturing result:
result=start+part_2.lstrip()+' :: '+part_1+end
return result
def multi_convert_py(s):
converts=s.count('convert')
for n in range(converts):
s=convert_py(s)
return s
Notes:
Unlike the solution based on the re module, which is presented in another answer - this version should not fail if there are more than two parameters in the 'convert' function in the given string. However, it will swap them only once, for example: convert(a,b, c) --> b, c : a
I am afraid that unforeseen cases may arise that will lead to failure. Please tell if you find any flaws
The task:
Swap the parameters of all the 'convert' functions in the given string. Parameters can contain any character, including nested 'convert' functions.
A solution based on the re module:
def convert_re(s):
import re
start,part_1,part_2,end=re.search(r'''
(.*?)
convert\(
([^,)(]+\(.+?\)[^,)(]*|[^,)(]+)
,
([^,)(]+\(.+?\)[^,)(]*|[^,)(]+)
\)
(.*)
''',s,re.X).groups()
result=start+part_2.lstrip()+' :: '+part_1+end
return result
def multi_convert_re(s):
converts=s.count('convert')
for n in range(converts):
s=convert_re(s)
return s
Discription of the 'convert_re' function:
Regular expression:
start is the first group with what comes before 'convert'
Then follows convert\() which has no group and contains the name of the function and the opening '('
part_1 is the second group ([^,)(]+\(.+?\)[^,)(]*|[^,)(]+). This should match the first parameter. It can be anything except - ,)(, or a function preceded by anything except ,)(, optionally followed by anything except ,)( and with anything inside (except a new line)
Then follows a comma ,, which has no group
part_2 is the third group and it acts like the second, but should catch everything what's left inside the external function
Then follows ), which has no group
end is the fourth group (.*) with what's left before the new line.
The resulting string is then created by swapping part_1 and part_2, putting ' :: ' between them, removing spaces on the left from part_2 and adding start to the beginning and end to the end.
Description of the 'multi_convert_re' function
Repeatedly calls 'convert_re' function until there are no "convert" left.
Notes:
N.B.: The code implies that the 'convert' function in the string has exactly two parameters.
The code works on the given examples, but I'm afraid there may still be unforeseen flaws when it comes to other examples. Please tell, if you find any flaws.
I have provided another solution presented in another answer that is not based on the re module. It may turn out that the results will be different.
Here's my solution based on #Иван-Балван's code. Breaking this structure into blocks makes further specification a lot easier than I previously thought and I'll be using this method for a lot of other operations as well.
# Check for balanced brackets
def checkBracket(my_string):
count = 0
for c in my_string:
if c == "(":
count+=1
elif c == ")":
count-=1
return count
# Modify the first convert in line
# Based on suggestions from stackoverflow.com/questions/73040953
def modifyConvert(l):
# find the location of convert()
count = l.index('convert(')
# select the group before convert() call
before = l[:count]
group=""
n1=0
n2=0
A=""
B=""
operate = False
operators = ["|", "<", ">", "="]
# look for A group before comma
for n1, i in enumerate(l[count+8:], start=len(before)+8):
# find current position in l
checkIndex = checkBracket(l[count+8:][:n1-len(before)-8])
if i == ',' and checkIndex == 0:
A = group
break
group += i
# look for B group after comma
group = ""
for n2, i in enumerate(l[n1+1:], start=n1+1):
checkIndex = checkBracket(l[count+n1-len(before):][:n2-n1+1])
if i == ',' and checkIndex == 0:
return l
elif checkIndex < 0:
B = group
break
group += i
# mark operators
if i in operators:
operate = True
# select the group after convert() call
after = l[n2+1:]
# (B) if it contains operators
if operate:
return before + "(" + B.lstrip() + ') :: ' + A + after
else:
return before + B.lstrip() + '::' + A + after
# Modify cast syntax with convert(a,b). return line.
def convertCast(l):
# Call helper for nested cases
i = l.count('convert(')
while i>0:
i -= 1
l = modifyConvert(l)
return l
Recently I was solving a problem in Codewars and got stuck. The link of the problem link
Basically what it is asking for is :
You are given a string for example :
"example(unwanted thing)example"
Your task is to remove everything inside the parentheses as well as the parentheses themselves.
The example above would return:
"exampleexample"
Don't worry about other brackets like "[]" and "{}" as these will never appear.
There can be multiple parentheses.
The parentheses can be nested.
Some other test cases are given below :
test.assert_equals(remove_parentheses("hello example (words(more words) here) something"), "hello example something")
test.assert_equals(remove_parentheses("(first group) (second group) (third group)"), " ")
I looked up online and I found some solution involving Regex, but I wanted to solve this problem without Regex.
Till now I have tried similar solutions as given below :
def remove_parentheses(s):
while s.find('(') != -1 or s.find(')') != -1 :
f = s.find('(')
l = s.find(')')
s = s[:f] + s [l+1:]
return s
But when I try to run this snippet, I get Execution Timed Out.
You just need to track the number of open parentheses (the nested depth, technically) to see whether the current character should be included in the output.
def remove_parentheses(s):
parentheses_count = 0
output = ""
for i in s:
if i=="(":
parentheses_count += 1
elif i==")":
parentheses_count -= 1
else:
if parentheses_count == 0:
output += i
return output
print(remove_parentheses("hello example (words(more words) here) something"))
print(remove_parentheses("(first group) (second group) (third group)"))
Use Stack to check whether '(' is closed or not.
If the length of Stack is not zero, that means that parentheses are still open, So you have to ignore the characters.
The code below will pass all the test cases.
def remove_parentheses(s):
stack = []
answer = []
for character in s:
if(character == '('):
stack.append('(')
continue
if(character == ')'):
stack.pop()
continue
if(len(stack) == 0):
answer.append(character)
return "".join(answer)
The reason for your code to have Execution Timed out is because it is stuck in an infinity loop. Since s = s[:f] + s [l+1:] doesn't remove the parentheses properly, such as
a nested example hello example (words(more words) here) something
your code will locate the first ( and the first ) and return hello example here) something on the first loop, which will lead to incorrect result in the next loop as one of your ( is removed.
To be honest, an approach like this is not ideal as it is difficult to understand and read since you have to dry run the index in the loop one by one. You may continue to debug this code and fix the indexing, such as only search the nearest/enclosed closing bracket according to your first located (, which will make it even more harder to read but get the job done.
For me, I would personally suggest you to look up regular expression, or what is often referred as regex,
a very simple algorithm that builds on regex is
import re
def remove_parentheses(s):
s = re.sub("\(.{1,25}\)", "", s)
return s
def f(s):
pairs = []
output = ''
for i, v in enumerate(s):
if "(" == v:
pairs.append(1)
if ")" == v:
pairs.pop()
continue
if len(pairs) ==0:
output +=v
return output
Can be achieved easily if we use a recursive function.. Try this out.
def rmp(st):
if st.find('(') == -1 or st.find(')') == -1: return st
else :
i=st.rindex('(')
j=st[i+1:].index(')')
return rmp(st[:i] + st[i+1+j+1:])
Here are a few cases I tested...
print(rmp("hello example (words(more words) here) something"))
print(rmp("(first group) (second group) (third group)"))
print(rmp("This does(n't) work (so well)"))
print(rmp("(1233)qw()"))
print(rmp("(1(2(3(4(5(6(7(8))))))))abcdqw(hkfjfj)"))
And the results are..
hello example something
This does work
qw
abcdqw
Basically, I have a list of special characters. I need to split a string by a character if it belongs to this list and exists in the string. Something on the lines of:
def find_char(string):
if string.find("some_char"):
#do xyz with some_char
elif string.find("another_char"):
#do xyz with another_char
else:
return False
and so on. The way I think of doing it is:
def find_char_split(string):
char_list = [",","*",";","/"]
for my_char in char_list:
if string.find(my_char) != -1:
my_strings = string.split(my_char)
break
else:
my_strings = False
return my_strings
Is there a more pythonic way of doing this? Or the above procedure would be fine? Please help, I'm not very proficient in python.
(EDIT): I want it to split on the first occurrence of the character, which is encountered first. That is to say, if the string contains multiple commas, and multiple stars, then I want it to split by the first occurrence of the comma. Please note, if the star comes first, then it will be broken by the star.
I would favor using the re module for this because the expression for splitting on multiple arbitrary characters is very simple:
r'[,*;/]'
The brackets create a character class that matches anything inside of them. The code is like this:
import re
results = re.split(r'[,*;/]', my_string, maxsplit=1)
The maxsplit argument makes it so that the split only occurs once.
If you are doing the same split many times, you can compile the regex and search on that same expression a little bit faster (but see Jon Clements' comment below):
c = re.compile(r'[,*;/]')
results = c.split(my_string)
If this speed up is important (it probably isn't) you can use the compiled version in a function instead of having it re compile every time. Then make a separate function that stores the actual compiled expression:
def split_chars(chars, maxsplit=0, flags=0, string=None):
# see note about the + symbol below
c = re.compile('[{}]+'.format(''.join(chars)), flags=flags)
def f(string, maxsplit=maxsplit):
return c.split(string, maxsplit=maxsplit)
return f if string is None else f(string)
Then:
special_split = split_chars(',*;/', maxsplit=1)
result = special_split(my_string)
But also:
result = split_chars(',*;/', my_string, maxsplit=1)
The purpose of the + character is to treat multiple delimiters as one if that is desired (thank you Jon Clements). If this is not desired, you can just use re.compile('[{}]'.format(''.join(chars))) above. Note that with maxsplit=1, this will not have any effect.
Finally: have a look at this talk for a quick introduction to regular expressions in Python, and this one for a much more information packed journey.