Removing a list of characters in string

You can use str.translate():

s.translate(None, ",!.;")

Example:

>>> s = "asjo,fdjk;djaso,oio!kod.kjods;dkps"
>>> s.translate(None, ",!.;")
'asjofdjkdjasooiokodkjodsdkps'

If you are using python3 and looking for the translate solution - the function was changed and now takes 1 parameter instead of 2.

That parameter is a table (can be dictionary) where each key is the Unicode ordinal (int) of the character to find and the value is the replacement (can be either a Unicode ordinal or a string to map the key to).

Here is a usage example:

>>> list = [',', '!', '.', ';']
>>> s = "This is, my! str,ing."
>>> s.translate({ord(x): '' for x in list})
'This is my string'

You can use the translate method.

s.translate(None, '!.;,')

''.join(c for c in myString if not c in badTokens)

Why not a simple loop?

for i in replace_list:
    string = string.replace(i, '')

Also, avoid naming lists 'list'. It overrides the built-in function list.

Another approach using regex:

''.join(re.split(r'[.;!?,]', s))

you could use something like this

def replace_all(text, dic):
  for i, j in dic.iteritems():
    text = text.replace(i, j)
  return text

This code is not my own and comes from here its a great article and dicusses in depth doing this

simple way,

import re
str = 'this is string !    >><< (foo---> bar) @-tuna-#   sandwich-%-is-$-* good'

// condense multiple empty spaces into 1
str = ' '.join(str.split()

// replace empty space with dash
str = str.replace(" ","-")

// take out any char that matches regex
str = re.sub('[!@#$%^&*()_+<>]', '', str)

output:

this-is-string--foo----bar--tuna---sandwich--is---good

Remove *%,&@! from below string:

s = "this is my string,  and i will * remove * these ** %% "
new_string = s.translate(s.maketrans('','','*%,&@!'))
print(new_string)

# output: this is my string  and i will  remove  these  

In Python 3.8, this works for me:

s.translate(s.maketrans(dict.fromkeys(',!.;', '')))

Also an interesting topic on removal UTF-8 accent form a string converting char to their standard non-accentuated char:

What is the best way to remove accents in a python unicode string?

code extract from the topic:

import unicodedata

def remove_accents(input_str):
    nkfd_form = unicodedata.normalize('NFKD', input_str)
    return u"".join([c for c in nkfd_form if not unicodedata.combining(c)])

Perhaps a more modern and functional way to achieve what you wish:

>>> subj = 'A.B!C?'
>>> list = set([',', '!', '.', ';', '?'])
>>> filter(lambda x: x not in list, subj)
'ABC'

please note that for this particular purpose it's quite an overkill, but once you need more complex conditions, filter comes handy

How about this - a one liner.

reduce(lambda x,y : x.replace(y,"") ,[',', '!', '.', ';'],";Test , ,  !Stri!ng ..")

i think this is simple enough and will do!

list = [",",",","!",";",":"] #the list goes on.....

theString = "dlkaj;lkdjf'adklfaj;lsd'fa'dfj;alkdjf" #is an example string;
newString="" #the unwanted character free string
for i in range(len(TheString)):
    if theString[i] in list:
        newString += "" #concatenate an empty string.
    else:
        newString += theString[i]

this is one way to do it. But if you are tired of keeping a list of characters that you want to remove, you can actually do it by using the order number of the strings you iterate through. the order number is the ascii value of that character. the ascii number for 0 as a char is 48 and the ascii number for lower case z is 122 so:

theString = "lkdsjf;alkd8a'asdjf;lkaheoialkdjf;ad"
newString = ""
for i in range(len(theString)):
     if ord(theString[i]) < 48 or ord(theString[i]) > 122: #ord() => ascii num.
         newString += ""
     else:
        newString += theString[i]

These days I am diving into scheme, and now I think am good at recursing and eval. HAHAHA. Just share some new ways:

first ,eval it

print eval('string%s' % (''.join(['.replace("%s","")'%i for i in replace_list])))

second , recurse it

def repn(string,replace_list):
    if replace_list==[]:
        return string
    else:
        return repn(string.replace(replace_list.pop(),""),replace_list)

print repn(string,replace_list)

Hey ,don't downvote. I am just want to share some new idea.

I am thinking about a solution for this. First I would make the string input as a list. Then I would replace the items of list. Then through using join command, I will return list as a string. The code can be like this:

def the_replacer(text):
    test = []    
    for m in range(len(text)):
        test.append(text[m])
        if test[m]==','\
        or test[m]=='!'\
        or test[m]=='.'\
        or test[m]=='\''\
        or test[m]==';':
    #....
            test[n]=''
    return ''.join(test)

This would remove anything from the string. What do you think about that?

Here is a more_itertools approach:

import more_itertools as mit


s = "A.B!C?D_E@F#"
blacklist = ".!?_@#"

"".join(mit.flatten(mit.split_at(s, pred=lambda x: x in set(blacklist))))
# 'ABCDEF'

Here we split upon items found in the blacklist, flatten the results and join the string.

Python 3, single line list comprehension implementation.

from string import ascii_lowercase # 'abcdefghijklmnopqrstuvwxyz'
def remove_chars(input_string, removable):
  return ''.join([_ for _ in input_string if _ not in removable])

print(remove_chars(input_string="Stack Overflow", removable=ascii_lowercase))
>>> 'S O'

Why not utilize this simple function:

def remove_characters(str, chars_list):
    for char in chars_list:
        str = str.replace(char, '')
  
    return str

Use function:

print(remove_characters('A.B!C?', ['.', '!', '?']))

Output:

ABC