Best way to replace multiple characters in a string?

Here is a python3 method using str.translate and str.maketrans:

s = "abc&def#ghi"
print(s.translate(str.maketrans({'&': '\&', '#': '\#'})))

The printed string is abc\&def\#ghi.

Late to the party, but I lost a lot of time with this issue until I found my answer.

Short and sweet, translate is superior to replace. If you're more interested in funcionality over time optimization, do not use replace.

Also use translate if you don't know if the set of characters to be replaced overlaps the set of characters used to replace.

Case in point:

Using replace you would naively expect the snippet "1234".replace("1", "2").replace("2", "3").replace("3", "4") to return "2344", but it will return in fact "4444".

Translation seems to perform what OP originally desired.

You may consider writing a generic escape function:

def mk_esc(esc_chars):
    return lambda s: ''.join(['\\' + c if c in esc_chars else c for c in s])

>>> esc = mk_esc('&#')
>>> print esc('Learn & be #1')
Learn \& be \#1

This way you can make your function configurable with a list of character that should be escaped.

For Python 3.8 and above, one can use assignment expressions

[text := text.replace(s, f"\\{s}") for s in "&#" if s in text];

Although, I am quite unsure if this would be considered "appropriate use" of assignment expressions as described in PEP 572, but looks clean and reads quite well (to my eyes). The semicolon at the end suppresses output if you run this in a REPL.

This would be "appropriate" if you wanted all intermediate strings as well. For example, (removing all lowercase vowels):

text = "Lorem ipsum dolor sit amet"
intermediates = [text := text.replace(i, "") for i in "aeiou" if i in text]

['Lorem ipsum dolor sit met',
 'Lorm ipsum dolor sit mt',
 'Lorm psum dolor st mt',
 'Lrm psum dlr st mt',
 'Lrm psm dlr st mt']

On the plus side, it does seem (unexpectedly?) faster than some of the faster methods in the accepted answer, and seems to perform nicely with both increasing strings length and an increasing number of substitutions.

The code for the above comparison is below. I am using random strings to make my life a bit simpler, and the characters to replace are chosen randomly from the string itself. (Note: I am using ipython's %timeit magic here, so run this in ipython/jupyter).

import random, string

def make_txt(length):
    "makes a random string of a given length"
    return "".join(random.choices(string.printable, k=length))

def get_substring(s, num):
    "gets a substring"
    return "".join(random.choices(s, k=num))

def a(text, replace): # one of the better performing approaches from the accepted answer
    for i in replace:
        if i in text:
             text = text.replace(i, "")

def b(text, replace):
    _ = (text := text.replace(i, "") for i in replace if i in text) 


def compare(strlen, replace_length):
    "use ipython / jupyter for the %timeit functionality"

    times_a, times_b = [], []

    for i in range(*strlen):
        el = make_txt(i)
        et = get_substring(el, replace_length)

        res_a = %timeit -n 1000 -o a(el, et) # ipython magic

        el = make_txt(i)
        et = get_substring(el, replace_length)
        
        res_b = %timeit -n 1000 -o b(el, et) # ipython magic

        times_a.append(res_a.average * 1e6)
        times_b.append(res_b.average * 1e6)
        
    return times_a, times_b

#----run
t2 = compare((2*2, 1000, 50), 2)
t10 = compare((2*10, 1000, 50), 10)

How about this?

def replace_all(dict, str):
    for key in dict:
        str = str.replace(key, dict[key])
    return str

then

print(replace_all({"&":"\&", "#":"\#"}, "&#"))

output

\&\#

similar to answer

Using reduce which is available in python2.7 and python3.* you can easily replace mutiple substrings in a clean and pythonic way.

# Lets define a helper method to make it easy to use
def replacer(text, replacements):
    return reduce(
        lambda text, ptuple: text.replace(ptuple[0], ptuple[1]), 
        replacements, text
    )

if __name__ == '__main__':
    uncleaned_str = "abc&def#ghi"
    cleaned_str = replacer(uncleaned_str, [("&","\&"),("#","\#")])
    print(cleaned_str) # "abc\&def\#ghi"

In python2.7 you don't have to import reduce but in python3.* you have to import it from the functools module.

advanced way using regex

import re
text = "hello ,world!"
replaces = {"hello": "hi", "world":" 2020", "!":"."}
regex = re.sub("|".join(replaces.keys()), lambda match: replaces[match.string[match.start():match.end()]], text)
print(regex)

>>> a = '&#'
>>> print a.replace('&', r'\&')
\&#
>>> print a.replace('#', r'\#')
&\#
>>> 

You want to use a 'raw' string (denoted by the 'r' prefixing the replacement string), since raw strings to not treat the backslash specially.

Maybe a simple loop for chars to replace:

a = '&#'

to_replace = ['&', '#']

for char in to_replace:
    a = a.replace(char, "\\"+char)

print(a)

>>> \&\#

This will help someone looking for a simple solution.

def replacemany(our_str, to_be_replaced:tuple, replace_with:str):
    for nextchar in to_be_replaced:
        our_str = our_str.replace(nextchar, replace_with)
    return our_str

os = 'the rain in spain falls mainly on the plain ttttttttt sssssssssss nnnnnnnnnn'
tbr = ('a','t','s','n')
rw = ''

print(replacemany(os,tbr,rw))

Output:

he ri i pi fll mily o he pli