Best way to replace multiple characters in a string?

Question

I need to replace some characters as follows: & ➔ \&, # ➔ \#, ...

I coded as follows, but I guess there should be some better way. Any hints?

strs = strs.replace('&', '\&')
strs = strs.replace('#', '\#')
...

Answer 1

Multiple replace with .translate() function

# Original string
original_string = "12#4526&18##0"

translation_dict = {'&': r'\&',
                    '#': r'\#'}

# Create a translation table using the dictionary
translation_table = str.maketrans(translation_dict)

# Translate the string
modified_string = original_string.translate(translation_table)

print(modified_string)  # Output: 12\#4526\&18\#\#0

Answer 2

Example is given below for the or condition, it will delete all ' and , from the given string. pass as many characters as you want separated by |

import re
test = re.sub("('|,)","",str(jsonAtrList))

Before:

After:

Answer 3

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)

Answer 4

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

Answer 5

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)

Answer 6

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

Answer 7

Maybe a simple loop for chars to replace:

a = '&#'

to_replace = ['&', '#']

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

print(a)

>>> \&\#

Answer 8

Replacing two characters

I timed all the methods in the current answers along with one extra.

With an input string of abc&def#ghi and replacing & -> \& and # -> \#, the fastest way was to chain together the replacements like this: text.replace('&', '\&').replace('#', '\#').

Timings for each function:

• a) 1000000 loops, best of 3: 1.47 μs per loop
• b) 1000000 loops, best of 3: 1.51 μs per loop
• c) 100000 loops, best of 3: 12.3 μs per loop
• d) 100000 loops, best of 3: 12 μs per loop
• e) 100000 loops, best of 3: 3.27 μs per loop
• f) 1000000 loops, best of 3: 0.817 μs per loop
• g) 100000 loops, best of 3: 3.64 μs per loop
• h) 1000000 loops, best of 3: 0.927 μs per loop
• i) 1000000 loops, best of 3: 0.814 μs per loop

Here are the functions:

def a(text):
    chars = "&#"
    for c in chars:
        text = text.replace(c, "\\" + c)


def b(text):
    for ch in ['&','#']:
        if ch in text:
            text = text.replace(ch,"\\"+ch)


import re
def c(text):
    rx = re.compile('([&#])')
    text = rx.sub(r'\\\1', text)


RX = re.compile('([&#])')
def d(text):
    text = RX.sub(r'\\\1', text)


def mk_esc(esc_chars):
    return lambda s: ''.join(['\\' + c if c in esc_chars else c for c in s])
esc = mk_esc('&#')
def e(text):
    esc(text)


def f(text):
    text = text.replace('&', '\&').replace('#', '\#')


def g(text):
    replacements = {"&": "\&", "#": "\#"}
    text = "".join([replacements.get(c, c) for c in text])


def h(text):
    text = text.replace('&', r'\&')
    text = text.replace('#', r'\#')


def i(text):
    text = text.replace('&', r'\&').replace('#', r'\#')

Timed like this:

python -mtimeit -s"import time_functions" "time_functions.a('abc&def#ghi')"
python -mtimeit -s"import time_functions" "time_functions.b('abc&def#ghi')"
python -mtimeit -s"import time_functions" "time_functions.c('abc&def#ghi')"
python -mtimeit -s"import time_functions" "time_functions.d('abc&def#ghi')"
python -mtimeit -s"import time_functions" "time_functions.e('abc&def#ghi')"
python -mtimeit -s"import time_functions" "time_functions.f('abc&def#ghi')"
python -mtimeit -s"import time_functions" "time_functions.g('abc&def#ghi')"
python -mtimeit -s"import time_functions" "time_functions.h('abc&def#ghi')"
python -mtimeit -s"import time_functions" "time_functions.i('abc&def#ghi')"

---

Replacing 17 characters

Here's similar code to do the same but with more characters to escape (\`*_{}>#+-.!$):

def a(text):
    chars = "\\`*_{}[]()>#+-.!$"
    for c in chars:
        text = text.replace(c, "\\" + c)


def b(text):
    for ch in ['\\','`','*','_','{','}','[',']','(',')','>','#','+','-','.','!','$','\'']:
        if ch in text:
            text = text.replace(ch,"\\"+ch)


import re
def c(text):
    rx = re.compile('([&#])')
    text = rx.sub(r'\\\1', text)


RX = re.compile('([\\`*_{}[]()>#+-.!$])')
def d(text):
    text = RX.sub(r'\\\1', text)


def mk_esc(esc_chars):
    return lambda s: ''.join(['\\' + c if c in esc_chars else c for c in s])
esc = mk_esc('\\`*_{}[]()>#+-.!$')
def e(text):
    esc(text)


def f(text):
    text = text.replace('\\', '\\\\').replace('`', '\`').replace('*', '\*').replace('_', '\_').replace('{', '\{').replace('}', '\}').replace('[', '\[').replace(']', '\]').replace('(', '\(').replace(')', '\)').replace('>', '\>').replace('#', '\#').replace('+', '\+').replace('-', '\-').replace('.', '\.').replace('!', '\!').replace('$', '\$')


def g(text):
    replacements = {
        "\\": "\\\\",
        "`": "\`",
        "*": "\*",
        "_": "\_",
        "{": "\{",
        "}": "\}",
        "[": "\[",
        "]": "\]",
        "(": "\(",
        ")": "\)",
        ">": "\>",
        "#": "\#",
        "+": "\+",
        "-": "\-",
        ".": "\.",
        "!": "\!",
        "$": "\$",
    }
    text = "".join([replacements.get(c, c) for c in text])


def h(text):
    text = text.replace('\\', r'\\')
    text = text.replace('`', r'\`')
    text = text.replace('*', r'\*')
    text = text.replace('_', r'\_')
    text = text.replace('{', r'\{')
    text = text.replace('}', r'\}')
    text = text.replace('[', r'\[')
    text = text.replace(']', r'\]')
    text = text.replace('(', r'\(')
    text = text.replace(')', r'\)')
    text = text.replace('>', r'\>')
    text = text.replace('#', r'\#')
    text = text.replace('+', r'\+')
    text = text.replace('-', r'\-')
    text = text.replace('.', r'\.')
    text = text.replace('!', r'\!')
    text = text.replace('$', r'\$')


def i(text):
    text = text.replace('\\', r'\\').replace('`', r'\`').replace('*', r'\*').replace('_', r'\_').replace('{', r'\{').replace('}', r'\}').replace('[', r'\[').replace(']', r'\]').replace('(', r'\(').replace(')', r'\)').replace('>', r'\>').replace('#', r'\#').replace('+', r'\+').replace('-', r'\-').replace('.', r'\.').replace('!', r'\!').replace('$', r'\$')

Here's the results for the same input string abc&def#ghi:

• a) 100000 loops, best of 3: 6.72 μs per loop
• b) 100000 loops, best of 3: 2.64 μs per loop
• c) 100000 loops, best of 3: 11.9 μs per loop
• d) 100000 loops, best of 3: 4.92 μs per loop
• e) 100000 loops, best of 3: 2.96 μs per loop
• f) 100000 loops, best of 3: 4.29 μs per loop
• g) 100000 loops, best of 3: 4.68 μs per loop
• h) 100000 loops, best of 3: 4.73 μs per loop
• i) 100000 loops, best of 3: 4.24 μs per loop

And with a longer input string (## *Something* and [another] thing in a longer sentence with {more} things to replace$):

• a) 100000 loops, best of 3: 7.59 μs per loop
• b) 100000 loops, best of 3: 6.54 μs per loop
• c) 100000 loops, best of 3: 16.9 μs per loop
• d) 100000 loops, best of 3: 7.29 μs per loop
• e) 100000 loops, best of 3: 12.2 μs per loop
• f) 100000 loops, best of 3: 5.38 μs per loop
• g) 10000 loops, best of 3: 21.7 μs per loop
• h) 100000 loops, best of 3: 5.7 μs per loop
• i) 100000 loops, best of 3: 5.13 μs per loop

Adding a couple of variants:

def ab(text):
    for ch in ['\\','`','*','_','{','}','[',']','(',')','>','#','+','-','.','!','$','\'']:
        text = text.replace(ch,"\\"+ch)


def ba(text):
    chars = "\\`*_{}[]()>#+-.!$"
    for c in chars:
        if c in text:
            text = text.replace(c, "\\" + c)

With the shorter input:

• ab) 100000 loops, best of 3: 7.05 μs per loop
• ba) 100000 loops, best of 3: 2.4 μs per loop

With the longer input:

• ab) 100000 loops, best of 3: 7.71 μs per loop
• ba) 100000 loops, best of 3: 6.08 μs per loop

So I'm going to use ba for readability and speed.

Addendum

Prompted by haccks in the comments, one difference between ab and ba is the if c in text: check. Let's test them against two more variants:

def ab_with_check(text):
    for ch in ['\\','`','*','_','{','}','[',']','(',')','>','#','+','-','.','!','$','\'']:
        if ch in text:
            text = text.replace(ch,"\\"+ch)

def ba_without_check(text):
    chars = "\\`*_{}[]()>#+-.!$"
    for c in chars:
        text = text.replace(c, "\\" + c)

Times in μs per loop on Python 2.7.14 and 3.6.3, and on a different machine from the earlier set, so cannot be compared directly.

╭────────────╥──────┬───────────────┬──────┬──────────────────╮
│ Py, input  ║  ab  │ ab_with_check │  ba  │ ba_without_check │
╞════════════╬══════╪═══════════════╪══════╪══════════════════╡
│ Py2, short ║ 8.81 │    4.22       │ 3.45 │    8.01          │
│ Py3, short ║ 5.54 │    1.34       │ 1.46 │    5.34          │
├────────────╫──────┼───────────────┼──────┼──────────────────┤
│ Py2, long  ║ 9.3  │    7.15       │ 6.85 │    8.55          │
│ Py3, long  ║ 7.43 │    4.38       │ 4.41 │    7.02          │
└────────────╨──────┴───────────────┴──────┴──────────────────┘

We can conclude that:

• Those with the check are up to 4x faster than those without the check

• ab_with_check is slightly in the lead on Python 3, but ba (with check) has a greater lead on Python 2

• However, the biggest lesson here is Python 3 is up to 3x faster than Python 2! There's not a huge difference between the slowest on Python 3 and fastest on Python 2!

Answer 9

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.

Answer 10

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.

Answer 11

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.

Answer 12

FYI, this is of little or no use to the OP but it may be of use to other readers (please do not downvote, I'm aware of this).

As a somewhat ridiculous but interesting exercise, wanted to see if I could use python functional programming to replace multiple chars. I'm pretty sure this does NOT beat just calling replace() twice. And if performance was an issue, you could easily beat this in rust, C, julia, perl, java, javascript and maybe even awk. It uses an external 'helpers' package called pytoolz, accelerated via cython (cytoolz, it's a pypi package).

from cytoolz.functoolz import compose
from cytoolz.itertoolz import chain,sliding_window
from itertools import starmap,imap,ifilter
from operator import itemgetter,contains
text='&hello#hi&yo&'
char_index_iter=compose(partial(imap, itemgetter(0)), partial(ifilter, compose(partial(contains, '#&'), itemgetter(1))), enumerate)
print '\\'.join(imap(text.__getitem__, starmap(slice, sliding_window(2, chain((0,), char_index_iter(text), (len(text),))))))

I'm not even going to explain this because no one would bother using this to accomplish multiple replace. Nevertheless, I felt somewhat accomplished in doing this and thought it might inspire other readers or win a code obfuscation contest.

Answer 13

Simply chain the replace functions like this

strs = "abc&def#ghi"
print strs.replace('&', '\&').replace('#', '\#')
# abc\&def\#ghi

If the replacements are going to be more in number, you can do this in this generic way

strs, replacements = "abc&def#ghi", {"&": "\&", "#": "\#"}
print "".join([replacements.get(c, c) for c in strs])
# abc\&def\#ghi

Answer 14

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.

Answer 15

>>> 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.

Answer 16

>>> string="abc&def#ghi"
>>> for ch in ['&','#']:
...   if ch in string:
...      string=string.replace(ch,"\\"+ch)
...
>>> print string
abc\&def\#ghi

Answer 17

Are you always going to prepend a backslash? If so, try

import re
rx = re.compile('([&#])')
#                  ^^ fill in the characters here.
strs = rx.sub('\\\\\\1', strs)

It may not be the most efficient method but I think it is the easiest.