How to Replace Multiple Values at Once

Replace multiple strings with multiple other strings

As an answer to:

looking for an up-to-date answer

If you are using "words" as in your current example, you might extend the answer of Ben McCormick using a non capture group and add word boundaries \b at the left and at the right to prevent partial matches.

\b(?:cathy|cat|catch)\b

\b A word boundary to prevent a partial match
(?: Non capture group
- cathy|cat|catch match one of the alternatives
) Close non capture group
\b A word boundary to prevent a partial match

Example for the original question:

let str = "I have a cat, a dog, and a goat.";
const mapObj = {
  cat: "dog",
  dog: "goat",
  goat: "cat"
};
str = str.replace(/\b(?:cat|dog|goat)\b/gi, matched => mapObj[matched]);
console.log(str);

How to replace multiple substrings of a string?

Here is a short example that should do the trick with regular expressions:

import re

rep = {"condition1": "", "condition2": "text"} # define desired replacements here

# use these three lines to do the replacement
rep = dict((re.escape(k), v) for k, v in rep.iteritems()) 
#Python 3 renamed dict.iteritems to dict.items so use rep.items() for latest versions
pattern = re.compile("|".join(rep.keys()))
text = pattern.sub(lambda m: rep[re.escape(m.group(0))], text)

For example:

>>> pattern.sub(lambda m: rep[re.escape(m.group(0))], "(condition1) and --condition2--")
'() and --text--'

How to replace multiple values at once

A possible solution using match:

old <- 1:8
new <- c(2,4,6,8,1,3,5,7)

x[x %in% old] <- new[match(x, old, nomatch = 0)]

which gives:

> x
[1] 8 4 0 5 1 5 7 9

What this does:

Create two vectors: old with the values that need to be replaced and new with the corresponding replacements.
Use match to see where values from x occur in old. Use nomatch = 0 to remove the NA's. This results in an indexvector of the position in old for the x values
This index vector can then be used to index new.
Only assign the values from new to the positions of x that are present in old: x[x %in% old]

How to use str.replace to replace multiple pairs at once?

You can create a dictionary and pass it to the function replace() without needing to chain or name the function so many times.

replacers = {',':'','.':'','-':'','ltd':'limited'} #etc....
df1['CompanyA'] = df1['CompanyA'].replace(replacers)

Replace multiple characters in one replace call

If you want to replace multiple characters you can call the String.prototype.replace() with the replacement argument being a function that gets called for each match. All you need is an object representing the character mapping that you will use in that function.

For example, if you want a replaced with x, b with y, and c with z, you can do something like this:

const chars = {'a':'x','b':'y','c':'z'};
let s = '234abc567bbbbac';
s = s.replace(/[abc]/g, m => chars[m]);
console.log(s);

Output: 234xyz567yyyyxz

Pandas replace multiple values at once

try .replace({}, regex=True) method:

replacements = {
   'CITY': {
      r'(D.*DO|DOLLARD.*)': 'DOLLARD-DES-ORMEAUX',
      r'I[lL]*[eE]*.*': 'ILLE Perot'}
}

df.replace(replacements, regex=True, inplace=True)

print(df)

Output:

    SURNAME           ADDRESS                 CITY
0    Jenson    252 Des Ch├¬nes  DOLLARD-DES-ORMEAUX
1      Jean         236 Gouin  DOLLARD-DES-ORMEAUX
2     Denis   993 Boul. Gouin  DOLLARD-DES-ORMEAUX
3  Bradford   1690 Dollard #7  DOLLARD-DES-ORMEAUX
4   Alisson    115 Du Buisson           ILLE Perot
5     Abdul  9877 Boul. Gouin          Pierrefonds
6    O'Neil      5 Du College           ILLE Perot
7     Bundy   7345 Sherbrooke           ILLE Perot
8     Darcy   8671 Anthony #2           ILLE Perot
9     Adams       845 Georges          Pierrefonds

Replace multiple strings at once

You could extend the String object with your own function that does what you need (useful if there's ever missing functionality):

String.prototype.replaceArray = function(find, replace) {
  var replaceString = this;
  for (var i = 0; i < find.length; i++) {
    replaceString = replaceString.replace(find[i], replace[i]);
  }
  return replaceString;
};

For global replace you could use regex:

String.prototype.replaceArray = function(find, replace) {
  var replaceString = this;
  var regex; 
  for (var i = 0; i < find.length; i++) {
    regex = new RegExp(find[i], "g");
    replaceString = replaceString.replace(regex, replace[i]);
  }
  return replaceString;
};

To use the function it'd be similar to your PHP example:

var textarea = $(this).val();
var find = ["<", ">", "\n"];
var replace = ["<", ">", "<br/>"];
textarea = textarea.replaceArray(find, replace);