46

Is there a far shorter way to write the following code?

my_string = my_string.replace('A', '1')
my_string = my_string.replace('B', '2')
my_string = my_string.replace('C', '3')
my_string = my_string.replace('D', '4')
my_string = my_string.replace('E', '5')

Note that I don't need those exact values replaced; I'm simply looking for a way to turn 5+ lines into fewer than 5

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8 Answers 8

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74

Looks like a good opportunity to use a loop:

mapping = { 'A':'1', 'B':'2', 'C':'3', 'D':'4', 'E':'5'}
for k, v in mapping.iteritems():
    my_string = my_string.replace(k, v)

A faster approach if you don't mind the parentheses would be:

mapping = [ ('A', '1'), ('B', '2'), ('C', '3'), ('D', '4'), ('E', '5') ]
for k, v in mapping:
    my_string = my_string.replace(k, v)
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6 Comments

It's odd; that's the way I'd write it, because the problem is all about mapping from one character to another. But really, you're not using the dict as a dict; you're using it as a list of tuples. So why not write a list of tuples in the first place?
Certainly a list of tuples would also work (and be faster). I'll edit the answer to include that option.
What if you will have this mapping { 'A':'1', '1':'A'} probably the result will not be the same as expected
A more concise way to define mapping = zip(list('ABCDE'),range(1,6))
For Python 3, use .items() instead of .iteritems()
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51

You can easily use string.maketrans() to create the mapping string to pass to str.translate():

import string
trans = string.maketrans("ABCDE","12345")
my_string = my_string.translate(trans)
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4 Comments

this is nice, but doesn't work in the case of multi-character patterns or replacements
Correct. I assumed from the question that it was character-based replacement.
This is actually the only correct way to replace simultaneously a list of characters. For example, if you do '<>'.replace('<', '>').replace('>','<') you will get '<<', but with translation you get '><'.
in python 3 it is str.maketrans("ABCDE", "12345")
17

If you want to get the wrong answer, slowly, then use string.replace in a loop. (Though it does work in this case of no overlap among the patterns and replacements.)

For the general case with possible overlaps or a long subject string, use re.sub:

import re

def multisub(subs, subject):
    "Simultaneously perform all substitutions on the subject string."
    pattern = '|'.join('(%s)' % re.escape(p) for p, s in subs)
    substs = [s for p, s in subs]
    replace = lambda m: substs[m.lastindex - 1]
    return re.sub(pattern, replace, subject)

>>> multisub([('hi', 'bye'), ('bye', 'hi')], 'hi and bye')
'bye and hi'

For the special case of 1-character patterns and 1- or 0-character replacements, use string.maketrans.

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3 Comments

+1 on the approach of using re.sub, -1 for saying that string.replace gives the wrong answer, since it's not clear exactly what should happen for overlapping matches. So net 0 :)
str.replace can give an arguably wrong answer. For dictionary-based solutions, consider the string "wow" and translation table {'w': 'foo', 'o': 'bar'}. Depending on the order that the dict iterates, you can end up with different results: "fbarbarbarfbarbar" or "foobarfoo". A function that gives different output for equal input is arguably broken.
Miles, yes, that's the kind of problem I meant. You could also run into overlaps in the keys alone, like 'hello' and 'hell'.
14

Also look into str.translate(). It replaces characters according to a mapping you provide for Unicode strings, or otherwise must be told what to replace each character from chr(0) to chr(255) with.

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Comments

9 
replaceDict = {'A':'1','B':'2','C':'3','D':'4','E':'5'}       
for key, replacement in replaceDict.items():  
  my_string = my_string.replace( key, replacement )
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Comments

7

I think it could be a little more efficient:

mapping = { 'A':'1', 'B':'2', 'C':'3', 'D':'4', 'E':'5'}
my_string = "".join([mapping[c] if c in mapping else c for c in my_string])

I suggest some benchmark with "timeit", with real cases in base of the lenght of "my_string".

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2 Comments

this is the best solution IMO if you have multiple replacements at once and if the replacements are not always single characters.
Finally someone used a dictionary as it it was intended. All others are O(n * m) but your method is O(n * log(m)) or better, where m is the size of the dictionary
2

You can do it in one line using Pandas.

import pandas as pd

my_string="A B C test"

my_string =pd.DataFrame([my_string])[0].replace(["A","B","C","D","E"],['1','2','3','4','5'],regex=True)[0]

print(my_string)
'1 2 3 test'
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Comments

1

One way I do it is with an associated array (a dictionary). Here is an example of the replacements I use when getting a file ready for deployment in LaTeX using regular expressions.

  import re
  def escapeTexString(string): # Returns TeX-friendly string
    rep = { # define desired replacements in this dictionary (mapping)
         '&': '\\&',
         '%': '\\%',
         '#': '\\#',
         '_': '\\_',
         '{': '\\{', # REGEX Special
         '}': '\\}', # REGEX Special
         '~': '\\char"007E{}', # LaTeX Special
         '$': '\\$', # REGEX Special
         '\\': '\\char"005C{}', # REGEX/LaTeX Special
         '^': '\\char"005E{}', # REGEX/LaTeX Special
         '"': '\\char"FF02{}'
        }
    # use these two lines to do the replacement (could be shortened to one line)
    pattern = re.compile("|".join(map(re.escape,rep.keys()))) # Create single pattern object (key to simultaneous replacement)
    new_string = pattern.sub(lambda match: rep[match.group(0)], string)
    return new_string
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1 Comment

Please read the other answers before posting. Your answer does exactly the same thing as Darius' answer from 8 years earlier.

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