Easiest way to replace a string using a dictionary of replacements?

Consider..

dict = {
'Спорт':'Досуг',
'russianA':'englishA'
}

s = 'Спорт russianA'

I’d like to replace all dict keys with their respective dict values in s.

Answers:

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Method 1

Using re:

import re

s = 'Спорт not russianA'
d = {
'Спорт':'Досуг',
'russianA':'englishA'
}

pattern = re.compile(r'b(' + '|'.join(d.keys()) + r')b')
result = pattern.sub(lambda x: d[x.group()], s)
# Output: 'Досуг not englishA'

This will match whole words only. If you don’t need that, use the pattern:

pattern = re.compile('|'.join(d.keys()))

Note that in this case you should sort the words descending by length if some of your dictionary entries are substrings of others.

Method 2

You could use the reduce function:

reduce(lambda x, y: x.replace(y, dict[y]), dict, s)

Method 3

Solution found here (I like its simplicity):

def multipleReplace(text, wordDict):
    for key in wordDict:
        text = text.replace(key, wordDict[key])
    return text

Method 4

one way, without re

d = {
'Спорт':'Досуг',
'russianA':'englishA'
}

s = 'Спорт russianA'.split()
for n,i in enumerate(s):
    if i in d:
        s[n]=d[i]
print ' '.join(s)

Method 5

Almost the same as ghostdog74, though independently created. One difference,
using d.get() in stead of d[] can handle items not in the dict.

>>> d = {'a':'b', 'c':'d'}
>>> s = "a c x"
>>> foo = s.split()
>>> ret = []
>>> for item in foo:
...   ret.append(d.get(item,item)) # Try to get from dict, otherwise keep value
... 
>>> " ".join(ret)
'b d x'

Method 6

I used this in a similar situation (my string was all in uppercase):

def translate(string, wdict):
    for key in wdict:
        string = string.replace(key, wdict[key].lower())
    return string.upper()

hope that helps in some way… 🙂

Method 7

With the warning that it fails if key has space, this is a compressed solution similar to ghostdog74 and extaneons answers:

d = {
'Спорт':'Досуг',
'russianA':'englishA'
}

s = 'Спорт russianA'

' '.join(d.get(i,i) for i in s.split())

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0