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

Question

Consider..

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

s = 'Спорт russianA'

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

Answer 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.

Answer 2

您可以使用reduce函数：

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

Answer 3

Solution found here (I like its simplicity):

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

Answer 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)

Answer 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'

Answer 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... :)

Answer 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())