[Django]-Iterative find/replace from a list of tuples in Python

by

6👍

✅

You could consider using re.sub:

import re
REPLACEMENTS = dict([('find1', 'replace1'),
                     ('find2', 'replace2'),
                     ('find3', 'replace3')])

def replacer(m):
    return REPLACEMENTS[m.group(0)]

x = 'find1, find2, find3'
r = re.compile('|'.join(REPLACEMENTS.keys()))
print r.sub(replacer, x)
👤mhawke

1👍

A couple notes:

  1. The boilerplate argument about premature optimization, benchmarking, bottlenecks, 100 is small, etc.
  2. There are cases where the different solutions will return different results. if y = [('one', 'two'), ('two', 'three')] and x = 'one' then mhawke’s solution gives you 'two' and Unknown’s gives 'three'.
  3. Testing this out in a silly contrived example mhawke’s solution was a tiny bit faster. It should be easy to try it with your data though.
👤jtb

0👍

x = 'find1, find2, find3'
y = [('find1', 'replace1'), ('find2', 'replace2'), ('find3', 'replace3')]

def processThis(str,lst):
    for find, replace in lst:
        str = str.replace(find, replace)

    return str

>>> processThis(x,y)
'replace1, replace2, replace3'
👤Unknown

0👍

s = reduce(lambda x, repl: str.replace(x, *repl), lst, s)
👤Glenn Maynard

0👍

Same answer as mhawke, enclosed with method str_replace

def str_replace(data, search_n_replace_dict):
    import re
    REPLACEMENTS = search_n_replace_dict

    def replacer(m):
        return REPLACEMENTS[m.group(0)]

    r = re.compile('|'.join(REPLACEMENTS.keys()))
    return r.sub(replacer, data)

Then we can call this method with example as below

s = "abcd abcd efgh efgh;;;;;; lkmnkd kkkkk"
d = dict({ 'abcd' : 'aaaa', 'efgh' : 'eeee', 'mnkd' : 'mmmm' })


print (s)
print ("\n")
print(str_replace(s, d))

output :

abcd abcd efgh efgh;;;;;; lkmnkd kkkkk


aaaa aaaa eeee eeee;;;;;; lkmmmm kkkkk
👤leela

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