Applying a dictionary of string replacements to a list of strings

Question

Say I have a list of strings and a dictionary specifying replacements:

E.g.

my_replacements = {'1/2': 'half', '1/4': 'quarter', '3/4': 'three quarters'}

and a list of strings, where each string can possibly include keys from the above dictionary, e.g:

['I own 1/2 bottle', 'Give me 3/4 of the profit']

How can I apply the replacements to the list? What would be a Pythonic way to do this?

Answer 1

a = ['I own 1/2 bottle', 'Give me 3/4 of the profit']
b = {'1/2': 'half', '1/4': 'quarter', '3/4': 'three quarters'}

def replace(x):
    for what, new in b.items(): # or iteritems in Python 2
        x = x.replace(what, new)
    return x

print(list(map(replace, a)))

Output:

['I own half bottle', 'Give me three quarters of the profit']

Answer 2

O(n) solution:

reps = {'1/2': 'half', '1/4': 'quarter', '3/4': 'three quarters'}
li = ['I own 1/2 bottle', 'Give me 3/4 of the profit']

map(lambda s: ' '.join([reps.get(w,w) for w in s.split()]),li)
Out[6]: ['I own half bottle', 'Give me three quarters of the profit']

#for those who don't like `map`, the list comp version:
[' '.join([reps.get(w,w) for w in sentence.split()]) for sentence in li]
Out[9]: ['I own half bottle', 'Give me three quarters of the profit']

The issue with making lots of replace calls in a loop is that it makes your algorithm O(n**2). Not a big deal when you have a replacement dict of length 3, but when it gets large, suddenly you have a really slow algorithm that doesn't need to be.

As noted in comments, this approach fundamentally depends on being able to tokenize based on spaces - thus, if you have any whitespace in your replacement keys (say, you want to replace a series of words) this approach will not work. However being able to replace only-words is a far more frequent operation than needing to replace groupings-of-words, so I disagree with the commenters who believe that this approach isn't generic enough.