Python：替换列表中特定单词的所有出现的快速方法？

Question

I have a list of words. It's pretty large (len(list) ~ 70,000). I'm currently using this code:

replacement = "bla"
for word in data:
    if (word in unique_words):
        word = replacement

This code take a while to perform the operation. Is there a quicker way to do this?

Answer 1

Use a set for unique_words . Sets are considerably faster than lists for determining if an item is in them (see Python Sets vs Lists ).

Also, it's only a stylistic issue but I think you should drop the brackets in the if . It looks cleaner.

Answer 2

The code you have posted doesn't actually do any replacement. Here is a snippet that does:

for key,word in enumerate(data):
   if word in unique_words:
       data[key] = replacement

Here's a more compact way:

new_list = [replacement if word in unique_words else word for word in big_list]

I think unique_words is an odd name for the variable considering its use, perhaps it should be search_list ?

Edit :

After your comment, perhaps this is better:

from collections import Counter
c = Counter(data)
only_once = [k for k,v in c.iteritems() if v == 1]

# Now replace all occurances of these words with something else

for k, v in enumerate(data):
    if v in only_once:
        data[k] = replacement