从列表（如白名单）中删除除某些单词外的所有单词

Question

Imagine a list that has random words:

words = ['elephant', 'dog', 'blue', 'sam', 'white', 'red', 'sun', 'moon']

And I want to remove all but the following words (like a whitelist):

colors = ['red', 'green', 'blue', 'orange', 'white']

And I want to produce the following list (order matters):

filtered = ['blue', 'white', 'red']

I've thought about something like this (which works fine):

filtered = filter (lambda a: a == 'red' or a == 'green' or a == 'blue' or a == 'orange' or a == 'white', words)

But is this really the best / most efficient way?

Answer 1

If you want to keep order and efficiently filter out non-colors, create a set of colors, so that in checking is faster and then you can just go thru all words and filter out non-colors

words = ['elephant', 'dog', 'blue', 'sam', 'white', 'red', 'sun', 'moon']
colors = set(['red', 'green', 'blue', 'orange', 'white'])
print [word for word in words if word in colors]

output:

['blue', 'white', 'red']

Answer 2

words = ['elephant', 'dog', 'blue', 'sam', 'white', 'red', 'sun', 'moon']
filterset = frozenset(['red', 'green', 'blue', 'orange', 'white'])
filtered = [x for x in words if x in filterset]

This solution has the advantage that even for a relatively large filterset it will be relatively fast, and it doesn't assume that the words list contains only unique entries.

You could leave the filterset as just your filterlist , but this will hurt performance, especially if the list is large.

Answer 3

filtered = filter(lambda a: a in whitelis, words)

should do the trick

this can also be written as a list comprehension

filtered = [x for x in letters if x in whitelist]

as mentioned below, you can use the set type to make sure that every word in the whitelist is unique. This is useful when your whitelist is not hardcoded, but somehow generated, for example from records in a database.

Answer 4

use set operations:

words = ['elephant', 'dog', 'blue', 'sam', 'white', 'red', 'sun', 'moon']
colors = ['red', 'green', 'blue', 'orange', 'white']
filtered = set(words).difference(colors)

Answer 5

Although list comprehension are often considered as more pythonic, I do like functional filter if we can write it without lambda :

>> filter(set(colors).__contains__, words)
['blue', 'white', 'red']