CODEWITHSUNDEEP
pythonlist
JPP
JPP

Reputation: 143

removing words from a list using list comprehension

I´m trying to remove unnecessary words (an, a, the) from a list

Test = ['a', 'an', 'the', 'love']
unWantedWords = ['a', 'an', 'the']
RD1 = [x for x in Test if x != unWantedWords]
print(RD1)
output ->['a', 'an', 'the', 'love']

what is wrong w/ this?

Upvotes: 1

Views: 1547

Answers (4)

Don
Don

Reputation: 17606

If you don't mind:

  1. removing duplicates
  2. preserving the original order

you can simply use 'set' (here is the core algorithm):

>>> Test = ['a', 'an', 'the', 'love']
>>> unWantedWords = ['a', 'an', 'the']
>>> print set(Test) - set(unWantedWords)
set(['love'])

>>> print list(set(Test) - set(unWantedWords))
['love']

This has the advantage of an optimized complexity.

Of course you can wrap this code in order to keep duplicates and order...

Upvotes: 2

shshank
shshank

Reputation: 2641

This is wrong.

RD1 = [x for x in Test if x != unWantedWords]

your condition of if x != unWantedWords checks if x is equal to the list unWantedWords, instead of checking if x exists or not in unWantedWords.

The condition always is true because x is a string and not a list. Therefore all your words are added to the list.

The correct idiom would be if x not in unWantedWords.

You can do, RD1 = [x for x in Test if x not in set(unWantedWords)]

Upvotes: 1

dejester
dejester

Reputation: 656

The problem is you are comparing the value x to the entire list unWantedWords. 

RD1 = [x for x in Test if x != unWantedWords]

Replace != with not in to check if x is... not in! 

RD1 = [x for x in Test if x not in unWantedWords]

Upvotes: 5

VIKASH JAISWAL
VIKASH JAISWAL

Reputation: 838

RD1 = [x for x in Test if x not in unWantedWords]

unWantedWords is an array and you are comapring your words with an array that's why it's not working.

Upvotes: 4

Related Questions