removing words from a list using list comprehension

Question

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?

Answer 1

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]

Answer 2

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

unWantedWords是一个数组，您将单词与一个数组共同映射，这就是它不起作用的原因。

Answer 3

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...

Answer 4

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)]