CODEWITHSUNDEEP
python
Allen S. Rout
Allen S. Rout

Reputation: 313

How can I filter items from a list in Python?

I have data naively collected from package dependency lists.

Depends: foo bar baz >= 5.2

I end up with 

 d = set(['foo','bar','baz','>=','5.2'])

I don't want the numerics and the operands.

In Perl I would

@new = grep {/^[a-z]+$/} @old

but I can't find a way to e.g. pass remove() a lambda, or something.

The closest I've come is ugly: 

[ item != None for item in [ re.search("^[a-zA-Z]+$",atom)   for atom in d] ]

which gets me a map of which values out of the set I want...if the order of the set is repeatable? I know that's not the case in Perl hashes.

I know how to iterate. :) I'm trying to do it the pythonesque Right Way

Upvotes: 21

Views: 47532

Answers (3)

Sharveash
Sharveash

Reputation: 51

What about using filter function In this example we are trying to only keep the even numbers in list_1.

 list_1 = [1, 2, 3, 4, 5]
 filtered_list = filter(lambda x : x%2 == 0, list_1)
 print(list(filtered_list))

This prints the new list which contains only the even numbers from list_1

Upvotes: 5

Kenan Banks
Kenan Banks

Reputation: 211982

No need for regular expressions here. Use str.isalpha. With and without list comprehensions:

my_list = ['foo','bar','baz','>=','5.2']

# With
only_words = [token for token in my_list if token.isalpha()]

# Without
only_words = filter(str.isalpha, my_list)

Personally I don't think you have to use a list comprehension for everything in Python, but I always get frowny-faced when I suggest map or filter answers.

Upvotes: 39

Ian Clelland
Ian Clelland

Reputation: 44132

How about 

d = set([item for item in d if re.match("^[a-zA-Z]+$",item)])

that gives you just the values you want, back in d (the order may be different, but that's the price you pay for using sets.

Upvotes: 2

Related Questions