Remove empty strings from a list of strings

I would use filter:

str_list = filter(None, str_list)
str_list = filter(bool, str_list)
str_list = filter(len, str_list)
str_list = filter(lambda item: item, str_list)

Python 3 returns an iterator from filter, so should be wrapped in a call to list()

str_list = list(filter(None, str_list))

Using a list comprehension is the most Pythonic way:

>>> strings = ["first", "", "second"]
>>> [x for x in strings if x]
['first', 'second']

If the list must be modified in-place, because there are other references which must see the updated data, then use a slice assignment:

strings[:] = [x for x in strings if x]

filter actually has a special option for this:

filter(None, sequence)

It will filter out all elements that evaluate to False. No need to use an actual callable here such as bool, len and so on.

It's equally fast as map(bool, ...)

>>> lstr = ['hello', '', ' ', 'world', ' ']
>>> lstr
['hello', '', ' ', 'world', ' ']

>>> ' '.join(lstr).split()
['hello', 'world']

>>> filter(None, lstr)
['hello', ' ', 'world', ' ']

Compare time

>>> from timeit import timeit
>>> timeit('" ".join(lstr).split()', "lstr=['hello', '', ' ', 'world', ' ']", number=10000000)
4.226747989654541
>>> timeit('filter(None, lstr)', "lstr=['hello', '', ' ', 'world', ' ']", number=10000000)
3.0278358459472656

Notice that filter(None, lstr) does not remove empty strings with a space ' ', it only prunes away '' while ' '.join(lstr).split() removes both.

To use filter() with white space strings removed, it takes a lot more time:

>>> timeit('filter(None, [l.replace(" ", "") for l in lstr])', "lstr=['hello', '', ' ', 'world', ' ']", number=10000000)
18.101892948150635