Remove empty strings from a list of strings

Question

I want to remove all empty strings from a list of strings in python.

My idea looks like this:

while \'\' in str_list:
    str_list.remove(\'\')


        

Answer 1

Keep in mind that if you want to keep the white spaces within a string, you may remove them unintentionally using some approaches. If you have this list

['hello world', ' ', '', 'hello'] what you may want ['hello world','hello']

first trim the list to convert any type of white space to empty string:

space_to_empty = [x.strip() for x in _text_list]

then remove empty string from them list

space_clean_list = [x for x in space_to_empty if x]

Answer 2

As reported by Aziz Alto filter(None, lstr) does not remove empty strings with a space ' ' but if you are sure lstr contains only string you can use filter(str.strip, lstr)

>>> lstr = ['hello', '', ' ', 'world', ' ']
>>> lstr
['hello', '', ' ', 'world', ' ']
>>> ' '.join(lstr).split()
['hello', 'world']
>>> filter(str.strip, lstr)
['hello', 'world']

Compare time on my pc

>>> from timeit import timeit
>>> timeit('" ".join(lstr).split()', "lstr=['hello', '', ' ', 'world', ' ']", number=10000000)
3.356455087661743
>>> timeit('filter(str.strip, lstr)', "lstr=['hello', '', ' ', 'world', ' ']", number=10000000)
5.276503801345825

The fastest solution to remove '' and empty strings with a space ' ' remains ' '.join(lstr).split().

As reported in a comment the situation is different if your strings contain spaces.

>>> lstr = ['hello', '', ' ', 'world', '    ', 'see you']
>>> lstr
['hello', '', ' ', 'world', '    ', 'see you']
>>> ' '.join(lstr).split()
['hello', 'world', 'see', 'you']
>>> filter(str.strip, lstr)
['hello', 'world', 'see you']

You can see that filter(str.strip, lstr) preserve strings with spaces on it but ' '.join(lstr).split() will split this strings.

Answer 3

Use filter:

newlist=filter(lambda x: len(x)>0, oldlist) 

The drawbacks of using filter as pointed out is that it is slower than alternatives; also, lambda is usually costly.

Or you can go for the simplest and the most iterative of all:

# I am assuming listtext is the original list containing (possibly) empty items
for item in listtext:
    if item:
        newlist.append(str(item))
# You can remove str() based on the content of your original list

this is the most intuitive of the methods and does it in decent time.

Answer 4

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))

Answer 5

Sum up best answers:

1. Eliminate emtpties WITHOUT stripping:

That is, all-space strings are retained:

slist = list(filter(None, slist))

PROs:

• simplest;
• fastest (see benchmarks below).

2. To eliminate empties after stripping ...

2.a ... when strings do NOT contain spaces between words:

slist = ' '.join(slist).split()

PROs:

• small code
• fast (BUT not fastest with big datasets due to memory, contrary to what @paolo-melchiorre results)

2.b ... when strings contain spaces between words?

slist = list(filter(str.strip, slist))

PROs:

• fastest;
• understandability of the code.

Benchmarks on a 2018 machine:

## Build test-data
#
import random, string
nwords = 10000
maxlen = 30
null_ratio = 0.1
rnd = random.Random(0)                  # deterministic results
words = [' ' * rnd.randint(0, maxlen)
         if rnd.random() > (1 - null_ratio)
         else
         ''.join(random.choices(string.ascii_letters, k=rnd.randint(0, maxlen)))
         for _i in range(nwords)
        ]

## Test functions
#
def nostrip_filter(slist):
    return list(filter(None, slist))

def nostrip_comprehension(slist):
    return [s for s in slist if s]

def strip_filter(slist):
    return list(filter(str.strip, slist))

def strip_filter_map(slist): 
    return list(filter(None, map(str.strip, slist))) 

def strip_filter_comprehension(slist):  # waste memory
    return list(filter(None, [s.strip() for s in slist]))

def strip_filter_generator(slist):
    return list(filter(None, (s.strip() for s in slist)))

def strip_join_split(slist):  # words without(!) spaces
    return ' '.join(slist).split()

## Benchmarks
#
%timeit nostrip_filter(words)
142 µs ± 16.8 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

%timeit nostrip_comprehension(words)
263 µs ± 19.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit strip_filter(words)
653 µs ± 37.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit strip_filter_map(words)
642 µs ± 36 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit strip_filter_comprehension(words)
693 µs ± 42.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit strip_filter_generator(words)
750 µs ± 28.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit strip_join_split(words)
796 µs ± 103 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Answer 6

For a list with a combination of spaces and empty values, use simple list comprehension -

>>> s = ['I', 'am', 'a', '', 'great', ' ', '', '  ', 'person', '!!', 'Do', 'you', 'think', 'its', 'a', '', 'a', '', 'joke', '', ' ', '', '?', '', '', '', '?']

So, you can see, this list has a combination of spaces and null elements. Using the snippet -

>>> d = [x for x in s if x.strip()]
>>> d
>>> d = ['I', 'am', 'a', 'great', 'person', '!!', 'Do', 'you', 'think', 'its', 'a', 'a', 'joke', '?', '?']