How to remove empty string in a list?

Question

For example I have a sentence

"He is so .... cool!"

Then I remove all the punctuation and make it in a list.

["He", "is", "so", "", "cool"]

How do I remove or ignore the empty string?

Answer 1

You can use filter , with None as the key function, which filters out all elements which are False ish (including empty strings)

>>> lst = ["He", "is", "so", "", "cool"]
>>> filter(None, lst)
['He', 'is', 'so', 'cool']

Note however, that filter returns a list in Python 2, but a generator in Python 3. You will need to convert it into a list in Python 3, or use the list comprehension solution.

False ish values include:

False
None
0
''
[]
()
# and all other empty containers

Answer 2

You can filter it like this

orig = ["He", "is", "so", "", "cool"]
result = [x for x in orig if x]

Or you can use filter . In python 3 filter returns a generator, thus list() turns it into a list. This works also in python 2.7

result = list(filter(None, orig))

Answer 3

You can use a list comprehension:

cleaned = [x for x in your_list if x]

Although I would use regex to extract the words:

>>> import re
>>> sentence = 'This is some cool sentence with,    spaces'
>>> re.findall(r'(\w+)', sentence)
['This', 'is', 'some', 'cool', 'sentence', 'with', 'spaces']

Answer 4

I'll give you the answer to the question you should have asked -- how to avoid the empty string altogether. I assume you do something like this to get your list:

>>> "He is so .... cool!".replace(".", "").split(" ")
['He', 'is', 'so', '', 'cool!']

The point is that you use .split(" ") to split on space characters. However, if you leave out the argument to split , this happens:

>>> "He is so .... cool!".replace(".", "").split()
['He', 'is', 'so', 'cool!']

Quoth the docs:

If sep is not specified or is None, a different splitting algorithm is applied: runs of consecutive whitespace are regarded as a single separator, and the result will contain no empty strings at the start or end if the string has leading or trailing whitespace.

So you really don't need to bother with the other answers (except Blender's, which is a totally different approach), because split can do the job for you!

Answer 5

>>> from string import punctuation
>>> text = "He is so .... cool!"
>>> [w.strip(punctuation) for w in text.split() if w.strip(punctuation)]
['He', 'is', 'so', 'cool']

Answer 6

You can filter out empty strings very easily using a list comprehension:

x = ["He", "is", "so", "", "cool"]
x = [str for str in x if str]
>>> ['He', 'is', 'so', 'cool']

Answer 7

Python 3 从filter返回一个iterator filter ，所以应该包含在对 list() 的调用中

str_list = list(filter(None, str_list)) # fastest

Answer 8

lst = ["He", "is", "so", "", "cool"]
lst = list(filter(str.strip, lst))

Answer 9

You can do this with a filter .

a = ["He", "is", "so", "", "cool"]
filter(lambda s: len(s) > 0, a)