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?
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Just a reminder that some english words contain punctuation. Depending on how you intend to use this, you may need to take this into consideration.Lauritz V. Thaulow– Lauritz V. Thaulow2013-04-19 08:08:18 +00:00Commented Apr 19, 2013 at 8:08
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9 Answers
You can use filter, with None as the key function, which filters out all elements which are Falseish (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.
Falseish values include:
False
None
0
''
[]
()
# and all other empty containers
4 Comments
Yuushi
Be careful however, as this will return a
list in Python 2 and a generator in Python 3.
tumultous_rooster
what happens if you have
['']?
Volatility
@MattO'Brien since the list has an element, it is non-empty, and therefore it is considered
Trueish.tumultous_rooster
A
False inside a True, ehYou 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))
Comments
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']
Comments
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!
answered Apr 19, 2013 at 7:47
Lauritz V. Thaulow
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Comments
>>> 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']
2 Comments
glglgl
In order to avoid calling
w.strip(punctuation) twice, l = [w.strip(punctuation) for w in text.split()] and then [w for w in l if w] wuld be preferable...
jamylak
@glglgl I'm okay with calling it twice in this case, but yes that is also an option. I would rather use a generator though
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']
Comments
Python 3 returns an iterator from filter, so should be wrapped in a call to list()
str_list = list(filter(None, str_list)) # fastest
Comments
lst = ["He", "is", "so", "", "cool"]
lst = list(filter(str.strip, lst))
Comments
You can do this with a filter.
a = ["He", "is", "so", "", "cool"]
filter(lambda s: len(s) > 0, a)
