Can't seem to remove all empty strings and blank strings from a list?

Question

I have a long list of information scraped from the Billboard Hot 100 page which I am trying to distill into just songs and artists. However, I can't seem to remove all the empty parts as well as points in the list that are just blank spaces. '' and ' '

I have tried everything I can find on here with regards to the filter function as well as other options people have suggested. My initial attempt was using the included code, which worked for the first part of the list but for some reason never finishes and always stops part way.

for y in parse:
    if y == "":
        parse.remove("")
    elif y == " ":
        parse.remove(" ")

I would expect to be receiving a list with no '' or ' ' by themselves, but only the first part of the list is affected.

You are modifying the list as you iterating over it, this will cause items to be skipped. You should use something like a list comprehension or filter for this. For example: parse = [p for p in parse if p not in ['', ' ']] — Mark
– Mark, Commented Apr 3, 2019 at 14:53
I guess this can do the job: clean_list = [row for row in parse if row.strip()] — accdias
– accdias, Commented Apr 3, 2019 at 14:53
Possible duplicate of Python: Removing spaces from list objects — sniperd
– sniperd, Commented Apr 3, 2019 at 14:54

accdias · Accepted Answer · 2019-04-03 14:58:28Z

4

This will do the job:

>>> dirty =['1',' ', '', '2','3','4']
>>> clean = [row for row in dirty if row.strip()]
>>> clean
['1', '2', '3', '4']
>>>

answered Apr 3, 2019 at 14:58

accdias

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2 Comments

DirtyBit Over a year ago

I see what you did there. dirty

SuperStew Over a year ago

the if row.strip() bit is clever

DirtyBit · Accepted Answer · 2019-04-04 06:27:03Z

2

Using filter with str.strip:

parse = ['a', ' ', 'b', 'c', '', 'd']

Python 2.x:

print(filter(str.strip, parse))

Python 3.x:

print(list(filter(str.strip, parse)))

OUTPUT:

['a', 'b', 'c', 'd']

EDIT:

To remove the empty spaces within the elements, let's say:

parse = ['a ', ' ', ' b', 'c', '', 'd ']

Using map() with filter and str.strip:

Python 2.x:

print(map(str.strip, filter(str.strip, parse)))

Python 3.x:

print(list(map(str.strip, filter(str.strip, parse))))

OUTPUT:

['a', 'b', 'c', 'd']

edited Apr 4, 2019 at 6:27

answered Apr 3, 2019 at 15:03

DirtyBit

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Comments

SuperStew · Accepted Answer · 2019-04-03 14:56:33Z

1

Dont modify the list you are iterating over.

Assuming parse is like

parse=[1,' ', '', 2,3,4]

you can do something like

parse_fix=[]
for y in parse:
    if y!='' and y!=' ': parse_fix.append(y)

then parse_fix will be the list you want

the short version might look like this

parse=[y for y in parse if y!='' and y!=' ']

edited Apr 3, 2019 at 14:56

answered Apr 3, 2019 at 14:54

SuperStew

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3 Comments

DirtyBit Over a year ago

Can't say because we're not sure of how exactly is the parse list in OP's case.

SuperStew Over a year ago

@DirtyBit well thats what the "assuming" bit was for

Bryan Over a year ago

This worked! With one caveat - it needs to use 'and' instead of 'or'. Still not sure I understand why just modifying the list doesn't work, but now I know.

Frenchy · Accepted Answer · 2019-04-03 15:01:37Z

0

you could use:

list1 = ["1 ", "", " ", "3", "  4", "       ",""]
list2 = [x for x in list1 if x.strip() ]

output list2:

['1 ', '3', '  4']

answered Apr 3, 2019 at 15:01

Frenchy

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2 Comments

accdias Over a year ago

That is exactly the answer I gave, @Frenchy.

Frenchy Over a year ago

yes ,just a problem of bandwith so i let you the paternity!! and i upvote to you

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