Remove empty string from list

As the OP specifically wants to get rid of empty strings, I think this is the best solution, it's clear and concise.

An equivalent generator solution would be icing on the cake, to prevent extra copies of the lists.

@PhilH I thought about that, but you can't get the length of generator expressions so it would make this much more complicated.

@Matt -- If we're interested in keeping py2k compatability, you may want to consider dividing by float(len(mylist)) -- just to make sure you do "true division" here, but +1 from me

@mgilson For some applications that may be what you want, for others you may need an integer average. Ill include it

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Ashwini Chaudhary · Accepted Answer · 2014-03-02 18:12:11Z

7

You can use filter():

filter() returns a list in Python 2 if we pass it a list and an iterator in Python 3. As suggested by @PhilH you can use itertools.ifilter() in Python 2 to get an iterator.

To get a list as output in Python 3 use list(filter(lambda x:x != '', lis))

In [29]: lis = [1, 2, 3, '', 4, 0]

In [30]: filter(lambda x:x != '', lis)
Out[30]: [1, 2, 3, 4, 0]

Note to filter any falsy value you can simply use filter(None, ...):

>>> lis = [1, 2, 3, '', 4, 0]
>>> filter(None, lis)
[1, 2, 3, 4]

edited Mar 2, 2014 at 18:12

answered Oct 30, 2012 at 12:49

Ashwini Chaudhary

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4 Comments

This will also remove 0 were it an element in the list.

Ashwini Chaudhary Over a year ago

I'd suggest using itertools.ifilter() as it is iterator based rather than creating a copy of the list just for an average.

@PhilH yes on python 2.x ifilter() should be preferred, on python 3.x filter() returns an iterator itself.

DSM Over a year ago

This will be a fair bit slower than a listcomp.

mgilson · Accepted Answer · 2012-10-30 13:22:02Z

2

The other answers show you how to create a new list with the desired element removed (which is the usual way to do this in python). However, there are occasions where you want to operate on a list in place -- Here's a way to do it operating on the list in place:

while True:
    try:
        mylist.remove('')
    except ValueError:
        break

Although I suppose it could be argued that you could do this with slice assignment and a list comprehension:

mylist[:] = [i for i in mylist if i != '']

And, as some have raised issues about memory usage and the wonders of generators:

mylist[:] = (i for i in mylist if i != '')

works too.

edited Oct 30, 2012 at 13:22

answered Oct 30, 2012 at 12:44

mgilson

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6 Comments

How about: idx = [idx for idx, val in enumerate(x) if val=='']; for i in reversed(idx): x.pop(i) ? - more efficient than .remove...

@JonClements -- You're right, it is more efficient, but it's also less explicit. (and performs slightly slower than the easier to read slice-assignment version for a list set up like lst = [1,2,3,'',4]*20)

If you explain how I could improve this answer, I'd be happy to edit (or if you have appropriate privileges yourself, feel free to edit)

Umm, well +1 from me anyway for alternate solutions - not sure why you got -1 though...

@JonClements -- I don't know either, that's how it goes sometimes :-). (I liked your clever use of reversed and pop by the way -- I would have added it if it did better than the in place version with slice assignment)

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Phil H · Accepted Answer · 2012-10-30 13:36:46Z

2

itertools.ifilterfalse(lambda x: x=='', myList)

This uses iterators, so it doesn't create copies of the list and should be more efficient both in time and memory, making it robust for long lists.

JonClements points out that this means keeping track of the length separately, so to show that process:

def ave(anyOldIterator):
    elementCount = 0
    runningTotal = 0
    for element in anyOldIterator:
        runningTotal += element
        elementCount += 1
    return runningTotal/elementCount

Or even better

def ave(anyOldIterator):
    idx = None
    runningTotal = 0
    for idx,element in enumerate(anyOldIterator):
        runningTotal += element
    return runningTotal/(idx+1)

Reduce:

def ave(anyOldIterator):
    pieces = reduce(lambda x,y: (y[0],x[1]+y[1]), enumerate(anyOldIterator))
    return pieces[1]/(pieces[0]+1)

Timeit on the average of range(0,1000) run 10000 times gives the list comprehension a time of 0.9s and the reduce version 0.16s. So it's already 5x faster before we add in filtering.

edited Oct 30, 2012 at 13:36

answered Oct 30, 2012 at 12:55

Phil H

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12 Comments

Unfortunately with the side effect that the length will need to be tracked separately to compute an average...

@JonClements, why would that make a difference? I'll illuminate in an edit.

@pistache how did you calculate the average from this in your speed test?

@pistache: surely the most appropriate timing is the calculation of an average? If you do it without any list() call as in my edit above, or using a reduce step, then it should be faster for large lists.

JonClements suggests this makes the length a problem - to be fair, that's not what I said - my statement the length will need to be tracked separately was referring to the fact that extra work would need to be done (as you have shown) as sum(blah)/len(blah) isn't possible

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Le Hoai Duc · Accepted Answer · 2015-01-06 03:15:12Z

1

You can use:

alist = ['',1,2]
new_alist = filter(None, alist)
new_alist_2 = filter(bool, alist)

Result:

new_alist = [1,2]
new_alist_2 = [1,2]

answered Jan 6, 2015 at 3:15

Le Hoai Duc

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Comments

Rag Sagar · Accepted Answer · 2012-10-30 12:37:23Z

-1

mylist = [1, 2, 3, '', 4]
newlist = []
for i in mylist:
    try:
        newlist.append(int(i))
    except ValueError:
        pass
avg = sum(newlist)/len(newlist)

edited Oct 30, 2012 at 12:37

answered Oct 30, 2012 at 12:30

Rag Sagar

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5 Comments

This is a little dangerous as 0 could get stripped out too.

Bryan Oakley Over a year ago

With the update, this becomes a good all-round solution, however it may be overkill for the asker's use case.

Rag Sagar Over a year ago

Updated to consider 0 as Lattyware said. But i too think it is an overkill.

Too complex for a case where a simple list comprehension will suffice.

Rag Sagar Over a year ago

By the time i was about answer Matt asnwered. Thought let it be there as an alternative.

pistache · Accepted Answer · 2012-10-30 13:12:02Z

-1

'' is equivalent to False. If we filter the 0 case out (because 0 is equivalent to False), we can use list comprehension :

[x for x in a if x or x == 0]

Or if we strictly want to filter out empty strings :

[x for x in a if x != '']

This may not be the fastest way.

Edit, added some bench results comparing with the other solutions (not for the sake of comparing myself to others, but I was curious too of what method was the fastest)

ragsagar>
6.81217217445
pistache>
1.0873541832
cerealy>
1.07090902328
Matt>
1.40736508369
Ashwini Chaudhary>
2.04662489891
Phil H (just the generator) >
0.935978889465
Phil H with list() >
3.58926296234

I made the script quickly, using timeit(), I used [0,1,2,0,3,4,'',5,8,0,'',4] as the list. I ran multiple tests, results did not vary.

NOTE: I'm not trying to put my solution on top using speed as a criteria. I know OP didn't specifically ask for speed, but I was curious and maybe some other are.

edited Oct 30, 2012 at 13:12

answered Oct 30, 2012 at 12:34

pistache

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11 Comments

Is there a reason you can't just do [x for x in a if a != '']?

Bryan Oakley Over a year ago

@Matt This potentially strips out [] and other similar things too, but to be honest, if you want that extra functionality, I'd go with ragsagar's updated answer as it'll handle more cases.

Rag Sagar Over a year ago

@Matt it is [x for x in a if x != '']

Comparing to false is decidedly different then comparing for equality to '', and obscures the actual intent (which is specifically to strip out '')