Remove empty strings from list
tec
technic.tec at gmail.com
Mon Sep 14 22:33:05 EDT 2009
Chris Rebert 写道:
> On Mon, Sep 14, 2009 at 6:49 PM, Helvin <helvinlui at gmail.com> wrote:
>> Hi,
>>
>> Sorry I did not want to bother the group, but I really do not
>> understand this seeming trivial problem.
>> I am reading from a textfile, where each line has 2 values, with
>> spaces before and between the values.
>> I would like to read in these values, but of course, I don't want the
>> whitespaces between them.
>> I have looked at documentation, and how strings and lists work, but I
>> cannot understand the behaviour of the following:
>> line = f.readline()
>> line = line.lstrip() # take away whitespace at the beginning of the
>> readline.
>> list = line.split(' ') # split the str line into a list
>>
>> # the list has empty strings in it, so now,
>> remove these empty strings
>> for item in list:
>> if item is ' ':
>> print 'discard these: ',item
>> index = list.index(item)
>> del list[index] # remove this item from the list
>> else:
>> print 'keep this: ',item
>> The problem is, when my list is : ['44', '', '', '', '', '',
>> '0.000000000\n']
>> The output is:
>> len of list: 7
>> keep this: 44
>> discard these:
>> discard these:
>> discard these:
>> So finally the list is: ['44', '', '', '0.000000000\n']
>> The code above removes all the empty strings in the middle, all except
>> two. My code seems to miss two of the empty strings.
>>
>> Would you know why this is occuring?
>
> Block quoting from http://effbot.org/zone/python-list.htm
> """
> Note that the for-in statement maintains an internal index, which is
> incremented for each loop iteration. This means that if you modify the
> list you’re looping over, the indexes will get out of sync, and you
> may end up skipping over items, or process the same item multiple
> times.
> """
>
> Thus why your code is skipping over some elements and not removing them.
> Moral: Don't modify a list while iterating over it. Use the loop to
> create a separate, new list from the old one instead.
or use filter
list=filter(lambda x: len(x)>0, list)
>
> Cheers,
> Chris
> --
> http://blog.rebertia.com
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