Some built-in to pad a list in python

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暗喜
暗喜 2020-11-27 14:06

I have a list of size < N and I want to pad it up to the size N with a value.

Certainly, I can use something like the following, but I feel that there sh

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  • 2020-11-27 14:38

    If you want to pad with None instead of '', map() does the job:

    >>> map(None,[1,2,3],xrange(7))
    
    [(1, 0), (2, 1), (3, 2), (None, 3), (None, 4), (None, 5), (None, 6)]
    
    >>> zip(*map(None,[1,2,3],xrange(7)))[0]
    
    (1, 2, 3, None, None, None, None)
    
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  • 2020-11-27 14:45

    gnibbler's answer is nicer, but if you need a builtin, you could use itertools.izip_longest (zip_longest in Py3k):

    itertools.izip_longest( xrange( N ), list )
    

    which will return a list of tuples ( i, list[ i ] ) filled-in to None. If you need to get rid of the counter, do something like:

    map( itertools.itemgetter( 1 ), itertools.izip_longest( xrange( N ), list ) )
    
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  • 2020-11-27 14:48

    To go off of kennytm:

    def pad(l, size, padding):
        return l + [padding] * abs((len(l)-size))
    
    >>> l = [1,2,3]
    >>> pad(l, 7, 0)
    [1, 2, 3, 0, 0, 0, 0]
    
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  • 2020-11-27 14:52

    There is no built-in function for this. But you could compose the built-ins for your task (or anything :p).

    (Modified from itertool's padnone and take recipes)

    from itertools import chain, repeat, islice
    
    def pad_infinite(iterable, padding=None):
       return chain(iterable, repeat(padding))
    
    def pad(iterable, size, padding=None):
       return islice(pad_infinite(iterable, padding), size)
    

    Usage:

    >>> list(pad([1,2,3], 7, ''))
    [1, 2, 3, '', '', '', '']
    
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