Slicing a dictionary by keys that start with a certain string

Question

This is pretty simple but I’d love a pretty, pythonic way of doing it. Basically, given a dictionary, return the subdictionary that contains only those keys that start with a certain string.

» d = {'Apple': 1, 'Banana': 9, 'Carrot': 6, 'Baboon': 3, 'Duck': 8, 'Baby': 2}
» print slice(d, 'Ba')
{'Banana': 9, 'Baby': 2, 'Baboon': 3}

This is fairly simple to do with a function:

def slice(sourcedict, string):
    newdict = {}
    for key in sourcedict.keys():
        if key.startswith(string):
            newdict[key] = sourcedict[key]
    return newdict

But surely there is a nicer, cleverer, more readable solution? Could a generator help here? (I never have enough opportunities to use those).

Asked By: Aphex

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Source

Answer 1

How about this:

in python 2.x :

def slicedict(d, s):
    return {k:v for k,v in d.iteritems() if k.startswith(s)}

In python 3.x :

def slicedict(d, s):
    return {k:v for k,v in d.items() if k.startswith(s)}

Answered By: Mark Byers

Answer 2

In functional style:

dict(filter(lambda item: item[0].startswith(string),sourcedict.iteritems()))

Answered By: seriyPS

Answer 3

In Python 3 use items() instead:

def slicedict(d, s):
    return {k:v for k,v in d.items() if k.startswith(s)}

Answered By: gc5

Slicing a dictionary by keys that start with a certain string

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