Python: filter dict to contain only certain keys?

Question

I've got a dict that has a whole bunch of entries. I'm only interested in a select few of them. Is there an easy way to prune all the other ones out?

Answer 1

Here's an example in python 2.6:

>>> a = {1:1, 2:2, 3:3}
>>> dict((key,value) for key, value in a.iteritems() if key == 1)
{1: 1}

The filtering part is the if statement.

This method is slower than delnan's answer if you only want to select a few of very many keys.

Answer 2

Constructing a new dict:

dict_you_want = { your_key: old_dict[your_key] for your_key in your_keys }

Uses dictionary comprehension, if you use a version which lacks them, make it dict((your_key, old_dict[your_key]) for ...). It's the same, though uglier.

Note that this, unlike jnnnnn's version, has stable performance (depends only on number of your_keys) for old_dicts of any size. Both in terms of speed and memory. Since this is a generator expression, it processes one item at a time, and it doesn't looks through all items of old_dict.

Removing everything in-place:

unwanted = set(your_dict) - set(keys)
for unwanted_key in unwanted: del your_dict[unwanted_key]

Answer 3

Given your original dictionary orig and the set of entries that you're interested in keys:

filtered = dict(zip(keys, [orig[k] for k in keys]))

which isn't as nice as delnan's answer, but should work in every Python version of interest. It is, however, fragile to each element of keys existing in your original dictionary.