Dictionary keys match on list; get key/value pair

Question

In python... I have a list of elements 'my_list', and a dictionary 'my_dict' where some keys match in 'my_list'.

I would like to search the dictionary and retrieve key/value pairs for the keys matching the 'my_list' elements.

I tried this...

    if any(x in my_dict for x in my_list):
          print set(my_list)&set(my_dict)

But it doesn't do the job.

Please don't name your variables `list` or `dict`. It confuses the heck out of people (and possibly then Python interpreter, too, if you're not careful). — Tim Pietzcker, Jun 28 '11 at 10:29

score 43 · Accepted Answer · edited Mar 17 '22 at 12:22

43

(I renamed list to my_list and dict to my_dict to avoid the conflict with the type names.)

For better performance, you should iterate over the list and check for membership in the dictionary:

for k in my_list:
    if k in my_dict:
        print(k, my_dict[k])

If you want to create a new dictionary from these key-value pairs, use

new_dict = {k: my_dict[k] for k in my_list if k in my_dict}

edited Mar 17 '22 at 12:22

mirekphd

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answered Jun 28 '11 at 10:28

Sven Marnach

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adding `my_dict.keys()` would make it much faster, as it would only search the `keys` in the `dict`. Or to search in values just use `my_dict.values()` – Deepak Harish Sep 16 '22 at 04:52

Felix Kling · Answer 2 · 2011-06-28T10:43:02.360

16

Don't use dict and list as variable names. They shadow the built-in functions. Assuming list l and dictionary d:

kv = [(k, d[k]) for k in l if k in d]

edited Jun 28 '11 at 10:43

answered Jun 28 '11 at 10:28

Felix Kling

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Rob Cowie · Answer 3 · 2011-06-28T10:36:58.840

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 new_dict = dict((k, v) for k, v in dict.iteritems() if k in list)

Turning list into a set set(list) may yield a noticeable speed increase

edited Jun 28 '11 at 10:36

answered Jun 28 '11 at 10:28

Rob Cowie

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Note that `if k in list` will be `O(n)`. – Felix Kling Jun 28 '11 at 10:30
Yeah; I was conveniently ignoring that for the sake of simplicity – Rob Cowie Jun 28 '11 at 10:32
1

hehe; I lost an upvote. Seems a bit harsh; There is merit in _deliberately_ providing a simple answer to a simple question. – Rob Cowie Jun 28 '11 at 10:35
dict.iteritems() would be preferable to dict.items(). – jd. Jun 28 '11 at 10:35
1

Rob: The point is that the efficient solution is just as simple. – Sven Marnach Jun 28 '11 at 10:37
In this case I agree, it is _almost_ as simple. Kind of depends on the context of the OP's question but given the paucity of details, I concede my answer should be ignored :) – Rob Cowie Jun 28 '11 at 10:41
my_dict.iteritems() is preferable in that it combines the key and valye lookup into one call. – david.barkhuizen Jan 31 '13 at 11:56

score 4 · Answer 4 · answered Aug 21 '16 at 14:07

4

Try This:

mydict = {'one': 1, 'two': 2, 'three': 3}
mykeys = ['three', 'one','ten']
newList={k:mydict[k] for k in mykeys if k in mydict}
print newList
{'three': 3, 'one': 1}

answered Aug 21 '16 at 14:07

Vikram Singh Chandel

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score 3 · Answer 5 · answered Jun 28 '11 at 10:28

3

What about print([kv for kv in dict.items() if kv[0] in list])

answered Jun 28 '11 at 10:28

Alexander Gessler

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score 1 · Answer 6 · answered Oct 12 '17 at 07:19

1

Here is a one line solution for that

{i:my_dict[i] for i in set(my_dict.keys()).intersection(set(my_list))}

answered Oct 12 '17 at 07:19

Palash Jhamb

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Dictionary keys match on list; get key/value pair

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