Copy items from dictionary into another dictionary based on keys

Question

I have a list

list_x = ['8', '4', '5', '6', '1']

and a dictionary

dict_1 = {'0': {'ID=3': 1, 'ID=7': 1},
 '1': {'ID=7': 2, 'ID=6': 2},
 '2': {'ID=7': 3, 'ID=5': 1},
 '3': {'ID=7': 2, 'ID=1': 1},
 '4': {'ID=3': 3, 'ID=5': 1, 'ID=4': 1},
 '5': {'ID=5': 1, 'ID=2': 3, 'ID=4': 2, 'ID=7': 1, 'ID=8': 1},
 '6': {'ID=0': 5, 'ID=3': 7, 'ID=1': 1},
 '7': {'ID=1': 1, 'ID=8': 1, 'ID=7': 1},
 '8': {'ID=6': 2, 'ID=0': 2, 'ID=2': 2}}

and I want to create a new dictionary dict_2 in which all items of dict_1 are stored that have a key that exists in list_x. How can I do that? I had trouble doing it with two nested for loops since I could access the keys but I could not copy the entire item.

You can use `dict_2 = {k: v for k, v in dict_1.items() if k in list_x}` — ChrisOram, Aug 12 '21 at 15:29
Can there be a value in ```list_x``` that is not a key of ```dict_1``` ? — Metapod, Aug 12 '21 at 15:30

score 2 · Answer 1 · 2021-08-12T15:44:47.050

2

Try something like this

list_x = ['8', '4', '5', '6', '1']


dict_1 = {'0': {'ID=3': 1, 'ID=7': 1},
 '1': {'ID=7': 2, 'ID=6': 2},
 '2': {'ID=7': 3, 'ID=5': 1},
 '3': {'ID=7': 2, 'ID=1': 1},
 '4': {'ID=3': 3, 'ID=5': 1, 'ID=4': 1},
 '5': {'ID=5': 1, 'ID=2': 3, 'ID=4': 2, 'ID=7': 1, 'ID=8': 1},
 '6': {'ID=0': 5, 'ID=3': 7, 'ID=1': 1},
 '7': {'ID=1': 1, 'ID=8': 1, 'ID=7': 1},
 '8': {'ID=6': 2, 'ID=0': 2, 'ID=2': 2}}
d={k:dict_1.get(k).copy() for k in list_x if k in dict_1}
print(d)

edited Aug 12 '21 at 15:44

answered Aug 12 '21 at 15:31

@metapod, thanks. Added the control statement – Aug 12 '21 at 15:38
If any value from ```dict_1``` is changed, it will be changed in ```dict_2```too ! You should get the nested dict by ```copy```: ```k:dict_1.get(k).copy()``` – Metapod Aug 12 '21 at 15:43

score 2 · Accepted Answer · answered Aug 12 '21 at 15:43

You can use operator.itemgetter for this pretty easily:

itemgetter(*list_x)(dict_1)

All code below:

from operator import itemgetter

list_x = ['8', '4', '5', '6', '1']
dict_1 = {'0': {'ID=3': 1, 'ID=7': 1},
 '1': {'ID=7': 2, 'ID=6': 2},
 '2': {'ID=7': 3, 'ID=5': 1},
 '3': {'ID=7': 2, 'ID=1': 1},
 '4': {'ID=3': 3, 'ID=5': 1, 'ID=4': 1},
 '5': {'ID=5': 1, 'ID=2': 3, 'ID=4': 2, 'ID=7': 1, 'ID=8': 1},
 '6': {'ID=0': 5, 'ID=3': 7, 'ID=1': 1},
 '7': {'ID=1': 1, 'ID=8': 1, 'ID=7': 1},
 '8': {'ID=6': 2, 'ID=0': 2, 'ID=2': 2}}
 
print(dict(zip(list_x, (itemgetter(*list_x)(dict_1)))))

{'8': {'ID=6': 2, 'ID=0': 2, 'ID=2': 2},
 '4': {'ID=3': 3, 'ID=5': 1, 'ID=4': 1},
 '5': {'ID=5': 1, 'ID=2': 3, 'ID=4': 2, 'ID=7': 1, 'ID=8': 1},
 '6': {'ID=0': 5, 'ID=3': 7, 'ID=1': 1},
 '1': {'ID=7': 2, 'ID=6': 2}}

score 0 · Answer 3 · answered Aug 12 '21 at 15:36

The simple way is to use a dictionary comprehension, as so:

dict_2 = {k: v for k, v in dict_1.items() if k in list_x}

A note should be made, however, on the linking of the two dicts since they are collections which contain mutable items. If you want a completely separate result (when one changes the other stays the same) you can use deepcopy from the copy module:

from copy import deepcopy

list_x = ['8', '4', '5', '6', '1']

dict_1 = {'0': {'ID=3': 1, 'ID=7': 1},
 '1': {'ID=7': 2, 'ID=6': 2},
 '2': {'ID=7': 3, 'ID=5': 1},
 '3': {'ID=7': 2, 'ID=1': 1},
 '4': {'ID=3': 3, 'ID=5': 1, 'ID=4': 1},
 '5': {'ID=5': 1, 'ID=2': 3, 'ID=4': 2, 'ID=7': 1, 'ID=8': 1},
 '6': {'ID=0': 5, 'ID=3': 7, 'ID=1': 1},
 '7': {'ID=1': 1, 'ID=8': 1, 'ID=7': 1},
 '8': {'ID=6': 2, 'ID=0': 2, 'ID=2': 2}}

dict_2 = {k: v for k, v in deepcopy(dict_1).items() if k in list_x}

Now when doing an assignment such as dict_1['4']['ID=3'] = 1000, dict_2 will not be changed.

y_159 · Answer 4 · 2021-08-12T16:05:31.473

0

try this:

dict_1 = {'0': {'ID=3': 1, 'ID=7': 1},
 '1': {'ID=7': 2, 'ID=6': 2},
 '2': {'ID=7': 3, 'ID=5': 1},
 '3': {'ID=7': 2, 'ID=1': 1},
 '4': {'ID=3': 3, 'ID=5': 1, 'ID=4': 1},
 '5': {'ID=5': 1, 'ID=2': 3, 'ID=4': 2, 'ID=7': 1, 'ID=8': 1},
 '6': {'ID=0': 5, 'ID=3': 7, 'ID=1': 1},
 '7': {'ID=1': 1, 'ID=8': 1, 'ID=7': 1},
 '8': {'ID=6': 2, 'ID=0': 2, 'ID=2': 2}}
 

list_x = ['8', '4', '5', '6', '1']
 
dict_2 = {}
for key, value in dict_1.items():
    if key in list_x:
        dict_2[key] = dict_1[key].copy()
         
         
print(dict_2)

Verify the above output.

See how to iterate dictionary items from here , here and here

To understand why copy() is used, see this post.

edited Aug 12 '21 at 16:05

answered Aug 12 '21 at 15:42

y_159

458
3
15

1

As dicts are mutable, you should use ```copy```: ```dict_2[key] = dict_1[key].copy()```. Otherwise, changing ```dict_1``` will change ```dict_2``` too – Metapod Aug 12 '21 at 15:45
not in the above answer. you can verify it. – y_159 Aug 12 '21 at 15:48
I did: ```list_x = ['1']; dict_1['1']['ID=6'] = 456 ;print(dict_2); >> {'1': {'ID=7': 2, 'ID=6': 456}} ``` Because ```dict_1[X]``` is a dict. (I didn't mention the part with your algorithm, too lazy :) But it's of course done before the re-assignation) – Metapod Aug 12 '21 at 15:50

Copy items from dictionary into another dictionary based on keys

4 Answers4