通过将其与现有列表进行比较来更新嵌套字典值

Question

How to make changes in nested dictionary value by comparing its key value with an existing list.

Eg.: I have list name as common:

common=['abc','def','xyz','etc']
mydict={'abc':{'a':10,'b':20,'c':10}, 'rat':{'r':10,'a':20,'t':10} , 'etc':{'e':10,'t':20,'c':50}}

Now I want to change the value of 'b' to 50 because it exists in mydict key value same as I want to change the value of 't' to 10 I don't want any changes in 'rat' because it does not exist in list common.

common=['abc','def','xyz','etc']

mydict={'abc':{'a':10,'b':20,'c':10}, 'rat':{'r':10,'a':20,'t':10} , 'etc':{'e':10,'t':20,'c':50}}

input:

common=['abc','def','xyz','etc']

mydict={'abc':{'a':10,'b':20,'c':10}, 'rat':{'r':10,'a':20,'t':10} , 'etc':{'e':10,'t':20,'c':50}}

expected output:

mynewdict={'abc':{'a':10,'b':'50,'c':10}, 'rat':{'r':10,'a':20,'t':10} , 'etc':{'e':10,'t':10,'c':50}}

Answer 1

You can use indexing into a nested dictionary and compare the values of its keys.

common=['abc','def','xyz','etc']
mydict={'abc':{'a':10,'b':20,'c':10}, 'rat':{'r':10,'a':20,'t':10} , 'etc':{'e':10,'t':20,'c':50}}

for items in mydict:
    if items in common:
        inneritem= mydict[items]
        if 'b' in inneritem:
            inneritem['b']=50
        if 't' in inneritem:
            inneritem['t']=10

print(mydict)

Answer 2

You can do it with a comprehension for one update at a time. Multiple updates would make this approach quite cumbersome unless you don't mind adding the "t" key to sub-dictionaries that don't have it.:

newdict={k:{**d,'t':10} if k in common and 't' in d else d for k,d in mydict.items()}