How to get and modify the value of nested dictionary?

Question

I have this kind of dictionary in python:

x = {'test': {1: 2, 2: 4, 3: 5},
     'this': {1: 2, 2: 3, 7: 6},
     'is': {1: 2},
     'something': {90: 2,92:3}}

I want to modify all of the value in the key by whatever value I want. Let's say 100, the methods I tried are below:

counter = 1
print(x)
for key,anotherKey in x.items():
    while counter not in x[key]:
        counter+=1
    while counter in x[key]:
        x[key][counter] = 100
        counter+=1
    counter =0

Which got the result below:

{'test': {1: 100, 2: 100, 3: 100},
 'this': {1: 100, 2: 100, 7: 6},
 'is': {1: 100},
 'something': {90: 100,92: 3}}

I know why this is happening it's because the loop doesn't consider if the differences is more than 1 which in this case in 'this' : where the differences from 2 to 7 is more than 1. However I don't know how to solve this.

Did one of the 2 solutions below help? Feel free to accept one if it did (green tick on left), or ask for clarification. — jpp
– jpp, Commented May 11, 2018 at 13:09

jpp · Accepted Answer · 2018-05-09 15:48:52Z

4

You can iterate via a nested for loop:

x = {'test': {1: 2, 2: 4, 3: 5},
     'this': {1: 2, 2: 3, 7: 6},
     'is': {1: 2},
     'something': {90: 2,92:3}}

for a in x:
    for b in x[a]:
         x[a][b] = 100

print(x)

{'is': {1: 100},
 'something': {90: 100, 92: 100},
 'test': {1: 100, 2: 100, 3: 100},
 'this': {1: 100, 2: 100, 7: 100}}

Or for a new dictionary you can use a dictionary comprehension:

res = {a: {b: 100 for b in x[a]} for a in x}

edited May 9, 2018 at 15:48

answered May 9, 2018 at 15:43

jpp

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Comments

Abdul Fatah · Accepted Answer · 2018-05-09 15:49:12Z

-1

You can use dictionary comprehension in this way:

{a: {b:100 for b in d} for (a,d) in x.items()}

answered May 9, 2018 at 15:49

Abdul Fatah

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