How to recursively check a nested dictionary?

Question

0.00/5 (No votes)

See more:

How would I recursively check a dictionary which contains nested dictionaries.

So the dictionary would go like this:

dict = {"1": {"1.1": {"x", "x", "x"},
              "1.2": {"y", "y", "y"}},
        "2": {"2.1": {"z", "z", "z"}},
        "3": {"3.1": {"a"}}}

How would I loop through the entire dict?
I am searching the dict by keys so if I put in like 1.2, it would write out everything until it finds that key.

1, 1.1, 1.2

What I have tried:

Ok so to go throught the first layer I do the normal for loop:

def recursion(item):
    for i in stiki:
        print(recursion(i))

What would be the correct way to recursivly call a dictionary like that?

Posted 16-Dec-20 6:15am

Member 14631234

Updated 1-Jan-21 9:40am

Add a Solution

Comments

Richard MacCutchan 16-Dec-20 12:53pm

The key is to know when your dictionary contains normal entries, and when it contains further dictionaries.

2 solutions

Add a Solution

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Patrice T · Answer 1 · 2020-12-16T08:14:00

You need to study the technique.
Just like with any recursive algorithm, at any point, you need to know if there is matter to recursion, in your case, is there a sub-directory.

Python

def recursion(item):
    for i in stiki: # what is stiki here ? I would have expected item instead.
        print(recursion(i))

Recursion (computer science) - Wikipedia[^]

Mr Bobcat · Answer 2 · 2021-01-01T09:40:00

def recursion(my_dict):
	for x, y in my_dict.items():
		if isinstance(y, dict):
			recursion(y)
		else:
			print(y)

isinstance() checks if first argument is of type specified in the second argument, in this case it checks if y is of type 'dict', if it is then continue with the recursion, or else print it.

How to recursively check a nested dictionary?

2 solutions

Solution 1

Solution 2

Add your solution here

Preview 0

How to recursively check a nested dictionary?

2 solutions

Solution 1

Solution 2

Add your solution here

Preview 0

Existing Members

...or Join us