How to create dynamic nested dictionaries in Python

Question

I have a list as follows:

list_name=["node1","node2","node3",...,"nodeN"]

I have a dictionary as follows:

dictionary_name:{
  "node1":{
    "node2":true}
   }
}

I would like to set node2 to equal the following:

 "node2"={"node3":{"node4":{...."nodeN"=true}...}}}

I have been unsuccessfully lost in for loops for a while now.

Has anyone got any suggestions?

Thanks in advance.

Are you trying to learn about creating nested dictionaries or are you trying to model a chain of nodes in python? — Niko Fohr
– Niko Fohr, Commented Dec 11, 2020 at 17:36
Here is the larger problem: stackoverflow.com/questions/65257032/… — Charmalade
– Charmalade, Commented Dec 11, 2020 at 19:12

Sumit Chaturvedi · Accepted Answer · 2020-12-11 17:26:46Z

4

from functools import reduce

list_name=["node1","node2","node3","nodeN"]
print(reduce(lambda k, v: {v: k}, reversed(list_name), True))

Output:

{'node1': {'node2': {'node3': {'nodeN': True}}}}

answered Dec 11, 2020 at 17:26

Sumit Chaturvedi

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3 Comments

Charmalade Over a year ago

Thanks you for your help!

Charmalade Over a year ago

How would I attach the output to my initial dictionary_name?

Sumit Chaturvedi Over a year ago

dictionary_name = reduce(lambda k, v: {v: k}, reversed(list_name), True)

Ian Wilson · Accepted Answer · 2020-12-11 21:04:50Z

2

root = d = {'node1': {'node2': True}}
n = 10
keys = ["node{}".format(i) for i in range(1, n+1)]
for k in keys[:-1]:
    # Overwrite the True with a dictionary or create a dictionary,
    # not sure the point of the True values in this problem.
    if k not in d or not isinstance(d[k], dict):
        d[k] = {}
    # Recurse into the nested dictionary
    d = d[k]
# Set the final key to True.
d[keys[-1]] = True
print(root)

edited Dec 11, 2020 at 21:04

answered Dec 11, 2020 at 17:24

Ian Wilson

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5 Comments

Charmalade Over a year ago

Sorry, I am probably being dense, but when I run the above and print d all I get is {'node10': True} without the nesting.

Ian Wilson Over a year ago

Yeah sorry I didn't keep a reference to the original dictionary but that should be fixed if you keep a root reference like I added.

Charmalade Over a year ago

Works! Thanks for your help :)

Charmalade Over a year ago

A quick query with regards to your solution. Say for example the entries in list_name are random strings instead of "node1", "node2" etc, how would I go about the same problem?

Ian Wilson Over a year ago

I moved the keys into a variable. Should be replaceable with random list.

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