update values in nested dict based on key

Question

I have a nested dictionary and I want to update the value of an item based on its key value. For example, I have the following dictionary and I want to set the value of every occurrence of an item with key=='p' to be 1.

my_dict = {'p': 0, 'nested_dict': {'p': 0, 'a': 2}}

For a regular dictionary (non-nested) the update method provides a simple one-liner solution:

my_dict.update((x, 1) for x, y in my_dict.items() if x=='p')

I'm looking for a similar solution for the case of nested dictionary

we are talking arbitrarily nested or is the max depth known? — Ma0
– Ma0, Commented Jun 21, 2021 at 7:55
arbitrary depth, I think this implies applying some kind of recursion — itamar kanter
– itamar kanter, Commented Jun 21, 2021 at 8:02

Ma0 · Accepted Answer · 2021-06-21 09:01:16Z

2

If the max depth of the dictionary is known and constant and the 'p's always appear at the same depth this can be solved without recursion. In all other cases, a recursive approach is needed. Based on the example problem and your comments, I will assume that the second one is true. Here is a recursive solution:

def setPto1(dct):
    return {k: setPto1(v) if isinstance(v, dict) else [v, 1][k=='p'] for k, v in dct.items()}

So you basically loop through the key-value pairs with a recursive dictionary comprehension. You were not far off with your approach. If there is something you do not understand about it, leave a comment.

If [v, 1][k=='p'] makes you feel uneasy, you can replace it with the more straight-forward (1 if k=='p' else v). They are the same thing.

edited Jun 21, 2021 at 9:01

answered Jun 21, 2021 at 8:09

Ma0

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6 Comments

Patrick Artner Over a year ago

[v, 1][k=='p'] looks fancy-numpy-ish?

Ma0 Over a year ago

fancy sure, but I would not say numpyish; it is just basic boolean indexing. numpy has a much more powerful boolean indexing engine that operates similarly to filter.

itamar kanter Over a year ago

Thanks!, personally, I like this [v, 1][k=='p'] style

Ma0 Over a year ago

Its a matter of taste as long as you understand it. Note however that it might be a bit more costly in terms of computer resources since Python needs to build the [v, 1] list. Feel free to accept the answer if it solves your problem.

deponovo Over a year ago

@PatrickArtner This syntax, though requiring some adaptation, is quite nice. Since which version of python is it available? What's the exact name for this syntax?

|

deponovo · Accepted Answer · 2021-06-21 08:31:19Z

Maybe this could get you started (you can still make it more generic, and it might have problems with repeated keys):

from functools import reduce

def decision_func(d, item, replace_on_item, replace_val):
    if item == replace_on_item:
        d[item] = replace_val
    else:
        return d[item]

my_dict = {'p': 0, 'nested_dict': {'p': 0, 'a': 2}}
reduce(lambda d, item: decision_func(d, item, 'p', ["some_other_val", 2]), ['nested_dict', 'p'], my_dict)
print(my_dict)

Output:

{'p': 0, 'nested_dict': {'p': ['some_other_val', 2], 'a': 2}}

What is done here is that you are going to run down the nesting tree by using the reduce function, and the decision_func is the one making the decision on whether to update a given entry or return the value for that key and continue looking for the next nesting level (basically a recursion prob). Note that as written above, this is not yet a bullet proof implementation.

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