Access dict hierarchically

Question

Want to access a dict via a String like dict[accessor] while acccessor is 'keyA.nestedKeyB'. Found something here with Lambda and overwriting the dict class ( https://stackoverflow.com/questions/39818669/dynamically-accessing-nested-dictionary-keys ) but unfortunetly there is no working example.

Code

data = { "keyA": { "nestedKeyB": "Hello" } }

print(data['keyA']['nestedKeyB'])  # prints 'Hello'

accessor = 'keyA.nestedKeyB'
print(data[accessor])  # KeyError

You cannot do this with a `dict`. You could write your own class that does this, but why? Why not just use the regular way? You can even very trivially use `'keyA.nestedKeyB'` to generate the keys using `'keyA.nestedKeyB'.split('.')` Note, you can't "overwrite" the dict class, and even if you could you really shouldn't — juanpa.arrivillaga, May 14 '20 at 00:14
@juanpa.arrivillaga: The dict can have different nested levels. For that reason my `accessor` could have different amount of keys. Because of that I need a way to access each level by the `accessor`. I do not understand your `split` example. — Dirk Schiller, May 14 '20 at 00:17
So what? Again, just *split* the accessors then loop over them and access, essentially like the answers in the question you already linked to... — juanpa.arrivillaga, May 14 '20 at 00:18

score 2 · Accepted Answer · answered May 14 '20 at 00:17

2

Since python dict doesn't support dot notations, you will have to convert the accessor to foo["bar"]["baz"] form.

Something as simple as,

def access(foo, accessor):
    keys = accessor.split('.')
    data = foo
    for key in keys:
        data = data[key]
    return data

answered May 14 '20 at 00:17

dumbPy

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@ZhdZilin. Post answer then – Mad Physicist May 14 '20 at 00:28

score 0 · Answer 2 · answered May 14 '20 at 00:35

import jsonpath
import json #in this sample, don't need import this.
data = { "keyA": { "nestedKeyB": "Hello" } }
node = jsonpath.jsonpath(data, "$.keyA.nestedKeyB")
print(node) # if node is not False, it's a list
print (node == False) # not found match result by this jsonpath? Then node == False

recommend you to use jsonpath library and JSONPath expression. You can work with more complex json data

Access dict hierarchically

Code

2 Answers2