Access nested dictionary through a variable

Question

I have the following setup: a function returns a dictionary with N timelines of equal size (100k points). The dictionary returns looks like:

timelines = dict()
timelines["Name1"] = dict()
timelines["Name1"]["Name2"] = dict()
timelines["Name1"]["Name3"] = dict()
timelines["Name1"]["Name2"]["a"] = # List of 100k points
timelines["Name1"]["Name2"]["b"] = # List of 100k points
timelines["Name1"]["Name2"]["c"] = # List of 100k points
timelines["Name1"]["Name3"]["b"] = # List of 100k points
timelines["Name1"]["Name2"]["c"] = # List of 100k points
timelines["Name1"]["a"] = # List of 100k points
timelines["Name1"]["b"] = # List of 100k points
timelines["Name2"] # and so on.

As you might have understood, the timelines (list of points) are not always stored in the same level. Sometimes I can access it with 1 key, sometimes with 2, sometimes with 5. Those keys will give me the labels for the plot and are necessary. My plan was to pass to the plot function a tuple of the keys.

Example:

T = ("Name1", "Name2", "b") 
# Will allow me to access the timeline:
timelines["Name1"]["Name2"]["b"]
# by doing:
timelines[T[0]][T[1]][T[2]]

In the example above, I wrote the dictionary path myself ([T[0]][T[1]][T[2]]), however how can I acess the right timeline with a tuple T of unknown size? How can I unpack the tuple into a dictionary path?

Thanks :)

If you access your dict many times, it may be beneficial to flatten it and compressing the keys to have easier access later. Read about it [here](https://stackoverflow.com/a/14692747/9225671) and [here](https://stackoverflow.com/a/6027615/9225671). — Ralf, Feb 14 '19 at 12:46

score 9 · Accepted Answer · answered Feb 14 '19 at 12:50

I would actually do it like this, this will be the fastest method more than likely

from functools import reduce
from operator import getitem

path = ['Name1', 'Name2', 'a']

reduce(getitem, path, dictionary)

Lambda calls can become expensive especially as data grows, not to mention getitem will be faster than any other method listed here because it's implemented purely in C

Thierry Lathuille · Answer 2 · 2019-02-14T12:38:09.630

2

You could use reduce

from functools import reduce

timelines = dict()
timelines["Name1"] = dict()
timelines["Name1"]["Name2"] = dict()
timelines["Name1"]["Name2"]["b"] = 'my data'

T = ("Name1", "Name2", "b") 

data = reduce(lambda d, key: d[key], T, timelines)

print(data)
# my data

The initial value is your timeline dict. Reduce iterates on your T tuple, and at each iteration, calls the lambda function. The lambda takes two arguments: the first one is the accumulated value, in this case the current dict d, and the second one is the key it got from the tuple during this iteration. The lambda returns d[key], which becomes the new accumulated value for the next iteration.

edited Feb 14 '19 at 12:38

answered Feb 14 '19 at 12:28

Thierry Lathuille

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Works great. Could you explain a bit what it does? I don't get this lambda statement. – Mathieu Feb 14 '19 at 12:32
it's a better approach to do `reduce(getitem, [list_of_keys], input_dict)` – gold_cy Feb 14 '19 at 12:47

score 1 · Answer 3 · answered Feb 14 '19 at 12:40

1

you can use a recursive approach

def access_my_dict(my_dict, tuple_keys):
    value = my_dict[tuple_keys[0]]
    if isinstance(value, dict):
        return access_my_dict(value, tuple_keys[0:])
    return value

answered Feb 14 '19 at 12:40

kederrac

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score 1 · Answer 4 · answered Feb 14 '19 at 12:44

Traverse the data structure like this:

timelines = dict()
timelines["Name1"] = dict()
timelines["Name2"] = dict()
timelines["Name1"]["Name2"] = dict()
timelines["Name1"]["Name3"] = dict()
timelines["Name1"]["Name2"]["a"] = []  # List of 100k points
timelines["Name1"]["Name2"]["b"] = [12, 67, 45]  # List of 100k points
timelines["Name1"]["Name2"]["c"] = [12, 45]  # List of 100k points
timelines["Name1"]["Name3"]["b"] = []  # List of 100k points
timelines["Name1"]["Name2"]["c"] = []  # List of 100k points
timelines["Name1"]["a"] = []  # List of 100k points
timelines["Name1"]["b"] = []  # List of 100k points
timelines["Name2"]["Name23"] = dict()
timelines["Name2"]["Name23"]["Z"] = []  # and so on.


def get_time_line(path):
    data = timelines
    for path_element in path:
        data = data[path_element]
    return data

path = ("Name1", "Name2", "b")
print(get_time_line(path))

Output:

[12, 67, 45]

Access nested dictionary through a variable

4 Answers4