Nested get with dotted path

Safely extract values from nested dictionaries using a dotted path and a default fallback.

dicts nested parsing default

Write a function `nested_get(data, path, default=None)` that takes a nested dictionary `data`, a string `path` of keys separated by dots, and an optional `default` value. It should traverse the dictionary following the keys in order. If the full path exists, return the value at that location. If at any point a key is missing (or the current value is not a dictionary), return `default` instead. An empty `path` should return `data` itself.

Constraints

- `data` is either a dictionary or any value (not necessarily a dict).
- `path` is a string; it may be empty.
- Dictionaries may be nested arbitrarily deep.
- Time complexity: O(k) where k is the number of dots in path (number of keys).

Example

>>> nested_get({'a': {'b': {'c': 42}}}, 'a.b.c')
42
>>> nested_get({'a': {'b': {'c': 42}}}, 'a.b.x')
None
>>> nested_get({'a': {'b': 1}}, 'a.b', 'missing')
1
>>> nested_get({}, '')  # returns {} (the data itself)

8 points ~10 min

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Hints

Use `path.split('.')` to get the list of keys; handle empty path separately.

Keep a reference to the current value as you loop through keys.

If the current value isn't a dictionary, you can't go deeper—return default.

If a key is missing (use `if key in current` or try/except), return default.

If you reach the end of the key list, return the current value (or default if None).

Test Results

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