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Resolve lookups without asking twice · Python scenario · D8LooP

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Resolve lookups without asking twice

An enrichment step looks each row's key up in a dimension that costs real money per call. The stream repeats keys constantly, and the obvious loop pays for every repeat. Write resolve_lookups(keys, source, cache_size). It returns the resolved values and how many times the source was actually consulted.

Function to write

resolve_lookups(keys: list, source: dict, cache_size: int) -> dict

A dict with the `results` in input order and the number of `source_calls` made.

How to approach it

Count the source calls first, then make the number go down.

Sample cases

+ 2 held back until you submit

a repeated key and a miss

Four lookups, three distinct keys, three source calls — and the unknown key resolves to None.

Input

Argument 1

[
  'u1',
  'u1',
  'u2',
  'u3'
]

Argument 2

{
  'u1': 'Amir',
  'u2': 'Cara'
}

Argument 3

Returns

{
  'results': [
    'Amir',
    'Amir',
    'Cara',
    None
  ],
  'source_calls': 3
}

caching turned off

A cache size of zero is a real setting: every lookup reaches the source.

Input

Argument 1

[
  'u1',
  'u1'
]

Argument 2

{
  'u1': 'Amir',
  'u2': 'Cara'
}

Argument 3

Returns

{
  'results': [
    'Amir',
    'Amir'
  ],
  'source_calls': 2
}

nothing to resolve

No keys means no results and no source calls, rather than a division or an index error.

Input

Argument 1

[] (empty list)

Argument 2

{
  'u1': 'Amir',
  'u2': 'Cara'
}

Argument 3

Returns

{
  'results': [],
  'source_calls': 0
}

Constraints

1Return {'results': [...], 'source_calls': <count>}, with one result per key in input order.
2A key the source does not have resolves to None, and that is a result — asking again would be a second call for the same answer.
3With a cache, the source is consulted once per distinct key while that key is still cached.
4cache_size of 0 disables caching entirely. It is a real setting, not an error: it is how somebody measures what the cache is worth.
5A cache smaller than the working set evicts, so a key may be fetched more than once. Report what actually happened.
6No keys returns an empty list and zero calls.

Worked example

Resolve u1, u1, u2, u3 with a large cache. Four results come back — Amir, Amir, Cara, None — from three source calls. The u3 case is the one worth thinking about. It is not in the source, so the answer is None; caching that None is the difference between one wasted call and one per occurrence, and a cache that only stores successes is a cache that does nothing at all for a stream of unknown keys.

What this tests

That memoisation is a decorator over a function rather than a dict scattered through a loop, that the cache size is a real trade rather than a constant, and that the cost of a lookup is measured rather than assumed.

PYTHONresolve_lookups(keys: list, source: dict, cache_size: int) -> dict

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