Dictionaries as Tables

Dict Lookup Table

Build a code-to-name lookup dictionary from two parallel lists, then resolve a batch of query codes — including one that is absent — using a guarded membership test. The replay shows lookup growing one entry at a time and result accumulating found names and the fallback value.

By hand

The Pythonic way

Write lookup as a dict literal and replace the if/else guard with lookup.get(q, 'unknown'), which returns the fallback value in one expression without a separate membership test.

naive.py 
codes = ['r', 'g', 'b', 'y', 'w']
names = ['red', 'green', 'blue', 'yellow', 'white']
lookup = {}
for code, name in zip(codes, names):
    lookup[code] = name
queries = ['r', 'b', 'x', 'y']
result = []
for q in queries:
    if q in lookup:
        result.append(lookup[q])
    else:
        result.append('unknown')
print('RESULT:', result)
library.py 
lookup = {
    'r': 'red', 'g': 'green', 'b': 'blue',
    'y': 'yellow', 'w': 'white',
}
queries = ['r', 'b', 'x', 'y']
result = [lookup.get(q, 'unknown') for q in queries]
print('RESULT:', result)

RESULT: ['red', 'blue', 'unknown', 'yellow']

Implementation notes

  • d.get(key, default) is preferred over if key in d: d[key] when you only need the value and not the key's presence for branching; it is also slightly faster because it avoids a second hash lookup.
  • Dict lookup is O(1) average-case regardless of size, making it efficient for large tables where a linear list scan would be O(n).
  • Key order in Python dicts is insertion order (guaranteed since 3.7); the replay shows lookup entries appearing in the order they were inserted.