Python dict: Build and Update a Station Lookup

Start with a small, explicit task

Build a lookup from fictional station codes to display labels. Correct one label and add a station. Then compare the mapping’s size and ordered keys with the original. The codes are deliberately strings, including a numeric-looking code, to show that an identifier’s type affects lookup.

Run the complete example

This original example was checked with Python 3.12.14. Save it as python_dictionary.py and run it with Python 3.12. It uses only the standard library. The assertions check the stated results.

stations = {"N1": "North", "S2": "South", "03": "Workshop"}
stations["S2"] = "South entrance"
stations["E4"] = "East"
assert stations["S2"] == "South entrance"
assert len(stations) == 4
assert list(stations) == ["N1", "S2", "03", "E4"]
assert "03" in stations and 3 not in stations
assert stations.get("missing", "Unknown") == "Unknown"
print(stations["S2"])
print(list(stations))
print(stations["03"])

Explain the result

Replacing S2 changes its label but leaves the number of keys unchanged. Adding E4 increases the mapping to four entries. Iterating this dictionary produces keys in insertion order; updating S2 does not move it to the end. That order is not alphabetical sorting, so sort explicitly if a report requires a sorted display.

The string key 03 exists, while the integer 3 does not. The distinction matters for imported identifiers: converting a code to a number can change what record you are asking for. get supplies Unknown for an absent key in this example. Direct bracket lookup is more appropriate when a missing required key should be treated as an error rather than silently defaulted.

Avoid these mistakes

Dictionary keys need to be hashable; a mutable list is not a usable key. Also remember that repeated equal keys cannot store separate entries: the later assignment replaces the earlier value. If one station needs several labels, model the value as a collection instead of repeating the same key.