Real records sort by more than one field. Sort by the least important field first, or hand the sort a tuple of keys, and the order becomes total.
About 3 min of reading
Before this new lesson, one small question about the lesson you just finished. Recall it, do not look it up.
Sort records by multiple fields.
Sort tuples by both fields.
Before you run it, say out loud what it will print. Then open the answer.
Sort by time then name.
Before you run it, say out loud what it will print. Then open the answer.
Sort ISO dates.
Before you run it, say out loud what it will print. Then open the answer.
Sort level descending then id.
Before you run it, say out loud what it will print. Then open the answer.
Before you write anything, put these lines back in the right order. Click a line to move it up one step. This support step is easier than writing code from scratch, and it is here to make the last example stick.
Read the first example one line at a time. A line that prints shows what appears on screen; the rest set things up quietly.
Returning a tuple from the key is the clean way to sort by several fields. Each field is a comparison level, left to right.
Sorting by one key at a time, least important first, works only because the sort is stable. Know the guarantee you are leaning on.
rows = [(3, "a"), (1, "b"), (3, "a"), (1, "c")]
print(sorted(rows))A tuple compares left to right. Ties in the first field fall through to the second.
rows = [{"n": "owl", "t": 9}, {"n": "bee", "t": 4}, {"n": "owl", "t": 2}]
rows.sort(key=lambda r: (r["t"], r["n"]))
print(rows)The key returns a tuple. Sort time first, then name, giving a total order.
rows = [(2, "b"), (1, "a"), (2, "a")]
rows.sort(key=lambda r: r[1])
rows.sort(key=lambda r: r[0])
print(rows)Stable sorts let you sort by the last key first, then the earlier one. The second sort keeps the first order on ties.
rows = ["2026-10-11", "2025-04-01", "2026-01-05"]
print(sorted(rows))ISO dates sort correctly as text because the biggest unit comes first. That is why the format is year first.
rows = [3, None, 1]
print(sorted(rows, key=lambda x: (x is None, x if x is not None else 0)))Push the missing values to the end with a flag in the key. Decide where they belong on purpose.
words = ["Owlww", "apple", "Bee"]
print(sorted(words, key=str.lower))Lower-casing in the key gives dictionary order. The default puts all capitals before all lowercase.
rows = [(1, "a"), (1, "a")]
print(sorted(rows))If two records tie on every key, the order between them is arbitrary but stable. Add a final tiebreaker for a real total order.
rows = [{"lvl": 2, "id": "b"}, {"lvl": 1, "id": "a"}, {"lvl": 2, "id": "a"}]
rows.sort(key=lambda r: (-r["lvl"], r["id"]))
print(rows)Negate a number to reverse just that field. Level descending, then id ascending.
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