Group By Length

Write a function that groups words by their length.

group_by_length(words)

Return a dict mapping each word length to a list of the words of that length, in the order they appeared in words.

Examples

group_by_length(["cat", "dog", "a", "bird", "ox", "at"])
// → {3: ["cat", "dog"], 1: ["a"], 4: ["bird"], 2: ["ox", "at"]}

group_by_length(["a", "b", "c"])
// → {1: ["a", "b", "c"]}

Walkthrough

Thinking it through

The tricky part isn't grouping by length — it's handling the first word of a given length, when there's no list to append to yet.

def group_by_length(words):
    groups = {}
    for word in words:
        groups.setdefault(len(word), []).append(word)
    return groups

.setdefault(key, default) is like .get()'s sibling, with one extra behavior: if key isn't in the dict yet, it stores default under that key (not just returns it) — and either way, it returns whatever ends up stored there. So groups.setdefault(len(word), []) does one of two things depending on whether that length has been seen before:

• First time seeing this length: no list exists yet, so setdefault creates a new empty list, stores it under groups[len(word)], and returns that same (now-stored) empty list.
• Seen this length before: the list already exists, so setdefault just returns the existing list, untouched.

Either way, you get back a real list that's already the one stored in groups, and .append(word) immediately adds to it. Without .setdefault(), you'd need an explicit check:

for word in words:
    length = len(word)
    if length not in groups:
        groups[length] = []
    groups[length].append(word)

Both versions do the same thing — .setdefault() just collapses the "does this key exist yet?" check and the list creation into one line.

Sign in to join the discussion.