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PYTHON STRUCTURES / CHAPTER 06 / Learn / LESSON 11

Group Count.

About 5–10 minutes

Counting with a dict beats nested loops:

counts = {} for word in ["a", "b", "a"]: counts[word] = counts.get(word, 0) + 1

get(word, 0) supplies 0 if missing. One pass O(n) instead of scanning all keys for every item.

Your lesson

words = ["a", "b", "a"]. Count with get in a dict. Print counts["a"] and counts["b"]. Expected: 2 and 1.

Learn: one pass with dict.

lesson_11.pyPYTHON

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AGGREGATE IDLE

01get loop

02a=2

03b=1

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