Dictionaries and loops: counting

What you will be able to do

After this lesson you can

• go through the keys, or the keys and values, of a dictionary,
• count letters and words with a dictionary,
• find the most common item, and
• find the key that has the highest value.

Looping over a dictionary

Counting with a dictionary

This is the most useful trick with dictionaries. The key is the thing that you count. The value is how many times you have seen it.

Predict first: how many times does each letter of banana appear? Then run it.

The line counts[letter] = counts.get(letter, 0) + 1 does the work. Read it from the right:

1. counts.get(letter, 0) gives the count so far, or 0 the first time we see the letter.
2. We add 1.
3. We store the new count under the same key. If the key is new, this adds the pair.

Follow the picture: after the first b, the dictionary is {"b": 1}. After the first a, {"b": 1, "a": 1}. And so on. After six rounds: b is 1, a is 3, n is 2.

Without get, you would have to write an if: if the letter is not a key, set it to 1, else add 1. The get makes it one line.

counts.get(ch, 0) is the trick: it gives 0 for a letter that has not been counted yet, so the first count becomes 1.

Counting words

The same idea counts words. The method split() cuts a sentence into a list of words, and the loop goes through that list.

Before you count words in a real text, think about clean-up. "The" and "the" are different keys. "dog." and "dog" are different keys. You will meet lower() and replace() for this in the lesson on working with text. For now, our sentence is already clean.

The pattern

You can count anything: the letters of a word, the words of a sentence, the marks of a class, the grades, the cities of a list. The steps are always the same.

1. Start with an empty dictionary: counts = {}.
2. Go through the items.
3. For each item: counts[item] = counts.get(item, 0) + 1.
4. Use the dictionary: show it, find the biggest, or look up one item.

Common mistakes

• Forgetting the 0 in get. counts.get(letter) + 1 gives a TypeError the first time, because None + 1 is not possible.
• Counting without cleaning. "The" and "the" are two words.
• Looping with for x in counts and expecting the pair. You get only the key. Use .items().
• Starting the "best" at 0 when all values may be negative.
• Changing the dictionary while looping over it.

Practice

Start with two quick questions. Each New numbers gives you a fresh one, so you can practise as long as you like.

Then the labs and the problems:

1. Warm-up. Open the lab Count the letters. You count the letters of a word, and the checker confirms the counts.
2. Build. Open the lab A word counter and the problem Count the letters.
3. Stretch. Open the problem The highest mark and who has it, and the problem Count the words.

Remember these

• for key in d: goes through the keys. for key, value in d.items(): gives both.
• Counting: counts[x] = counts.get(x, 0) + 1.
• The key is the thing you count. The value is how many times.
• To find the biggest value, remember the best key and the best value so far.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. How many keys does the counting dictionary of "banana" have?

2. What does counts.get("z", 0) give if "z" has not been counted?

3. What does for k, v in d.items() give in each round?