Counting & Frequency

1Program 1 — how many times does one value appear?

# how many sixes?

rolls = (3, 6, 2, 6, 4, 6, 1)
wanted = 6
count = 0

for r in rolls:
    if r == wanted:
        count = count + 1

print(wanted, 'appeared', count, 'times')

6 appeared 3 times

The built-in does the same job in one call, and it is what you would write unless the question says not to:

rolls = (3, 6, 2, 6, 4, 6, 1)

print(rolls.count(6))
print(rolls.count(5))

3
0

count() answers 0 rather than raising anything for a value that never appears — which is the opposite of index(), and worth remembering because the two look like a pair.

2Program 2 — counting the ones that pass a test

# two counters, one if…else

numbers = (12, 7, 30, 45, 8, 21, 64)
evens = 0
odds = 0

for n in numbers:
    if n % 2 == 0:
        evens = evens + 1
    else:
        odds = odds + 1

print('Even:', evens)
print('Odd: ', odds)

Even: 4
Odd:  3

The check that it worked: 4 + 3 = 7, which is len(numbers). Every item landed in exactly one of the two, which is what an else guarantees and two separate ifs would not.

count() cannot do this one. It matches a value, not a rule — there is no way to ask it for “everything even”.

# the same shape, on a pass mark

marks = (56, 28, 91, 33, 12, 78, 65)
passed = 0
failed = 0

for m in marks:
    if m >= 33:
        passed = passed + 1
    else:
        failed = failed + 1

print('Passed:', passed)
print('Failed:', failed)

Passed: 5
Failed: 2

3Program 3 — the frequency of every value

A counter holds one number. Here the answer is six numbers — one per face — and they are not known in advance: on a tuple of names or colours you would not know how many counters to make. What is needed is a box that can grow a new compartment whenever an unseen value turns up, with the value itself as the label. That is a dictionary.

# tally every value into a dictionary

rolls = (3, 6, 2, 6, 4, 6, 1, 2)
freq = {}

for r in rolls:
    if r in freq:
        freq[r] = freq[r] + 1
    else:
        freq[r] = 1

print(freq)

{3: 1, 6: 3, 2: 2, 4: 1, 1: 1}

4The same thing with get()

get() asks for a key and takes a fallback for when the key is missing. That collapses the whole if…else into one line:

# the same tally, in one line inside the loop

rolls = (3, 6, 2, 6, 4, 6, 1, 2)
freq = {}

for r in rolls:
    freq[r] = freq.get(r, 0) + 1

print(freq)

{3: 1, 6: 3, 2: 2, 4: 1, 1: 1}

Read freq.get(r, 0) as “the count for r, or 0 if there is not one yet”. On a face never seen it hands back 0 and 0 + 1 makes the first entry; on a face seen before it hands back the running count. Same answer, four lines shorter, and it is the version experienced Python programmers write.

It works on anything a tuple can hold, not just numbers:

# the keys can be strings just as easily

words = ('red', 'blue', 'red', 'green', 'blue', 'red')
freq = {}

for w in words:
    freq[w] = freq.get(w, 0) + 1

print(freq)

{'red': 3, 'blue': 2, 'green': 1}

words = ('red', 'blue', 'red', 'green', 'blue', 'red')
freq = {}

for w in words:
    freq[w] = freq.get(w, 0) + 1

print(freq)

5Program 4 — which value appears most often?

The champion program again, with count() doing the measuring:

# the value that turns up most often

rolls = (3, 6, 2, 6, 4, 6, 1, 2)

best = rolls[0]
best_count = rolls.count(rolls[0])

for r in rolls:
    if rolls.count(r) > best_count:
        best = r
        best_count = rolls.count(r)

print('Most common:', best, '-', best_count, 'times')

Most common: 6 - 3 times

The statistics module has a name for this value — the mode — and computes it in one call. That is the last chapter of this course.

6Recap