Counting with a Dictionary

1Program 1 — every character in a string

# how many of each character?

text = 'programming'
freq = {}

for ch in text:
    if ch in freq:
        freq[ch] = freq[ch] + 1
    else:
        freq[ch] = 1

print(freq)

{'p': 1, 'r': 2, 'o': 1, 'g': 2, 'a': 1, 'm': 2, 'i': 1, 'n': 1}

2Program 2 — printing the tally properly

print(freq) shows the raw dictionary, braces and all. A report walks it — and this is the loop the last chapter was building towards:

# walk the tally to print it as a report

text = 'programming'
freq = {}

for ch in text:
    freq[ch] = freq.get(ch, 0) + 1

for ch, n in freq.items():
    print(ch, '->', n)

p -> 1
r -> 2
o -> 1
g -> 2
a -> 1
m -> 2
i -> 1
n -> 1

Two loops, and they do quite different jobs. The first builds the dictionary, walking the string. The second reads it, walking the dictionary. Trying to print inside the first loop prints a running total after every letter, which is a different and much less useful thing.

The building loop has also shrunk to one line, using get() with a fallback of 0 — the shorter form from the previous chapter. It does exactly what the four-line if…else did.

3Program 3 — only the characters that repeat

# build the tally, then filter it while reading

text = 'programming'
freq = {}

for ch in text:
    freq[ch] = freq.get(ch, 0) + 1

for ch, n in freq.items():
    if n > 1:
        print(ch, 'appears', n, 'times')

r appears 2 times
g appears 2 times
m appears 2 times

4Program 4 — spaces skipped, case folded

On a real sentence, two things spoil the raw count: the spaces get counted, and 'M' and 'm' are tallied as different characters. Both are one line each:

# skip the spaces, treat 'M' and 'm' as the same letter

text = 'Mississippi River'
freq = {}

for ch in text.lower():
    if ch != ' ':
        freq[ch] = freq.get(ch, 0) + 1

print(freq)

{'m': 1, 'i': 5, 's': 4, 'p': 2, 'r': 2, 'v': 1, 'e': 1}

text.lower() in the loop header hands the loop a lowercased copy — the original text is untouched, because a string cannot be changed. The if ch != ' ': is the filter that keeps spaces out of the tally.

5Program 5 — counting a set you already know

Sometimes you know the keys in advance — the five vowels, the four grades. Then the dictionary can be built full of zeros before the loop, and the loop only ever adds:

# the keys are known in advance, so make them all at the start

text = 'she sells sea shells on the sea shore'
vowels = {'a': 0, 'e': 0, 'i': 0, 'o': 0, 'u': 0}

for ch in text:
    if ch in vowels:
        vowels[ch] = vowels[ch] + 1

for v, n in vowels.items():
    print(v, '->', n)

a -> 2
e -> 7
i -> 0
o -> 2
u -> 0

if ch in vowels: is doing double duty here — it is both the filter (ignore consonants and spaces) and the guarantee that no new key can ever appear.

6Program 6 — counting words instead of letters

Nothing about the tally cares what it is counting. split() cuts a sentence into a list of words, and the same loop runs over that instead:

# the same four lines, over words

line = 'the cat sat on the mat and the cat slept'
freq = {}

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

for w, n in freq.items():
    print(w, '->', n)

the -> 3
cat -> 2
sat -> 1
on -> 1
mat -> 1
and -> 1
slept -> 1

# which word turns up most often?

line = 'the cat sat on the mat and the cat slept'
freq = {}

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

top_word = ''
top_count = 0

for w, n in freq.items():
    if n > top_count:
        top_count = n
        top_word = w

print('Commonest word:', top_word, '-', top_count, 'times')

Commonest word: the - 3 times

The champion program, over .items(). Two variables move together inside one if, exactly as they did on the tuple of records — the shape does not change just because the data is a dictionary.

line = 'the cat sat on the mat and the cat slept'
freq = {}

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

top_word = ''
top_count = 0

for w, n in freq.items():
    if n > top_count:
        top_count = n
        top_word = w

print('Commonest word:', top_word, '-', top_count, 'times')

7Program 7 — the dice tally, in order

The frequency tuple from the previous chapter, now read out properly. A dictionary keeps insertion order, so the raw walk reports faces in the order they were first rolled; sorted() puts them in the order a reader expects:

# insertion order — the order each face was first seen

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

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

for face, n in freq.items():
    print('Face', face, 'came up', n, 'time(s)')

Face 3 came up 1 time(s)
Face 6 came up 3 time(s)
Face 2 came up 2 time(s)
Face 4 came up 1 time(s)
Face 1 came up 1 time(s)

# the same tally, walked in order of face

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

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

for face in sorted(freq):
    print('Face', face, 'came up', freq[face], 'time(s)')

Face 1 came up 1 time(s)
Face 2 came up 2 time(s)
Face 3 came up 1 time(s)
Face 4 came up 1 time(s)
Face 6 came up 3 time(s)

sorted(freq) hands back the keys in order, so the value comes from freq[face] rather than from a second loop variable. Face 5 is missing from both, because it was never rolled — the grow-as-you-go tally only ever holds what it has seen.

8Recap