CyberStars

Frequency Dictionary

We briefly saw in the dictionaries lesson how to count letters in a word. This pattern is so important that it deserves its own lesson. It's called a frequency dictionary (or histogram) and it's the go-to tool when the question is "how many times does each X appear?"

word = "mississippi"
freq = {}

for letter in word:
    if letter in freq:
        freq[letter] += 1
    else:
        freq[letter] = 1

print(freq)

Output {'m': 1, 'i': 4, 's': 4, 'p': 2}

Let's understand this line by line. We start with an empty dictionary. For each letter, we check: is it already in the dictionary? If yes, add 1 to its count. If not, create a new entry with count 1

This works for anything, not just letters. Counting words in a sentence

sentence = "I like pizza and I like burgers and I like tacos"
words = sentence.split(" ")
freq = {}

for word in words:
    if word in freq:
        freq[word] += 1
    else:
        freq[word] = 1

print(freq)

Output {'I': 3, 'like': 3, 'pizza': 1, 'and': 2, 'burgers': 1, 'tacos': 1}

Once we have a frequency dictionary, we can answer interesting questions

What's the most common element?

best_key = ""
best_count = 0
for key, count in freq.items():
    if count > best_count:
        best_count = count
        best_key = key
print(f"Most common: {best_key} ({best_count} times)")

What elements appear only once?

unique = []
for key, count in freq.items():
    if count == 1:
        unique.append(key)
print(f"Unique: {unique}")

Sort by frequency?

sorted_items = sorted(freq.items(), key=lambda x: x[1], reverse=True)
for item, count in sorted_items:
    print(f"{item}: {count}")

Don't worry about lambda for now, just know that it tells sorted() which value to sort by (in this case, the count)

A real-life use case: you're building an analytics system for CyberStars and you want to know which course is most popular

enrollments = ["python", "java", "python", "c", "python", "java", "c", "python", "java", "python"]
freq = {}
for course in enrollments:
    if course in freq:
        freq[course] += 1
    else:
        freq[course] = 1

for course, count in freq.items():
    print(f"{course}: {count} students")

Output

python: 5 students
java: 3 students
c: 2 students

Mission: Signal Frequencies

The station's antenna logs every signal it receives by channel letter in a list named signals. Mission Control wants to know how busy each channel is, and which one is busiest.

  1. Build a frequency dictionary counting how many times each channel appears.
  2. Print each channel and its count as channel: count, sorted alphabetically (loop over sorted(freq)).
  3. Find the most common channel and print Most common: followed by its letter.

Output

A: 3
B: 4
C: 1
D: 1
Most common: B