Python Dictionaries, Sets and Tuples Practice

Practise counting and grouping with dictionaries, comparing groups with sets, and packing and unpacking tuples.

Beginner9 problemsUpdated October 23, 2026

Review first: Dictionaries, Sets, Tuples

1.Word frequency
Beginner
Count how many times each word appears in text = "red blue red green blue red", then print each word and its count in alphabetical order.
Show hint
counts[word] = counts.get(word, 0) + 1 adds one, starting from 0 for a new word.
Show solution
Python
text = "red blue red green blue red" counts = {} for word in text.split(): counts[word] = counts.get(word, 0) + 1 for word in sorted(counts): print(word, counts[word])
Output
blue 2 green 1 red 3
collections.Counter(text.split()) does the same counting in one line, and .most_common() sorts by count.
2.Safe lookups
Beginner
Given phone_book = {"Asha": "98200 11111", "Ravi": "98300 22222"}, print the numbers for Asha and Meera. For a name that is not in the book, print not found instead of crashing. Then add Meera's number 98400 33333 and print how many entries there are.
Show hint
phone_book["Meera"] raises a KeyError; phone_book.get("Meera", "not found") does not.
Show solution
Python
phone_book = {"Asha": "98200 11111", "Ravi": "98300 22222"} for name in ["Asha", "Meera"]: print(name, phone_book.get(name, "not found")) phone_book["Meera"] = "98400 33333" print(len(phone_book), "entries")
Output
Asha 98200 11111 Meera not found 3 entries
3.Average marks per student
Beginner
Print each student's average mark, rounded to 1 decimal place:

marks = {"Asha": [88, 92, 79], "Ravi": [75, 64, 81], "Meera": [95, 90, 98]}
Show hint
Loop with .items() to get each name and list together. The average is sum(list) / len(list).
Show solution
Python
marks = {"Asha": [88, 92, 79], "Ravi": [75, 64, 81], "Meera": [95, 90, 98]} for name, scores in marks.items(): average = sum(scores) / len(scores) print(f"{name}: {average:.1f}")
Output
Asha: 86.3 Ravi: 73.3 Meera: 94.3
4.Group words by first letter
Beginner
Group words = ["apple", "banana", "avocado", "cherry", "blueberry", "apricot"] into a dictionary where each key is a first letter and each value is the list of words starting with it. Print the dictionary.
Show hint
groups.setdefault(letter, []) returns the existing list for that letter, or creates an empty one.
Show solution
Python
words = ["apple", "banana", "avocado", "cherry", "blueberry", "apricot"] groups = {} for word in words: groups.setdefault(word[0], []).append(word) print(groups)
Output
{'a': ['apple', 'avocado', 'apricot'], 'b': ['banana', 'blueberry'], 'c': ['cherry']}
Dictionaries keep insertion order, so the keys appear in the order their first word was seen. collections.defaultdict(list) is another common way to write this.
5.Compare two clubs
Beginner
Given chess = {"Asha", "Ravi", "Meera", "John"} and music = {"Meera", "Priya", "John", "Kiran"}, print (as sorted lists):
1. students in both clubs
2. students in either club
3. students only in chess
Show hint
& is intersection, | is union and - is difference. Wrap each result in sorted() because sets have no order.
Show solution
Python
chess = {"Asha", "Ravi", "Meera", "John"} music = {"Meera", "Priya", "John", "Kiran"} print(sorted(chess & music)) print(sorted(chess | music)) print(sorted(chess - music))
Output
['John', 'Meera'] ['Asha', 'John', 'Kiran', 'Meera', 'Priya', 'Ravi'] ['Asha', 'Ravi']
Printing a set directly can show the names in a different order each time the program runs, so sort before printing whenever the order matters.
6.Remove duplicates, keep order
Beginner
Remove the duplicates from visits = ["home", "about", "home", "blog", "about", "contact"] while keeping the first-seen order.
Show hint
list(set(visits)) loses the order. dict.fromkeys(visits) keeps it, because dictionary keys are unique and ordered.
Show solution
Python
visits = ["home", "about", "home", "blog", "about", "contact"] unique = list(dict.fromkeys(visits)) print(unique)
Output
['home', 'about', 'blog', 'contact']
7.Swap keys and values
Beginner
Turn codes = {"IN": "India", "JP": "Japan", "FR": "France"} into a dictionary that maps each country name to its code, and print the code for Japan.
Show hint
Loop over .items() and build a new dictionary with new[value] = key, or use a dictionary comprehension.
Show solution
Python
codes = {"IN": "India", "JP": "Japan", "FR": "France"} by_country = {country: code for code, country in codes.items()} print(by_country) print(by_country["Japan"])
Output
{'India': 'IN', 'Japan': 'JP', 'France': 'FR'} JP
This only works safely when the values are unique. If two keys shared a value, the later one would overwrite the earlier one in the new dictionary.
8.Unpack tuples
Beginner
Given points = [(2, 3), (-1, 5), (4, -2)], print each point as x=2, y=3. Then swap the variables a = 1 and b = 2 in a single line and print them.
Show hint
for x, y in points: unpacks each tuple. a, b = b, a swaps two variables.
Show solution
Python
points = [(2, 3), (-1, 5), (4, -2)] for x, y in points: print(f"x={x}, y={y}") a = 1 b = 2 a, b = b, a print(a, b)
Output
x=2, y=3 x=-1, y=5 x=4, y=-2 2 1
9.Tuples cannot change
Beginner
Create rgb = (255, 128, 0), try to change its first value to 0, and catch the error to print its message. Then create a new tuple with the first value changed and print it.
Show hint
Assigning to rgb[0] raises a TypeError. Build a new tuple with (0,) + rgb[1:].
Show solution
Python
rgb = (255, 128, 0) try: rgb[0] = 0 except TypeError as error: print("Error:", error) new_rgb = (0,) + rgb[1:] print(new_rgb)
Output
Error: 'tuple' object does not support item assignment (0, 128, 0)
The comma makes a tuple, not the brackets, so (0,) is a one-item tuple, while (0) is just the number 0.

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