Dictionary creation and access in Swift

Introduction

Dictionaries help you store pairs of related information, like names and phone numbers. You can quickly find a value by using its key.

When you want to store a list of items with labels, like a phone book.

When you need to look up information fast using a unique key.

When you want to group related data together, like a person's details.

When you want to update or add new information easily by key.

Syntax

Swift

var dictionaryName: [KeyType: ValueType] = [key1: value1, key2: value2]

// Access value by key
let value = dictionaryName[key]

Keys must be unique and hashable (like strings or numbers).

Accessing a key returns an optional value because the key might not exist.

Examples

This creates a dictionary with names as keys and ages as values. Then it gets Alice's age.

Swift

var ages: [String: Int] = ["Alice": 30, "Bob": 25]
let aliceAge = ages["Alice"]

Swift can infer the types. This dictionary stores countries and their capitals.

Swift

var capitals = ["France": "Paris", "Japan": "Tokyo"]
let capitalOfJapan = capitals["Japan"]

This creates an empty dictionary and adds a key-value pair later.

Swift

var emptyDict = [String: String]()
emptyDict["key"] = "value"

Sample Program

This program creates a dictionary of fruits and their colors. It safely accesses the color of Apple and Orange, showing how to handle missing keys.

Swift

import Foundation

var fruitsColors: [String: String] = ["Apple": "Red", "Banana": "Yellow", "Grape": "Purple"]

if let appleColor = fruitsColors["Apple"] {
    print("The color of Apple is \(appleColor).")
} else {
    print("Apple color not found.")
}

// Trying to access a key that does not exist
if let orangeColor = fruitsColors["Orange"] {
    print("The color of Orange is \(orangeColor).")
} else {
    print("Orange color not found.")
}

Important Notes

Always use optional binding (if let) when accessing dictionary values to avoid errors if the key is missing.

You can add or update values by assigning to a key, like dict["newKey"] = newValue.

Dictionaries do not keep order, so the order of items is not guaranteed.

Summary

Dictionaries store data in key-value pairs for quick lookup.

Use square brackets with a key to get or set values.

Accessing a key returns an optional, so handle missing keys safely.

Practice

1. What is a dictionary in Swift used for?

easy

A. To store data in key-value pairs for quick lookup

B. To store a list of values without keys

C. To perform mathematical calculations

D. To create user interface elements

Solution

1. Step 1: Understand dictionary purpose

   Dictionaries store data as pairs of keys and values, allowing fast access by key.

2. Step 2: Compare options

   Only To store data in key-value pairs for quick lookup correctly describes this key-value storage purpose.

3. Final Answer:

   To store data in key-value pairs for quick lookup -> Option A

4. Quick Check:

   Dictionaries = key-value storage [OK]

Hint: Remember: dictionaries pair keys with values for fast access [OK]

Common Mistakes:

• Confusing dictionaries with arrays (which have no keys)
• Thinking dictionaries perform calculations
• Assuming dictionaries create UI elements

2. Which of the following is the correct way to create a dictionary in Swift?

easy

A. let dict = ("apple", 1), ("banana", 2)

B. let dict = ["apple": 1, "banana": 2]

C. let dict = {"apple": 1, "banana": 2}

D. let dict = ["apple", 1, "banana", 2]

Solution

1. Step 1: Identify Swift dictionary syntax

   Swift dictionaries use square brackets with key-value pairs separated by colons.

2. Step 2: Check each option

   let dict = ["apple": 1, "banana": 2] uses correct syntax: ["key": value, ...]. Others use tuples, braces, or arrays incorrectly.

3. Final Answer:

   let dict = ["apple": 1, "banana": 2] -> Option B

4. Quick Check:

   Dictionary syntax = [key: value] [OK]

Hint: Use square brackets with colons for key-value pairs [OK]

Common Mistakes:

• Using parentheses instead of square brackets
• Using curly braces which define closures
• Listing keys and values without colons

3. What will be the output of this Swift code?

let fruits = ["apple": 3, "banana": 5]
print(fruits["banana"] ?? 0)

medium

A. 5

B. nil

C. banana

D. 0

Solution

1. Step 1: Access dictionary value by key

   fruits["banana"] looks up the value for key "banana", which is 5.

2. Step 2: Use nil-coalescing operator

   The ?? 0 means if the key is missing (nil), print 0; here key exists, so print 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   fruits["banana"] = 5 [OK]

Hint: Use ?? to provide default if key missing [OK]

Common Mistakes:

• Expecting nil instead of default value
• Printing the key instead of the value
• Confusing optional with direct value

4. Identify the error in this Swift code snippet:

var scores = ["Alice": 90, "Bob": 85]
scores["Charlie"] = 95
print(scores["Charlie"])

medium

A. Cannot add new key-value pairs to a dictionary

B. Dictionary keys must be integers

C. Syntax error in dictionary declaration

D. Missing unwrap of optional when printing

Solution

1. Step 1: Check dictionary modification

   Adding a new key-value pair to a var dictionary is allowed, so no error there.

2. Step 2: Understand optional access

   Accessing scores["Charlie"] returns an optional Int?; printing it directly shows Optional(95), not 95.

3. Step 3: Identify missing unwrap

   To print the value without Optional(...), unwrap it safely or use ?? to provide default.

4. Final Answer:

   Missing unwrap of optional when printing -> Option D

5. Quick Check:

   Dictionary access returns optional, unwrap needed [OK]

Hint: Remember dictionary access returns optional, unwrap before use [OK]

Common Mistakes:

• Thinking keys must be integers
• Assuming dictionary is immutable by default
• Ignoring optional nature of dictionary values

5. Given a dictionary let items = ["pen": 3, "notebook": 0, "eraser": 2], which Swift code correctly creates a new dictionary containing only items with quantity greater than zero?

hard

A. let filtered = items.compactMap { $0.value > 0 }

B. let filtered = items.filter { $0.key > 0 }

C. let filtered = items.filter { $0.value > 0 }

D. let filtered = items.map { $0.value > 0 }

Solution

1. Step 1: Understand filtering dictionary by value

   We want to keep pairs where value > 0, so filter with condition on $0.value.

2. Step 2: Analyze each option

   let filtered = items.filter { $0.value > 0 } uses filter with correct condition. let filtered = items.filter { $0.key > 0 } incorrectly compares key (a String) to 0. Options A and C use map/compactMap which transform values, not filter pairs.

3. Final Answer:

   let filtered = items.filter { $0.value > 0 } -> Option C

4. Quick Check:

   Filter dictionary by value > 0 = let filtered = items.filter { $0.value > 0 } [OK]

Hint: Use filter with $0.value condition to keep matching pairs [OK]

Common Mistakes:

• Filtering by key instead of value
• Using map instead of filter for selection
• Using compactMap which changes type, not filters