Logical indexing in MATLAB - Step-by-Step Execution

Concept Flow - Logical indexing

Create array A

↓

Create logical mask L

↓

Use L to select elements from A

↓

Output selected elements

↓

End

Logical indexing uses a true/false mask to pick elements from an array.

Execution Sample

MATLAB

A = [10, 20, 30, 40, 50];
L = A > 25;
B = A(L);

Selects elements from A that are greater than 25 using logical indexing.

Execution Table

Step	Variable	Value	Action	Output
1	A	[10 20 30 40 50]	Create array A
2	L	[false false true true true]	Compare A > 25 to create logical mask
3	B	[30 40 50]	Use L to select elements from A	[30 40 50]
4	-	-	End of execution

💡 All steps completed; logical indexing selected elements > 25.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
A	undefined	[10 20 30 40 50]	[10 20 30 40 50]	[10 20 30 40 50]	[10 20 30 40 50]
L	undefined	undefined	[false false true true true]	[false false true true true]	[false false true true true]
B	undefined	undefined	undefined	[30 40 50]	[30 40 50]

Key Moments - 3 Insights

Why does the logical mask L have true only for some elements?

Because L is created by comparing each element of A to 25, only elements greater than 25 get true (see step 2 in execution_table).

What happens if we use L to index A?

Only elements where L is true are selected from A, producing B with those elements (see step 3 in execution_table).

Can logical indexing change the size of the output array?

Yes, the output array B only contains elements where L is true, so it can be smaller than A (see step 3 output).

Visual Quiz - 3 Questions

Look at the execution_table at step 2, what is the value of L?

A [true true false false false]

B [false false true true true]

C [true false true false true]

D [false true false true false]

💡 Hint

Check the 'Value' column for L at step 2 in the execution_table.

At which step does B get its value assigned?

A Step 3

B Step 1

C Step 2

D Step 4

💡 Hint

Look for when B is created and assigned in the execution_table.

If A was [5 15 25 35 45], what would be the value of L after step 2?

A [false true true true true]

B [true true true false false]

C [false false false true true]

D [true false false false true]

💡 Hint

Compare each element of A to 25 to create L as in step 2.

Concept Snapshot

Logical indexing in MATLAB:
- Create a logical mask by comparing array elements.
- Use mask to select elements: B = A(L).
- Output contains only elements where mask is true.
- Useful for filtering arrays easily.

Full Transcript

Logical indexing in MATLAB lets you pick elements from an array by using a true/false mask. First, you create an array A. Then you make a logical mask L by comparing A to a condition, like A > 25. This mask has true where the condition is met and false elsewhere. Using A(L) selects only the elements where L is true. For example, if A = [10 20 30 40 50], then L = A > 25 gives [false false true true true]. Using B = A(L) results in B = [30 40 50]. This method is simple and powerful for filtering data.

Practice

1. What does logical indexing in MATLAB allow you to do with an array A and a logical array L of the same size?

easy

A. Sort the elements of A based on L

B. Select elements of A where L is true

C. Change the size of A to match L

D. Convert A into a logical array

Solution

1. Step 1: Understand logical indexing concept

   Logical indexing uses a logical array to pick elements from another array where the logical array is true.

2. Step 2: Apply to arrays A and L

   Elements of A are selected only where L has true values, filtering A.

3. Final Answer:

   Select elements of A where L is true -> Option B

4. Quick Check:

   Logical indexing = select elements [OK]

Hint: Logical array true picks elements from original array [OK]

Common Mistakes:

• Thinking logical indexing changes array size arbitrarily
• Confusing logical indexing with sorting
• Assuming logical indexing converts data types

2. Which of the following is the correct syntax to select elements of vector v that are greater than 5 using logical indexing?

easy

A. v(v > 5)

B. v > 5(v)

C. v{v > 5}

D. v[v > 5]

Solution

1. Step 1: Identify correct logical indexing syntax

   In MATLAB, logical indexing uses parentheses with a logical condition inside, like v(condition).

2. Step 2: Check each option

   v(v > 5) uses v(v > 5), which is correct syntax. The other options use invalid indexing forms.

3. Final Answer:

   v(v > 5) -> Option A

4. Quick Check:

   Logical indexing syntax = v(condition) [OK]

Hint: Use parentheses with condition inside for logical indexing [OK]

Common Mistakes:

• Using curly braces or square brackets for logical indexing
• Placing condition outside parentheses
• Confusing logical indexing with cell or array indexing

3. Given A = [3, 7, 2, 9, 5], what is the result of A(A > 4)?

medium

A. [2, 5]

B. [3, 7, 2, 9, 5]

C. [7, 9]

D. [7, 9, 5]

Solution

1. Step 1: Create logical array for condition A > 4

   Check each element: 3>4?false, 7>4?true, 2>4?false, 9>4?true, 5>4?true. Logical array = [false, true, false, true, true].

2. Step 2: Select elements of A where logical array is true

   Elements at true positions: 7, 9, 5.

3. Final Answer:

   [7, 9, 5] -> Option D

4. Quick Check:

   Elements > 4 = [7, 9, 5] [OK]

Hint: Filter elements by condition, then pick true positions [OK]

Common Mistakes:

• Including elements not greater than 4
• Confusing logical indexing with logical AND operation
• Returning original array instead of filtered

4. What is wrong with this MATLAB code snippet?

A = [1, 4, 6, 8];
L = [true, false];
B = A(L);

medium

A. Logical array L size does not match A size

B. Logical array L must contain only numeric values

C. Cannot use logical indexing on row vectors

D. Syntax error in logical indexing

Solution

1. Step 1: Check sizes of A and L

   A has 4 elements, L has 2 elements. Logical indexing requires L to be same size as A.

2. Step 2: Identify error cause

   Since sizes differ, MATLAB will give an error when using L to index A.

3. Final Answer:

   Logical array L size does not match A size -> Option A

4. Quick Check:

   Logical array size must match data size [OK]

Hint: Logical array must be same size as indexed array [OK]

Common Mistakes:

• Assuming logical array can be smaller
• Thinking logical array must be numeric
• Believing logical indexing syntax is wrong here

5. You have a matrix M = [2, 5, 8; 1, 7, 3; 4, 6, 9]. How do you replace all elements less than 5 with zero using logical indexing?

hard

A. M(M <= 5) = 0;

B. M(M > 5) = 0;

C. M(M < 5) = 0;

D. M(M == 5) = 0;

Solution

1. Step 1: Identify elements less than 5 in M

   Elements less than 5 are 2, 1, 3, 4.

2. Step 2: Use logical indexing to assign zero to these elements

   Using M(M < 5) = 0; replaces all elements less than 5 with zero.

3. Final Answer:

   M(M < 5) = 0; -> Option C

4. Quick Check:

   Replace elements < 5 with 0 using M(M < 5) = 0 [OK]

Hint: Use condition inside parentheses to assign new values [OK]

Common Mistakes:

• Replacing wrong elements by using > or == instead of <
• Using incorrect logical operators
• Trying to assign without logical indexing