String searching (contains, strfind) in MATLAB - Step-by-Step Execution

Concept Flow - String searching (contains, strfind)

Start with string S

↓

Search for substring sub

↓

Return indices of sub

↓

End

The program starts with a main string and a substring to find. It checks if the substring exists inside the main string. If yes, it returns the positions; if no, it returns empty.

Execution Sample

MATLAB

S = 'hello world';
sub = 'lo';
idx = strfind(S, sub);
found = contains(S, sub);

This code searches for 'lo' inside 'hello world' using strfind and contains.

Execution Table

Step	Variable	Value	Action	Result
1	S	'hello world'	Initialize main string	'hello world'
2	sub	'lo'	Initialize substring to find	'lo'
3	idx	[4]	Call strfind(S, sub)	Returns [4] because 'lo' starts at position 4
4	found	true	Call contains(S, sub)	Returns true because 'lo' is in S
5	found	true	Assign contains result to found	found = true

💡 Search complete: strfind found index 4, contains returned true

Variable Tracker

Variable	Start	After strfind	After contains	Final
S	'hello world'	'hello world'	'hello world'	'hello world'
sub	'lo'	'lo'	'lo'	'lo'
idx	[]	[4]	[4]	[4]
found	false	false	true	true

Key Moments - 2 Insights

Why does strfind return [4] but contains returns true?

strfind returns the starting index of the substring in the main string (4 means 'lo' starts at position 4). contains returns a logical true/false if the substring exists anywhere. See execution_table rows 3 and 4.

What happens if the substring is not found?

strfind returns an empty array [], and contains returns false. This is shown by the exit_note and would be reflected in idx and found variables.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of idx after step 3?

A [4]

B []

C [1]

D true

💡 Hint

Check the 'Value' column for idx at step 3 in the execution_table.

At which step does the variable 'found' become true?

A Step 2

B Step 5

C Step 4

D Step 3

💡 Hint

Look at the 'found' variable changes in execution_table rows 4 and 5.

If sub = 'xyz' (not in S), what would strfind return?

A [4]

B [1]

C []

D true

💡 Hint

Recall from key_moments that strfind returns empty array if substring not found.

Concept Snapshot

String searching in MATLAB:
- Use strfind(S, sub) to get indices where sub starts in S
- Use contains(S, sub) to get true/false if sub is in S
- strfind returns empty [] if not found
- contains returns false if not found
- Both help find substrings inside strings

Full Transcript

This visual execution shows how MATLAB searches for a substring inside a string using strfind and contains. We start with a main string S and a substring sub. strfind returns the starting index of sub in S or empty if not found. contains returns true if sub is anywhere in S, otherwise false. The execution table traces variable values step-by-step. Key moments clarify why strfind returns indices and contains returns logical values. The quiz tests understanding of these steps.

Practice

1. Which MATLAB function returns a logical true or false indicating if a substring exists within a string?

easy

A. contains

B. strfind

C. findstr

D. substring

Solution

1. Step 1: Understand the purpose of contains

   contains returns true or false if a substring is found inside a string.

2. Step 2: Compare with strfind

   strfind returns positions, not logical true/false.

3. Final Answer:

   contains -> Option A

4. Quick Check:

   Logical check for substring = contains [OK]

Hint: Use contains for true/false substring check [OK]

Common Mistakes:

• Confusing strfind with contains
• Thinking strfind returns true/false
• Using findstr which is outdated

2. Which of the following is the correct syntax to find the starting indices of substring 'cat' in string str = 'concatenate' using strfind?

easy

A. contains(str, 'cat')

B. strfind('cat', str)

C. strfind(str, 'cat')

D. strfind(str, cat)

Solution

1. Step 1: Recall strfind syntax

   The first argument is the main string, second is the substring to find.

2. Step 2: Check each option

   strfind(str, 'cat') correctly uses strfind(str, 'cat'). strfind('cat', str) reverses arguments. contains(str, 'cat') uses contains, not strfind. strfind(str, cat) misses quotes around 'cat'.

3. Final Answer:

   strfind(str, 'cat') -> Option C

4. Quick Check:

   strfind(mainString, substring) = strfind(str, 'cat') [OK]

Hint: First argument main string, second substring in strfind [OK]

Common Mistakes:

• Swapping arguments in strfind
• Forgetting quotes around substring
• Using contains instead of strfind for indices

3. What is the output of the following MATLAB code?

str = 'abracadabra';
pos = strfind(str, 'ra');
disp(pos);

medium

A. [3 10]

B. [2 9]

C. [4 11]

D. []

Solution

1. Step 1: Identify substring positions in 'abracadabra'

   Look for 'ra' in the string: 'abracadabra'. The substring 'ra' starts at index 3 and 10.

2. Step 2: Confirm output of strfind

   strfind returns starting indices of matches, so output is [3 10].

3. Final Answer:

   [3 10] -> Option A

4. Quick Check:

   Positions of 'ra' = [3 10] [OK]

Hint: Count string indices carefully for strfind output [OK]

Common Mistakes:

• Counting indices starting at 0 instead of 1
• Confusing substring length with position
• Expecting logical output instead of indices

4. Identify the error in this MATLAB code snippet:

str = 'hello world';
pos = strfind('world', str);
disp(pos);

medium

A. disp cannot display arrays

B. Arguments of strfind are reversed

C. strfind cannot search for 'world'

D. Missing semicolon after strfind

Solution

1. Step 1: Check strfind argument order

   strfind expects the main string first, then the substring. Here, arguments are reversed.

2. Step 2: Understand impact of reversed arguments

   Searching 'world' inside 'hello world' is correct, but reversed arguments cause wrong or empty output.

3. Final Answer:

   Arguments of strfind are reversed -> Option B

4. Quick Check:

   strfind(mainString, substring) correct order [OK]

Hint: Main string first, substring second in strfind [OK]

Common Mistakes:

• Swapping arguments in strfind
• Assuming disp can't show arrays
• Thinking semicolon is mandatory for correctness

5. You want to find all positions of the substring 'ana' in str = 'banana bandana' and check if it exists. Which code correctly does this?

hard

A. positions = strfind('ana', str); exists = contains('ana', str);

B. positions = contains(str, 'ana'); exists = strfind(str, 'ana');

C. positions = find(str, 'ana'); exists = contains(str, 'ana');

D. positions = strfind(str, 'ana'); exists = contains(str, 'ana');

Solution

1. Step 1: Use strfind to get positions

   strfind(str, 'ana') returns all starting indices of 'ana' in the string.

2. Step 2: Use contains to check existence

   contains(str, 'ana') returns true if 'ana' is found anywhere in str.

3. Step 3: Verify options

   positions = strfind(str, 'ana'); exists = contains(str, 'ana'); correctly uses both functions. positions = contains(str, 'ana'); exists = strfind(str, 'ana'); swaps roles. positions = strfind('ana', str); exists = contains('ana', str); reverses arguments. positions = find(str, 'ana'); exists = contains(str, 'ana'); uses invalid function find for strings.

4. Final Answer:

   positions = strfind(str, 'ana'); exists = contains(str, 'ana'); -> Option D

5. Quick Check:

   Positions with strfind, existence with contains [OK]

Hint: Use strfind for positions, contains for true/false [OK]

Common Mistakes:

• Swapping arguments in strfind or contains
• Using find instead of strfind
• Confusing output types of contains and strfind