String Manipulation

This lesson covers string manipulation techniques as required by OCR J277 Section 2.3. String manipulation is the process of examining, modifying, and extracting parts of strings. OCR requires you to know specific string operations including length, substring, upper/lower case conversion, concatenation, and ASCII conversion.

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String Basics

A string is a sequence of characters. Each character in a string has a position (index), starting from index 0.

word = "COMPUTER"
// Index: 0=C, 1=O, 2=M, 3=P, 4=U, 5=T, 6=E, 7=R

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Length

The .length property returns the number of characters in a string.

OCR Pseudocode:

word = "Hello"
print(word.length)    // Output: 5

Python:

word = "Hello"
print(len(word))    # Output: 5

Note: the length counts all characters including spaces and punctuation. "Hi there!".length = 9.

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Substring

The .substring(start, end) method extracts a portion of a string from the start index up to (but not including) the end index.

OCR Pseudocode:

word = "COMPUTER"
print(word.substring(0, 3))    // Output: COM
print(word.substring(3, 6))    // Output: PUT
print(word.substring(4, 8))    // Output: UTER

Python:

word = "COMPUTER"
print(word[0:3])    # Output: COM
print(word[3:6])    # Output: PUT
print(word[4:8])    # Output: UTER

Example	Start	End	Result	Characters Extracted
"COMPUTER".substring(0, 3)	0	3	"COM"	Indices 0, 1, 2
"COMPUTER".substring(3, 6)	3	6	"PUT"	Indices 3, 4, 5
"COMPUTER".substring(6, 8)	6	8	"ER"	Indices 6, 7

OCR Exam Tip: The .substring(start, end) method uses zero-based indexing and the end index is EXCLUSIVE (not included). This is a very common source of errors. For "HELLO".substring(1, 4), you get characters at indices 1, 2, 3 — which is "ELL", not "ELLO".

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Upper and Lower Case Conversion

OCR Pseudocode:

word = "Hello World"
print(word.upper)    // Output: HELLO WORLD
print(word.lower)    // Output: hello world

Python:

word = "Hello World"
print(word.upper())    # Output: HELLO WORLD
print(word.lower())    # Output: hello world

Common Use: Case-Insensitive Comparison

userInput = input("Enter yes or no: ")
if userInput.lower == "yes" then
    print("You chose yes")
endif

user_input = input("Enter yes or no: ")
if user_input.lower() == "yes":
    print("You chose yes")

OCR Exam Tip: In OCR pseudocode, .upper and .lower do NOT use brackets. In Python, .upper() and .lower() DO use brackets. Make sure you use the correct syntax for whichever language you are writing in.

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Concatenation

Concatenation means joining two or more strings together using the + operator.

OCR Pseudocode:

firstName = "John"
lastName = "Smith"
fullName = firstName + " " + lastName
print(fullName)    // Output: John Smith

Python:

first_name = "John"
last_name = "Smith"
full_name = first_name + " " + last_name
print(full_name)    # Output: John Smith

When concatenating strings with numbers, convert the number to a string first:

age = 15
message = "You are " + str(age) + " years old"
print(message)    // Output: You are 15 years old

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ASCII Conversion

Every character has a numerical ASCII code. OCR pseudocode provides ASC() and CHR() functions for conversion.

Function	OCR Pseudocode	Python	Description
Character to ASCII code	ASC('A')	ord('A')	Returns the ASCII value: 65
ASCII code to character	CHR(65)	chr(65)	Returns the character: 'A'

Common ASCII Values

Character	ASCII Code
'0' to '9'	48 to 57
'A' to 'Z'	65 to 90
'a' to 'z'	97 to 122
Space	32

OCR Pseudocode example — Caesar cipher shift:

letter = "H"
asciiValue = ASC(letter)          // 72
shifted = asciiValue + 3          // 75
newLetter = CHR(shifted)          // "K"
print(newLetter)                  // Output: K

Python:

letter = "H"
ascii_value = ord(letter)         # 72
shifted = ascii_value + 3         # 75
new_letter = chr(shifted)         # "K"
print(new_letter)                 # Output: K

The transformation flow that turns a single character into a shifted character can be drawn as a string manipulation pipeline:

flowchart LR
    A[letter = 'H'] --> B[ASC letter]
    B --> C[asciiValue = 72]
    C --> D[asciiValue + 3]
    D --> E[shifted = 75]
    E --> F[CHR shifted]
    F --> G[newLetter = 'K']
    G --> H[print newLetter]

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Worked Example: Reverse a String

OCR Pseudocode:

original = input("Enter a word: ")
reversed = ""
for i = original.length - 1 to 0 step -1
    reversed = reversed + original.substring(i, i + 1)
next i
print(reversed)

Python:

original = input("Enter a word: ")
reversed_str = ""
for i in range(len(original) - 1, -1, -1):
    reversed_str += original[i]
print(reversed_str)

# Or more Pythonically:
print(original[::-1])

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Summary of String Operations

Operation	OCR Pseudocode	Python
Length	str.length	len(str)
Substring	str.substring(start, end)	str[start:end]
Upper case	str.upper	str.upper()
Lower case	str.lower	str.lower()
Concatenation	str1 + str2	str1 + str2
Char to ASCII	ASC(char)	ord(char)
ASCII to char	CHR(code)	chr(code)

OCR Exam Tip: String manipulation questions are very common on Paper 2. The most frequently tested operations are .substring(), .length, and ASC()/CHR(). Practise these until you can use them confidently and quickly.

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Worked Example: Extracting Initials from a Full Name

The user enters their full name, and the program outputs the initials in upper case separated by dots (for example: "john samuel smith" becomes "J.S.S.").

OCR reference language pseudocode:

fullName = input("Enter your full name: ")
initials = ""
initials = initials + fullName.substring(0, 1).upper + "."

for i = 1 to fullName.length - 1
    current = fullName.substring(i, i + 1)
    previous = fullName.substring(i - 1, i)
    if previous == " " AND current != " " then
        initials = initials + current.upper + "."
    endif
next i

print(initials)

Step-by-step trace for input "john samuel smith":