Calculate number of differences between two NSStrings - iphone

Question

How can I calculate the number of differences between two NSStrings.
Example:
NSString 1 = "this is a string"
NSString 2 = "Tihs isa string"
should return: 4 (one for the capital "T", one for the "i", the "h" and for the missing space)

Answer 1

What you're looking for is the Levenshtein Distance.
An implementation in Objective-C:
------------------------------------------------------------------------
//
// NSString-Levenshtein.h
//
// Created by Rick Bourner on Sat Aug 09 2003.
// rick#bourner.com
#interface NSString(Levenshtein)
// calculate the smallest distance between all words in stringA and stringB
- (float) compareWithString: (NSString *) stringB;
// calculate the distance between two string treating them each as a
// single word
- (float) compareWithWord: (NSString *) stringB;
// return the minimum of a, b and c
- (int) smallestOf: (int) a andOf: (int) b andOf: (int) c;
#end
--------------------------------------------------------------------
//
// NSString-Levenshtein.m
//
// Created by Rick Bourner on Sat Aug 09 2003.
// Rick#Bourner.com
#import "NSString-Levenshtein.h"
#implementation NSString(Levenshtein)
// calculate the mean distance between all words in stringA and stringB
- (float) compareWithString: (NSString *) stringB
{
float averageSmallestDistance = 0.0;
float smallestDistance;
float distance;
NSMutableString * mStringA = [[NSMutableString alloc] initWithString: self];
NSMutableString * mStringB = [[NSMutableString alloc] initWithString: stringB];
// normalize
[mStringA replaceOccurrencesOfString:#"\n"
withString: #" "
options: NSLiteralSearch
range: NSMakeRange(0, [mStringA length])];
[mStringB replaceOccurrencesOfString:#"\n"
withString: #" "
options: NSLiteralSearch
range: NSMakeRange(0, [mStringB length])];
NSArray * arrayA = [mStringA componentsSeparatedByString: #" "];
NSArray * arrayB = [mStringB componentsSeparatedByString: #" "];
NSEnumerator * emuA = [arrayA objectEnumerator];
NSEnumerator * emuB;
NSString * tokenA = NULL;
NSString * tokenB = NULL;
// O(n*m) but is there another way ?!?
while ( tokenA = [emuA nextObject] ) {
emuB = [arrayB objectEnumerator];
smallestDistance = 99999999.0;
while ( tokenB = [emuB nextObject] )
if ( (distance = [tokenA compareWithWord: tokenB] ) < smallestDistance )
smallestDistance = distance;
averageSmallestDistance += smallestDistance;
}
[mStringA release];
[mStringB release];
return averageSmallestDistance / [arrayA count];
}
// calculate the distance between two string treating them eash as a
// single word
- (float) compareWithWord: (NSString *) stringB
{
// normalize strings
NSString * stringA = [NSString stringWithString: self];
[stringA stringByTrimmingCharactersInSet:
[NSCharacterSet whitespaceAndNewlineCharacterSet]];
[stringB stringByTrimmingCharactersInSet:
[NSCharacterSet whitespaceAndNewlineCharacterSet]];
stringA = [stringA lowercaseString];
stringB = [stringB lowercaseString];
// Step 1
int k, i, j, cost, * d, distance;
int n = [stringA length];
int m = [stringB length];
if( n++ != 0 && m++ != 0 ) {
d = malloc( sizeof(int) * m * n );
// Step 2
for( k = 0; k < n; k++)
d[k] = k;
for( k = 0; k < m; k++)
d[ k * n ] = k;
// Step 3 and 4
for( i = 1; i < n; i++ )
for( j = 1; j < m; j++ ) {
// Step 5
if( [stringA characterAtIndex: i-1] ==
[stringB characterAtIndex: j-1] )
cost = 0;
else
cost = 1;
// Step 6
d[ j * n + i ] = [self smallestOf: d [ (j - 1) * n + i ] + 1
andOf: d[ j * n + i - 1 ] + 1
andOf: d[ (j - 1) * n + i -1 ] + cost ];
}
distance = d[ n * m - 1 ];
free( d );
return distance;
}
return 0.0;
}
// return the minimum of a, b and c
- (int) smallestOf: (int) a andOf: (int) b andOf: (int) c
{
int min = a;
if ( b < min )
min = b;
if( c < min )
min = c;
return min;
}
#end
Author of the source above: Rick Bourner, http://www.merriampark.com/ldobjc.htm