What is the smallest code I can use to count the number of occurrences of the newline character in a file with objective-c / cocoa touch?
Thanks!
Both of the other answers are correct, but with the caveat that they require loading the entire file into memory to work.
The way around that is to load the file incrementally using an NSFileHandle. Something like this:
NSFileHandle * file = [NSFileHandle fileHandleForReadingAtPath:pathToFile];
NSUInteger chunkSize = 1024;
NSData * chunk = [file readDataOfLength:chunkSize];
NSUInteger numberOfNewlines = 0;
while ([chunk length] > 0) {
const unichar * bytes = (const unichar *)[chunk bytes];
for (int index = 0; index < [chunk length]; ++index) {
unichar character = (unichar)bytes[index];
if ([[NSCharacterSet newlineCharacterSet] characterIsMember:character]) {
numberOfNewlines++;
}
}
chunk = [file readDataOfLength:chunkSize];
}
This should get you going:
NSString *fileContents = [NSString stringWithContentsOfFile:file encoding:encoding error:&error];
NSUInteger newlineCount = [fileContents numberOfOccurrencesOfString:#"\n"];
#interface NSString ()
- (NSUInteger)numberOfOccurrencesOfString:(NSString *)aString;
- (NSUInteger)numberOfOccurrencesOfChar:(char)aChar;
#end
#implementation NSString ()
- (NSUInteger)numberOfOccurrencesOfString:(NSString *)aString {
NSRange range = [self rangeOfString:aString];
NSUInteger length = [self length];
NSUInteger count = 0;
while (range.location != NSNotFound) {
range = [self rangeOfString:aString options:0 range:NSMakeRange(range.location + range.length, length - range.location - range.length)];
count++;
}
return count;
}
- (NSUInteger)numberOfOccurrencesOfChar:(char)aChar {
const char *cString = [self cStringUsingEncoding:NSUTF8StringEncoding];
NSUInteger stringLength = strlen(cString);
NSUInteger count = 0;
for (int i = 0; i < stringLength; i++) {
if (cString[i] == aChar) {
count++;
}
}
return count;
}
#end
While "numberOfOccurrencesOfString:" allocates no additional memory and supports string needles,
"numberOfOccurrencesOfChar:" allocates an autoreleased c-string copy of the NSString and searches for a single char. ""
As you were asking for a count of newlines (hence single chars) I figured a quick benchmark might be good for this particular purpose:
So I took a test string of length 2486813 containing total of 78312 '\n'. (I basically took a variation of OSX's words file) and…
…ran [testString numberOfOccurrencesOfString:#"\n"] 100 times: 19.35s
…ran [testString numberOfOccurrencesOfChar:'\n'] 100 times: 6.91s
(Setup: 2.2GHz Core 2 Duo MacBook Pro, run on a single thread)
[Edit: small bug fixed; made second snippet into category string method.]
You can scan through the string using SubstringWithRange:
Count up the number of times \n appears.
smallest you say? That automatically turns this question into code golf
FILE*f=fopen(path,"r");
int i,c;
while(1+c)i+=(c=fgetc(f))==10;
printf("%i",i);
(please don't ever actually use this code)
If you want to stay within Cocoa/CocoaTouch, you can use NSRegularExpression for this:
NSString *theString = [NSString stringWithContentsOfFile:filePath encoding:NSUTF8StringEncoding error:&error];
NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:#"\\n" options:NSRegularExpressionCaseInsensitive error:&error];
NSUInteger numLines = [regex numberOfMatchesInString:theString options:0 range:NSMakeRange(0, [theString length])] + 1;
Related
If I have this:
int toSplit = 208;
What's the best way to split it so I get:
2
0
8
Method would be like this
- (NSMutableArray *) toCharArray : (NSString *) str
{
NSMutableArray *characters = [[NSMutableArray alloc] initWithCapacity:[str length]];
for (int i=0; i < [str length]; i++)
{
NSString *ichar = [NSString stringWithFormat:#"%c", [str characterAtIndex:i]];
[characters addObject:ichar];
}
return characters;
}
do {
int digit = toSplit % 10;
toSplit /= 10;
printf(#"%i", digit);
} while (toSplit > 0);
I hope it's not a homework.
EDIT: it's backwards so it's not a valid answer... However, leaving it here because it can still be useful for others.
Sort it into a string then pull it apart. So like this:
int toSplit = 208;
NSString *string = [NSString stringWithFormat: #"%i", toSplit];
NSMutableString *splitApartString = [[NSMutableString alloc] init];
for (int i = 0; i < [string length]; i++) {
NSString *substring = [string substringFromIndex: i];
[splitApartString appendFormat: #"%#\n", substring];
}
NSLog(#"%#", splitApartString); //YAY!
So what this does, is puts this int into a string, splits it apart, then iterates through each character and gets a string out of that character. Then it appends that substring to a NEW string.
Another alternative, is instead of getting a substring just get the char and use the %c operator. Also if you take a look at this code you will see this output:
> 2
> 0
> 8
> // extra space here
You could just add a condition to check if i is the string length - 1 and not add a space or you could just remove the last character!
I'm not familiar with Objective-C syntax, but something like:
void split(int toSplit)
{
if (!toSplit)
{
return;
}
split(toSplit / 10);
int digit = toSplit % 10;
printf(#"%i", digit);
}
I am trying to replace all characters except last 4 in a String with *'s.
In objective-c there is a method in NSString class replaceStringWithCharactersInRange: withString: where I would give it range (0,[string length]-4) ) with string #"*". This is what it does: 123456789ABCD is modified to *ABCD while I am looking to make ********ABCD.
I understand that it replaced range I specified with string object. How to accomplish this ?
NSError *error;
NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:#"\\d" options:NSRegularExpressionCaseInsensitive error:&error];
NSString *newString = [regex stringByReplacingMatchesInString:string options:0 range:NSMakeRange(0, [string length]) withTemplate:#"*"];
This looks like a simple problem... get the first part string and return it with the last four characters appended to it.
Here is a function that returns the needed string :
-(NSString *)neededStringWithString:(NSString *)aString {
// if the string has less than or 4 characters, return nil
if([aString length] <= 4) {
return nil;
}
NSUInteger countOfCharToReplace = [aString length] - 4;
NSString *firstPart = #"*";
while(--countOfCharToReplace) {
firstPart = [firstPart stringByAppendingString:#"*"];
}
// range for the last four
NSRange lastFourRange = NSMakeRange([aString length] - 4, 4);
// return the combined string
return [firstPart stringByAppendingString:
[aString substringWithRange:lastFourRange]];
}
The most unintuitive part in Cocoa is creating the repeating stars without some kind of awkward looping. stringByPaddingToLength:withString:startingAtIndex: allows you to create a repeating string of any length you like, so once you have that, here's a simple solution:
NSInteger starUpTo = [string length] - 4;
if (starUpTo > 0) {
NSString *stars = [#"" stringByPaddingToLength:starUpTo withString:#"*" startingAtIndex:0];
return [string stringByReplacingCharactersInRange:NSMakeRange(0, starUpTo) withString:stars];
} else {
return string;
}
I'm not sure why the accepted answer was accepted, since it only works if everything but last 4 is a digit. Here's a simple way:
NSMutableString * str1 = [[NSMutableString alloc]initWithString:#"1234567890ABCD"];
NSRange r = NSMakeRange(0, [str1 length] - 4);
[str1 replaceCharactersInRange:r withString:[[NSString string] stringByPaddingToLength:r.length withString:#"*" startingAtIndex:0]];
NSLog(#"%#",str1);
You could use [theString substringToIndex:[theString length]-4] to get the first part of the string and then combine [theString length]-4 *'s with the second part. Perhaps their is an easier way to do this..
NSMutableString * str1 = [[NSMutableString alloc]initWithString:#"1234567890ABCD"];
[str1 replaceCharactersInRange:NSMakeRange(0, [str1 length] - 4) withString:#"*"];
NSLog(#"%#",str1);
it works
The regexp didn't work on iOS7, but perhaps this helps:
- (NSString *)encryptString:(NSString *)pass {
NSMutableString *secret = [NSMutableString new];
for (int i=0; i<[pass length]; i++) {
[secret appendString:#"*"];
}
return secret;
}
In your case you should stop replacing the last 4 characters. Bit crude, but gets the job done
I have been googling so much on how to do this, but how would I reverse a NSString? Ex:hi would become: ih
I am looking for the easiest way to do this.
Thanks!
#Vince I made this method:
- (IBAction)doneKeyboard {
// first retrieve the text of textField1
NSString *myString = field1.text;
NSMutableString *reversedString = [NSMutableString string];
NSUInteger charIndex = 0;
while(myString && charIndex < [myString length]) {
NSRange subStrRange = NSMakeRange(charIndex, 1);
[reversedString appendString:[myString substringWithRange:subStrRange]];
charIndex++;
}
// reversedString is reversed, or empty if myString was nil
field2.text = reversedString;
}
I hooked up that method to textfield1's didendonexit. When I click the done button, it doesn't reverse the text, the UILabel just shows the UITextField's text that I entered. What is wrong?
Block version.
NSString *myString = #"abcdefghijklmnopqrstuvwxyz";
NSMutableString *reversedString = [NSMutableString stringWithCapacity:[myString length]];
[myString enumerateSubstringsInRange:NSMakeRange(0,[myString length])
options:(NSStringEnumerationReverse | NSStringEnumerationByComposedCharacterSequences)
usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
[reversedString appendString:substring];
}];
// reversedString is now zyxwvutsrqponmlkjihgfedcba
Write a simple loop to do that:
// myString is "hi"
NSMutableString *reversedString = [NSMutableString string];
NSInteger charIndex = [myString length];
while (charIndex > 0) {
charIndex--;
NSRange subStrRange = NSMakeRange(charIndex, 1);
[reversedString appendString:[myString substringWithRange:subStrRange]];
}
NSLog(#"%#", reversedString); // outputs "ih"
In your case:
// first retrieve the text of textField1
NSString *myString = textField1.text;
NSMutableString *reversedString = [NSMutableString string];
NSInteger charIndex = [myString length];
while (myString && charIndex > 0) {
charIndex--;
NSRange subStrRange = NSMakeRange(charIndex, 1);
[reversedString appendString:[myString substringWithRange:subStrRange]];
}
// reversedString is reversed, or empty if myString was nil
textField2.text = reversedString;
jano’s answer is correct. Unfortunately, it creates a lot of unnecessary temporary objects. Here is a much faster (more complicated) implementation that basically does the same thing, but uses memcpy and unichar buffers to keep memory allocations to a minimum.
- (NSString *)reversedString
{
NSUInteger length = [self length];
if (length < 2) {
return self;
}
unichar *characters = calloc(length, sizeof(unichar));
unichar *reversedCharacters = calloc(length, sizeof(unichar));
if (!characters || !reversedCharacters) {
free(characters);
free(reversedCharacters);
return nil;
}
[self getCharacters:characters range:NSMakeRange(0, length)];
NSUInteger i = length - 1;
NSUInteger copiedCharacterCount = 0;
// Starting from the end of self, copy each composed character sequence into reversedCharacters
while (copiedCharacterCount < length) {
NSRange characterRange = [self rangeOfComposedCharacterSequenceAtIndex:i];
memcpy(reversedCharacters + copiedCharacterCount, characters + characterRange.location, characterRange.length * sizeof(unichar));
i = characterRange.location - 1;
copiedCharacterCount += characterRange.length;
}
free(characters);
NSString *reversedString = [[NSString alloc] initWithCharactersNoCopy:reversedCharacters length:length freeWhenDone:YES];
if (!reversedString) {
free(reversedCharacters);
}
return reversedString;
}
I tested this on 100,000 random multi-byte Unicode strings with lengths between 1 and 128. This version is about 4–5x faster than jano’s.
Enumerate substrings: 2.890528
MemCopy: 0.671090
Enumerate substrings: 2.840411
MemCopy: 0.662882
Test code is at https://gist.github.com/prachigauriar/9739805.
Update: I tried this again by simply converting to a UTF-32 buffer and reversing that.
- (NSString *)qlc_reversedStringWithUTF32Buffer
{
NSUInteger length = [self length];
if (length < 2) {
return self;
}
NSStringEncoding encoding = NSHostByteOrder() == NS_BigEndian ? NSUTF32BigEndianStringEncoding : NSUTF32LittleEndianStringEncoding;
NSUInteger utf32ByteCount = [self lengthOfBytesUsingEncoding:encoding];
uint32_t *characters = malloc(utf32ByteCount);
if (!characters) {
return nil;
}
[self getBytes:characters maxLength:utf32ByteCount usedLength:NULL encoding:encoding options:0 range:NSMakeRange(0, length) remainingRange:NULL];
NSUInteger utf32Length = utf32ByteCount / sizeof(uint32_t);
NSUInteger halfwayPoint = utf32Length / 2;
for (NSUInteger i = 0; i < halfwayPoint; ++i) {
uint32_t character = characters[utf32Length - i - 1];
characters[utf32Length - i - 1] = characters[i];
characters[i] = character;
}
return [[NSString alloc] initWithBytesNoCopy:characters length:utf32ByteCount encoding:encoding freeWhenDone:YES];
}
This is about 3–4x times faster than the memcpy version. The aforementioned gist has been updated with the latest version of the code.
Enumerate substrings: 2.168705
MemCopy: 0.488320
UTF-32: 0.150822
Enumerate substrings: 2.169655
MemCopy: 0.481786
UTF-32: 0.147534
Enumerate substrings: 2.248812
MemCopy: 0.505995
UTF-32: 0.154531
I thought I'd throw another version out there in case anyone's interested.. personally, I like the cleaner approach using NSMutableString but if performance is the highest priority this one is faster:
- (NSString *)reverseString:(NSString *)input {
NSUInteger len = [input length];
unichar *buffer = malloc(len * sizeof(unichar));
if (buffer == nil) return nil; // error!
[input getCharacters:buffer];
// reverse string; only need to loop through first half
for (NSUInteger stPos=0, endPos=len-1; stPos < len/2; stPos++, endPos--) {
unichar temp = buffer[stPos];
buffer[stPos] = buffer[endPos];
buffer[endPos] = temp;
}
return [[NSString alloc] initWithCharactersNoCopy:buffer length:len freeWhenDone:YES];
}
I also wrote a quick test as well to compare this with the more traditional NSMutableString method (which I also included below):
// test reversing a really large string
NSMutableString *string = [NSMutableString new];
for (int i = 0; i < 10000000; i++) {
int digit = i % 10;
[string appendFormat:#"%d", digit];
}
NSTimeInterval startTime = [[NSDate date] timeIntervalSince1970];
NSString *reverse = [self reverseString:string];
NSTimeInterval elapsedTime = [[NSDate date] timeIntervalSince1970] - startTime;
NSLog(#"reversed in %f secs", elapsedTime);
Results were:
using NSMutableString method (below) - "reversed in 3.720631 secs"
using unichar *buffer method (above) - "reversed in 0.032604 secs"
Just for reference, here's the NSMutableString method used for this comparison:
- (NSString *)reverseString:(NSString *)input {
NSUInteger len = [input length];
NSMutableString *result = [[NSMutableString alloc] initWithCapacity:len];
for (int i = len - 1; i >= 0; i--) {
[result appendFormat:#"%c", [input characterAtIndex:i]];
}
return result;
}
Use method with any objects: NSString,NSNumber,etc..:
NSLog(#"%#",[self reverseObject:#12345]);
NSLog(#"%#",[self reverseObject:#"Hello World"]);
Method:
-(NSString*)reverseObject:(id)string{
string = [NSString stringWithFormat:#"%#",string];
NSMutableString *endString = [NSMutableString new];
while ([string length]!=[endString length]) {
NSRange range = NSMakeRange([string length]-[endString length]-1, 1);
[endString appendString: [string substringWithRange:range]];
}
return endString;}
Log:
2014-04-16 11:20:25.312 TEST[23733:60b] 54321
2014-04-16 11:20:25.313 TEST[23733:60b] dlroW olleH
Swift 2.0:
1) let str = "Hello, world!"
let reversed = String(str.characters.reverse())
print(reversed)
In Short:
String("This is a test string.".characters.reverse())
2)
let string = "This is a test string."
let characters = string.characters
let reversedCharacters = characters.reverse()
let reversedString = String(reversedCharacters)
The short way :
String("This is a test string.".characters.reverse())
OR
let string = "This is a test string."
let array = Array(string)
let reversedArray = array.reverse()
let reversedString = String(reversedArray)
The short way :
String(Array("This is a test string.").reverse())
Tested on Play Ground:
import Cocoa
//Assigning a value to a String variable
var str = "Hello, playground"
//Create empty character Array.
var strArray:Character[] = Character[]()
//Loop through each character in the String
for character in str {
//Insert the character in the Array variable.
strArray.append(character)
}
//Create a empty string
var reversedStr:String = ""
//Read the array from backwards to get the characters
for var index = strArray.count - 1; index >= 0;--index {
//Concatenate character to String.
reversedStr += strArray[index]
}
The shorter version:
var str = “Hello, playground”
var reverseStr = “”
for character in str {
reverseStr = character + reverseStr
}
Would it be faster if you only iterated over half the string swapping the characters at each end? So for a 5 character string, you swap characters 1 + 5, then 2 + 4 and 3 doesn't need swapped with anything.
NSMutableString *reversed = [original mutableCopyWithZone:NULL];
NSUInteger i, length;
length = [reversed length];
for (i = 0; i < length / 2; i++) {
// Store the first character as we're going to replace with the character at the end
// in the example, it would store 'h'
unichar startChar = [reversed characterAtIndex:i];
// Only make the end range once
NSRange endRange = NSMakeRange(length - i, 1);
// Replace the first character ('h') with the last character ('i')
// so reversed now contains "ii"
[reversed replaceCharactersInRange:NSMakeRange(i, 1)
withString:[reversed subStringWithRange:endRange];
// Replace the last character ('i') with the stored first character ('h)
// so reversed now contains "ih"
[reversed replaceCharactersInRange:endRange
withString:[NSString stringWithFormat:#"%c", startChar]];
}
edit ----
Having done some tests, the answer is No, its about 6 times slower than the version that loops over everything. The thing that slows us down is creating the temporary NSStrings for the replaceCharactersInRange:withString method. Here is a method that creates only one NSString by manipulating the character data directly and seems a lot faster in simple tests.
NSUInteger length = [string length];
unichar *data = malloc(sizeof (unichar) * length);
int i;
for (i = 0; i < length / 2; i++) {
unichar startChar = [string characterAtIndex:i];
unichar endChar = [string characterAtIndex:(length - 1) - i];
data[i] = endChar;
data[(length - 1) - i] = startChar;
}
NSString *reversed = [NSString stringWithCharacters:data length:length];
free(data);
Reverse the string using recursion:
#implementation NSString (Reversed)
+ (NSString *)reversedStringFromString:(NSString *)string
{
NSUInteger count = [string length];
if (count <= 1) { // Base Case
return string;
} else {
NSString *lastLetter = [string substringWithRange:NSMakeRange(count - 1, 1)];
NSString *butLastLetter = [string substringToIndex:count - 1];
return [lastLetter stringByAppendingString:[self reversedStringFromString:butLastLetter]];
}
}
#end
Google is your friend:
-(NSString *) reverseString
{
NSMutableString *reversedStr;
int len = [self length];
// Auto released string
reversedStr = [NSMutableString stringWithCapacity:len];
// Probably woefully inefficient...
while (len > 0)
[reversedStr appendString:
[NSString stringWithFormat:#"%C", [self characterAtIndex:--len]]];
return reversedStr;
}
None of the answers seem to consider multibyte characters so here is my sample code. It assumes you only ever pass in a string longer than one character.
- (void)testReverseString:(NSString *)string
{
NSMutableString *rString = [NSMutableString new];
NSInteger extractChar = [string length] - 1;
while (extractChar >= 0)
{
NSRange oneCharPos = [string rangeOfComposedCharacterSequenceAtIndex:extractChar];
for (NSUInteger add = 0; add < oneCharPos.length; ++ add)
{
unichar oneChar = [string characterAtIndex:oneCharPos.location + add];
[rString appendFormat:#"%C", oneChar];
}
extractChar -= oneCharPos.length;
}
NSLog(#"%# becomes %#", string, encryptedString );
}
NSString into char utf32 (always 32 bits (unsigned int))
Reverse
char utf32 into NSString
+ (NSString *)reverseString3:(NSString *)str {
unsigned int *cstr, buf, len = [str length], i;
cstr = (unsigned int *)[str cStringUsingEncoding:NSUTF32LittleEndianStringEncoding];
for (i=0;i < len/2;i++) buf = cstr[i], cstr[i] = cstr[len -i-1], cstr[len-i-1] = buf;
return [[NSString alloc] initWithBytesNoCopy:cstr length:len*4 encoding:NSUTF32LittleEndianStringEncoding freeWhenDone:NO];
}
Example : Apple_is ---> si_elppA
NSMutableString *result = [NSMutableString stringWithString:#""];
for (long i = self.length - 1; i >= 0; i--) {
[result appendFormat:#"%c", [self characterAtIndex:i]];
}
return (NSString *)result;
Here is a collection of categories in Objective-C that will reverse both NSStrings and NSAttributedStrings (while preserving character attributes): TextFlipKit
For example:
NSString *example = #"Example Text";
NSString *reversed = example.tfk_reversed;
NSLog(#"Reversed: %#", reversed);
//prints 'Reversed: txeT elpmaxE'
Swift:
let string = "reverse"
let reversedStringCollection = string.characters.reversed()
for character in reversedStringCollection {
reversedString.append(character)
print(reversedString)
}
We can also achieve the reverse string as follows.
NSString *originalString = #"Hello";
NSString *reverseString;
for (NSUInteger index = originalString.length; index > 0; index--) {
char character = [originalString characterAtIndex:index];
reverseString = [reverseString stringByAppendingString:[NSString stringWithFormat:#"%c", character]];
}
or
NSString *originalString = #"Hello";
NSString *reverseString;
for (NSUInteger index = originalString.length; index > 0; index--) {
char *character = [originalString characterAtIndex:index];
reverseString = [reverseString stringByAppendingString:[NSString stringWithFormat:#"%s", character]];
}
Add a category to NSString so you can call reverse on any NSString in the future like this:
#import "NSString+Reverse.h"
#implementation NSString (Reverse)
-(NSString*)reverse {
char* cstring = (char*)[self UTF8String];
int length = [self length]-1;
int i=0;
while (i<=length) {
unichar tmp = cstring[i];
cstring[i] = cstring[length];
cstring[length] = tmp;
i++;
length--;
}
return [NSString stringWithCString:cstring encoding:NSUTF8StringEncoding];
}
#end
str=#"india is my countery";
array1=[[NSMutableArray alloc] init];
for(int i =0 ;i<[str length]; i++) {
NSString *singleCharacter = [NSString stringWithFormat:#"%c", [str characterAtIndex:i]];
[array1 addObject:singleCharacter];
}
NSMutableString* theString = [NSMutableString string];
for (int i=[array1 count]-1; i>=0;i--){
[theString appendFormat:#"%#",[array1 objectAtIndex:i]];
}
I have written a category ove that one :D
//NSString+Reversed.h
#import
//
// NSString+Reversed.h
// HTMLPageFormatter
// Created by beit46 on 21.06.13.
//
#interface NSString (Reversed)
- (NSString *)reversedString;
#end
//NSString+Reversed.m
//
// NSString+Reversed.m
// HTMLPageFormatter
// Created by beit46 on 21.06.13.
#import "NSString+Reversed.h"
#implementation NSString (Reversed)
- (NSString *)reversedString {
NSMutableString *reversedString = [NSMutableString stringWithCapacity:[self length]];
[self enumerateSubstringsInRange:NSMakeRange(0,[self length])
options:(NSStringEnumerationReverse | NSStringEnumerationByComposedCharacterSequences)
usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
[reversedString appendString:substring];
}];
return [reversedString copy];
}
#end
I have two simple solutions for that purpose:
+(NSString*)reverseString:(NSString *)str
{
NSMutableString* reversed = [NSMutableString stringWithCapacity:str.length];
for (int i = (int)str.length-1; i >= 0; i--){
[reversed appendFormat:#"%c", [str characterAtIndex:i]];
}
return reversed;
}
+(NSString*)reverseString2:(NSString *)str
{
char* cstr = (char*)[str UTF8String];
int len = (int)str.length;
for (int i = 0; i < len/2; i++) {
char buf = cstr[i];
cstr[i] = cstr[len-i-1];
cstr[len-i-1] = buf;
}
return [[NSString alloc] initWithBytes:cstr length:len encoding:NSUTF8StringEncoding];
}
Now, lets test it!
NSString* str = #"Objective-C is a general-purpose, object-oriented programming language that adds Smalltalk-style messaging to the C programming language";
NSLog(#"REV 1: %#", [Util reverseString:str]);
start = [NSDate date];
for (int i = 0 ; i < 1000; ++i)
[Util reverseString:str];
end = [NSDate date];
NSLog(#"Time per 1000 repeats: %f", [end timeIntervalSinceDate:start]);
NSLog(#"REV 2: %#", [Util reverseString2:str]);
start = [NSDate date];
for (int i = 0 ; i < 1000; ++i)
[Util reverseString2:str];
end = [NSDate date];
NSLog(#"Time per 1000 repeats: %f", [end timeIntervalSinceDate:start]);
Results:
ConsoleTestProject[68292:303] REV 1: egaugnal gnimmargorp C eht ot gnigassem elyts-klatllamS sdda taht egaugnal gnimmargorp detneiro-tcejbo ,esoprup-lareneg a si C-evitcejbO
ConsoleTestProject[68292:303] Time per 1000 repeats: 0.063880
ConsoleTestProject[68292:303] REV 2: egaugnal gnimmargorp C eht ot gnigassem elyts-klatllamS sdda taht egaugnal gnimmargorp detneiro-tcejbo ,esoprup-lareneg a si C-evitcejbO
ConsoleTestProject[68292:303] Time per 1000 repeats: 0.002038
And more chars result was:
ConsoleTestProject[68322:303] chars: 1982
ConsoleTestProject[68322:303] Time 1 per 1000 repeats: 1.014893
ConsoleTestProject[68322:303] Time 2 per 1000 repeats: 0.024928
The same text with above functions:
ConsoleTestProject[68366:303] Time 1 per 1000 repeats: 0.873574
ConsoleTestProject[68366:303] Time 2 per 1000 repeats: 0.019300
ConsoleTestProject[68366:303] Time 3 per 1000 repeats: 0.342735 <-Vladimir Gritsenko
ConsoleTestProject[68366:303] Time 4 per 1000 repeats: 0.584012 <- Jano
So, choose performance!
Here's what I want to do. I have 2 strings and I want to determine if one string is a permutation of another. I was thinking to simply remove the characters from string A from string B to determine if any characters are left. If no, then it passes.
However, I need to make sure that only 1 instance of each letter is removed (not all occurrences) unless there are multiple letters in the word.
An example:
String A: cant
String B: connect
Result: -o-nec-
Experimenting with NSString and NSScanner has yielded no results so far.
Hmmm, let's have a go:
NSString *stringA = #"cant";
NSString *stringB = #"connect";
NSUInteger length = [stringB length];
NSMutableCharacterSet *charsToRemove = [NSMutableCharacterSet characterSetWithCharactersInString:stringA];
unichar *buffer = calloc(length, sizeof(unichar));
[stringB getCharacters:buffer range:NSMakeRange(0, length)];
for (NSUInteger i = 0; i < length; i++)
{
if ([charsToRemove characterIsMember:buffer[i]])
{
[charsToRemove removeCharactersInRange:NSMakeRange(buffer[i], 1)];
buffer[i] = '-';
}
}
NSString *result = [NSString stringWithCharacters:buffer length:length];
free (buffer);
An inefficient yet simple way might be something like this (this is implemented as a category on NSString, but it could just as easily be a method or function taking two strings):
#implementation NSString(permutation)
- (BOOL)isPermutation:(NSString*)other
{
if( [self length] != [other length] ) return NO;
if( [self isEqualToString:other] ) return YES;
NSUInteger length = [self length];
NSCountedSet* set1 = [[[NSCountedSet alloc] initWithCapacity:length] autorelease];
NSCountedSet* set2 = [[[NSCountedSet alloc] initWithCapacity:length] autorelease];
for( int i = 0; i < length; i++ ) {
NSRange range = NSMakeRange(i, 1);
[set1 addObject:[self substringWithRange:range]];
[set2 addObject:[self substringWithRange:range]];
}
return [set1 isEqualTo:set2];
}
#end
This returns what your example asks for...
NSString* a = #"cant";
NSString* b = #"connect";
NSMutableString* mb = [NSMutableString stringWithString:b];
NSUInteger i;
for (i=0; i<[a length]; i++) {
NSString* theLetter = [a substringWithRange:NSMakeRange(i, 1)];
NSRange r = [mb rangeOfString:theLetter];
if (r.location != NSNotFound) {
[mb replaceCharactersInRange:r withString:#"-"];
}
}
NSLog(#"mb: %#", mb);
However, I wouldn't call that a permutation. To me a permutation would only hold true if all the characters from string "a" were contained by string "b". In your example, since the letter a in cant isn't in string b then I would say that cant is not a permutation of connect. With this definition I would use this:
-(BOOL)isString:(NSString*)firstString aPermutationOfString:(NSString*)secondString {
BOOL isPermutation = YES;
NSMutableString* mb = [NSMutableString stringWithString:secondString];
NSUInteger i;
for (i=0; i<[firstString length]; i++) {
NSString* theLetter = [firstString substringWithRange:NSMakeRange(i, 1)];
NSRange r = [mb rangeOfString:theLetter];
if (r.location != NSNotFound) {
[mb deleteCharactersInRange:r];
} else {
return NO;
}
}
return isPermutation;
}
How can I get the number of times an NSString (for example, #"cake") appears in a larger NSString (for example, #"Cheesecake, apple cake, and cherry pie")?
I need to do this on a lot of strings, so whatever method I use would need to be relatively fast.
Thanks!
This isn't tested, but should be a good start.
NSUInteger count = 0, length = [str length];
NSRange range = NSMakeRange(0, length);
while(range.location != NSNotFound)
{
range = [str rangeOfString: #"cake" options:0 range:range];
if(range.location != NSNotFound)
{
range = NSMakeRange(range.location + range.length, length - (range.location + range.length));
count++;
}
}
A regex like the one below should do the job without a loop interaction...
Edited
NSString *string = #"Lots of cakes, with a piece of cake.";
NSError *error = NULL;
NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:#"cake" options:NSRegularExpressionCaseInsensitive error:&error];
NSUInteger numberOfMatches = [regex numberOfMatchesInString:string options:0 range:NSMakeRange(0, [string length])];
NSLog(#"Found %i",numberOfMatches);
Only available on iOS 4.x and superiors.
was searching for a better method then mine but here's another example:
NSString *find = #"cake";
NSString *text = #"Cheesecake, apple cake, and cherry pie";
NSInteger strCount = [text length] - [[text stringByReplacingOccurrencesOfString:find withString:#""] length];
strCount /= [find length];
I would like to know which one is more effective.
And I made an NSString category for better usage:
// NSString+CountString.m
#interface NSString (CountString)
- (NSInteger)countOccurencesOfString:(NSString*)searchString;
#end
#implementation NSString (CountString)
- (NSInteger)countOccurencesOfString:(NSString*)searchString {
NSInteger strCount = [self length] - [[self stringByReplacingOccurrencesOfString:searchString withString:#""] length];
return strCount / [searchString length];
}
#end
simply call it by:
[text countOccurencesOfString:find];
Optional:
you can modify it to search case insensitive by defining options:
There are a couple ways you could do it. You could iteratively call rangeOfString:options:range:, or you could do something like:
NSArray * portions = [aString componentsSeparatedByString:#"cake"];
NSUInteger cakeCount = [portions count] - 1;
EDIT I was thinking about this question again and I wrote a linear-time algorithm to do the searching (linear to the length of the haystack string):
+ (NSUInteger) numberOfOccurrencesOfString:(NSString *)needle inString:(NSString *)haystack {
const char * rawNeedle = [needle UTF8String];
NSUInteger needleLength = strlen(rawNeedle);
const char * rawHaystack = [haystack UTF8String];
NSUInteger haystackLength = strlen(rawHaystack);
NSUInteger needleCount = 0;
NSUInteger needleIndex = 0;
for (NSUInteger index = 0; index < haystackLength; ++index) {
const char thisCharacter = rawHaystack[index];
if (thisCharacter != rawNeedle[needleIndex]) {
needleIndex = 0; //they don't match; reset the needle index
}
//resetting the needle might be the beginning of another match
if (thisCharacter == rawNeedle[needleIndex]) {
needleIndex++; //char match
if (needleIndex >= needleLength) {
needleCount++; //we completed finding the needle
needleIndex = 0;
}
}
}
return needleCount;
}
A quicker to type, but probably less efficient solution.
- (int)numberOfOccurencesOfSubstring:(NSString *)substring inString:(NSString*)string
{
NSArray *components = [string componentsSeparatedByString:substring];
return components.count-1; // Two substring will create 3 separated strings in the array.
}
Here is a version done as an extension to NSString (same idea as Matthew Flaschen's answer):
#interface NSString (my_substr_search)
- (unsigned) countOccurencesOf: (NSString *)subString;
#end
#implementation NSString (my_substring_search)
- (unsigned) countOccurencesOf: (NSString *)subString {
unsigned count = 0;
unsigned myLength = [self length];
NSRange uncheckedRange = NSMakeRange(0, myLength);
for(;;) {
NSRange foundAtRange = [self rangeOfString:subString
options:0
range:uncheckedRange];
if (foundAtRange.location == NSNotFound) return count;
unsigned newLocation = NSMaxRange(foundAtRange);
uncheckedRange = NSMakeRange(newLocation, myLength-newLocation);
count++;
}
}
#end
<somewhere> {
NSString *haystack = #"Cheesecake, apple cake, and cherry pie";
NSString *needle = #"cake";
unsigned count = [haystack countOccurencesOf: needle];
NSLog(#"found %u time%#", count, count == 1 ? #"" : #"s");
}
If you want to count words, not just substrings, then use CFStringTokenizer.
Here's another version as a category on NSString:
-(NSUInteger) countOccurrencesOfSubstring:(NSString *) substring {
if ([self length] == 0 || [substring length] == 0)
return 0;
NSInteger result = -1;
NSRange range = NSMakeRange(0, 0);
do {
++result;
range = NSMakeRange(range.location + range.length,
self.length - (range.location + range.length));
range = [self rangeOfString:substring options:0 range:range];
} while (range.location != NSNotFound);
return result;
}
Swift solution would be:
var numberOfSubstringAppearance = 0
let length = count(text)
var range: Range? = Range(start: text.startIndex, end: advance(text.startIndex, length))
while range != nil {
range = text.rangeOfString(substring, options: NSStringCompareOptions.allZeros, range: range, locale: nil)
if let rangeUnwrapped = range {
let remainingLength = length - distance(text.startIndex, rangeUnwrapped.endIndex)
range = Range(start: rangeUnwrapped.endIndex, end: advance(rangeUnwrapped.endIndex, remainingLength))
numberOfSubstringAppearance++
}
}
Matthew Flaschen's answer was a good start for me. Here is what I ended up using in the form of a method. I took a slightly different approach to the loop. This has been tested with empty strings passed to stringToCount and text and with the stringToCount occurring as the first and/or last characters in text.
I use this method regularly to count paragraphs in the passed text (ie. stringToCount = #"\r").
Hope this of use to someone.
- (int)countString:(NSString *)stringToCount inText:(NSString *)text{
int foundCount=0;
NSRange range = NSMakeRange(0, text.length);
range = [text rangeOfString:stringToCount options:NSCaseInsensitiveSearch range:range locale:nil];
while (range.location != NSNotFound) {
foundCount++;
range = NSMakeRange(range.location+range.length, text.length-(range.location+range.length));
range = [text rangeOfString:stringToCount options:NSCaseInsensitiveSearch range:range locale:nil];
}
return foundCount;
}
Example call assuming the method is in a class named myHelperClass...
int foundCount = [myHelperClass countString:#"n" inText:#"Now is the time for all good men to come to the aid of their country"];
for(int i =0;i<htmlsource1.length-search.length;i++){
range = NSMakeRange(i,search.length);
checker = [htmlsource1 substringWithRange:range];
if ([search isEqualToString:checker]) {
count++;
}
}
No built-in method. I'd suggest returning a c-string and using a common c-string style algorithm for substring counting... if you really need this to be fast.
If you want to stay in Objective C, this link might help. It describes the basic substring search for NSString. If you work with the ranges, adjust and count, then you'll have a "pure" Objective C solution... albeit, slow.
-(IBAction)search:(id)sender{
int maincount = 0;
for (int i=0; i<[self.txtfmainStr.text length]; i++) {
char c =[self.substr.text characterAtIndex:0];
char cMain =[self.txtfmainStr.text characterAtIndex:i];
if (c == cMain) {
int k=i;
int count=0;
for (int j = 0; j<[self.substr.text length]; j++) {
if (k ==[self.txtfmainStr.text length]) {
break;
}
if ([self.txtfmainStr.text characterAtIndex:k]==[self.substr.text characterAtIndex:j]) {
count++;
}
if (count==[self.substr.text length]) {
maincount++;
}
k++;
}
}
NSLog(#"%d",maincount);
}
}