I need to reverse my NSArray.
As an example:
[1,2,3,4,5] must become: [5,4,3,2,1]
What is the best way to achieve this?
This question is related to
objective-c
nsarray
Use enumerateObjectsWithOptions:NSEnumerationReverse usingBlock. Using @JohannesFahrenkrug's benchmark above, this completed 8x quicker than [[array reverseObjectEnumerator] allObjects];:
NSDate *methodStart = [NSDate date];
[anArray enumerateObjectsWithOptions:NSEnumerationReverse usingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
//
}];
NSDate *methodFinish = [NSDate date];
NSTimeInterval executionTime = [methodFinish timeIntervalSinceDate:methodStart];
NSLog(@"executionTime = %f", executionTime);
Here is a nice macro that will work for either NSMutableArray OR NSArray:
#define reverseArray(__theArray) {\
if ([__theArray isKindOfClass:[NSMutableArray class]]) {\
if ([(NSMutableArray *)__theArray count] > 1) {\
NSUInteger i = 0;\
NSUInteger j = [(NSMutableArray *)__theArray count]-1;\
while (i < j) {\
[(NSMutableArray *)__theArray exchangeObjectAtIndex:i\
withObjectAtIndex:j];\
i++;\
j--;\
}\
}\
} else if ([__theArray isKindOfClass:[NSArray class]]) {\
__theArray = [[NSArray alloc] initWithArray:[[(NSArray *)__theArray reverseObjectEnumerator] allObjects]];\
}\
}
To use just call: reverseArray(myArray);
DasBoot has the right approach, but there are a few mistakes in his code. Here's a completely generic code snippet that will reverse any NSMutableArray in place:
/* Algorithm: swap the object N elements from the top with the object N
* elements from the bottom. Integer division will wrap down, leaving
* the middle element untouched if count is odd.
*/
for(int i = 0; i < [array count] / 2; i++) {
int j = [array count] - i - 1;
[array exchangeObjectAtIndex:i withObjectAtIndex:j];
}
You can wrap that in a C function, or for bonus points, use categories to add it to NSMutableArray. (In that case, 'array' would become 'self'.) You can also optimize it by assigning [array count] to a variable before the loop and using that variable, if you desire.
If you only have a regular NSArray, there's no way to reverse it in place, because NSArrays cannot be modified. But you can make a reversed copy:
NSMutableArray * copy = [NSMutableArray arrayWithCapacity:[array count]];
for(int i = 0; i < [array count]; i++) {
[copy addObject:[array objectAtIndex:[array count] - i - 1]];
}
Or use this little trick to do it in one line:
NSArray * copy = [[array reverseObjectEnumerator] allObjects];
If you just want to loop over an array backwards, you can use a for/in loop with [array reverseObjectEnumerator], but it's likely a bit more efficient to use -enumerateObjectsWithOptions:usingBlock::
[array enumerateObjectsWithOptions:NSEnumerationReverse
usingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
// This is your loop body. Use the object in obj here.
// If you need the index, it's in idx.
// (This is the best feature of this method, IMHO.)
// Instead of using 'continue', use 'return'.
// Instead of using 'break', set '*stop = YES' and then 'return'.
// Making the surrounding method/block return is tricky and probably
// requires a '__block' variable.
// (This is the worst feature of this method, IMHO.)
}];
(Note: Substantially updated in 2014 with five more years of Foundation experience, a new Objective-C feature or two, and a couple tips from the comments.)
If all you want to do is iterate in reverse, try this:
// iterate backwards
nextIndex = (currentIndex == 0) ? [myArray count] - 1 : (currentIndex - 1) % [myArray count];
You can do the [myArrayCount] once and save it to a local variable (I think its expensive), but I’m also guessing that the compiler will pretty much do the same thing with the code as written above.
After reviewing the other's answers above and finding Matt Gallagher's discussion here
I propose this:
NSMutableArray * reverseArray = [NSMutableArray arrayWithCapacity:[myArray count]];
for (id element in [myArray reverseObjectEnumerator]) {
[reverseArray addObject:element];
}
As Matt observes:
In the above case, you may wonder if -[NSArray reverseObjectEnumerator] would be run on every iteration of the loop — potentially slowing down the code. <...>
Shortly thereafter, he answers thus:
<...> The "collection" expression is only evaluated once, when the for loop begins. This is the best case, since you can safely put an expensive function in the "collection" expression without impacting upon the per-iteration performance of the loop.
As for me, have you considered how the array was populated in the first place? I was in the process of adding MANY objects to an array, and decided to insert each one at the beginning, pushing any existing objects up by one. Requires a mutable array, in this case.
NSMutableArray *myMutableArray = [[NSMutableArray alloc] initWithCapacity:1];
[myMutableArray insertObject:aNewObject atIndex:0];
I don't know of any built in method. But, coding by hand is not too difficult. Assuming the elements of the array you are dealing with are NSNumber objects of integer type, and 'arr' is the NSMutableArray that you want to reverse.
int n = [arr count];
for (int i=0; i<n/2; ++i) {
id c = [[arr objectAtIndex:i] retain];
[arr replaceObjectAtIndex:i withObject:[arr objectAtIndex:n-i-1]];
[arr replaceObjectAtIndex:n-i-1 withObject:c];
}
Since you start with a NSArray then you have to create the mutable array first with the contents of the original NSArray ('origArray').
NSMutableArray * arr = [[NSMutableArray alloc] init];
[arr setArray:origArray];
Edit: Fixed n -> n/2 in the loop count and changed NSNumber to the more generic id due to the suggestions in Brent's answer.
Or the Scala-way:
-(NSArray *)reverse
{
if ( self.count < 2 )
return self;
else
return [[self.tail reverse] concat:[NSArray arrayWithObject:self.head]];
}
-(id)head
{
return self.firstObject;
}
-(NSArray *)tail
{
if ( self.count > 1 )
return [self subarrayWithRange:NSMakeRange(1, self.count - 1)];
else
return @[];
}
To update this, in Swift it can be done easily with:
array.reverse()
Some benchmarks
1. reverseObjectEnumerator allObjects
This is the fastest method:
NSArray *anArray = @[@"aa", @"ab", @"ac", @"ad", @"ae", @"af", @"ag",
@"ah", @"ai", @"aj", @"ak", @"al", @"am", @"an", @"ao", @"ap", @"aq", @"ar", @"as", @"at",
@"au", @"av", @"aw", @"ax", @"ay", @"az", @"ba", @"bb", @"bc", @"bd", @"bf", @"bg", @"bh",
@"bi", @"bj", @"bk", @"bl", @"bm", @"bn", @"bo", @"bp", @"bq", @"br", @"bs", @"bt", @"bu",
@"bv", @"bw", @"bx", @"by", @"bz", @"ca", @"cb", @"cc", @"cd", @"ce", @"cf", @"cg", @"ch",
@"ci", @"cj", @"ck", @"cl", @"cm", @"cn", @"co", @"cp", @"cq", @"cr", @"cs", @"ct", @"cu",
@"cv", @"cw", @"cx", @"cy", @"cz"];
NSDate *methodStart = [NSDate date];
NSArray *reversed = [[anArray reverseObjectEnumerator] allObjects];
NSDate *methodFinish = [NSDate date];
NSTimeInterval executionTime = [methodFinish timeIntervalSinceDate:methodStart];
NSLog(@"executionTime = %f", executionTime);
Result: executionTime = 0.000026
2. Iterating over an reverseObjectEnumerator
This is between 1.5x and 2.5x slower:
NSDate *methodStart = [NSDate date];
NSMutableArray *array = [NSMutableArray arrayWithCapacity:[anArray count]];
NSEnumerator *enumerator = [anArray reverseObjectEnumerator];
for (id element in enumerator) {
[array addObject:element];
}
NSDate *methodFinish = [NSDate date];
NSTimeInterval executionTime = [methodFinish timeIntervalSinceDate:methodStart];
NSLog(@"executionTime = %f", executionTime);
Result: executionTime = 0.000071
3. sortedArrayUsingComparator
This is between 30x and 40x slower (no surprises here):
NSDate *methodStart = [NSDate date];
NSArray *reversed = [anArray sortedArrayUsingComparator: ^(id obj1, id obj2) {
return [anArray indexOfObject:obj1] < [anArray indexOfObject:obj2] ? NSOrderedDescending : NSOrderedAscending;
}];
NSDate *methodFinish = [NSDate date];
NSTimeInterval executionTime = [methodFinish timeIntervalSinceDate:methodStart];
NSLog(@"executionTime = %f", executionTime);
Result: executionTime = 0.001100
So [[anArray reverseObjectEnumerator] allObjects] is the clear winner when it comes to speed and ease.
There is a much easier solution, if you take advantage of the built-in reverseObjectEnumerator method on NSArray, and the allObjects method of NSEnumerator:
NSArray* reversedArray = [[startArray reverseObjectEnumerator] allObjects];
allObjects is documented as returning an array with the objects that have not yet been traversed with nextObject, in order:
This array contains all the remaining objects of the enumerator in enumerated order.
Try this:
for (int i = 0; i < [arr count]; i++)
{
NSString *str1 = [arr objectAtIndex:[arr count]-1];
[arr insertObject:str1 atIndex:i];
[arr removeObjectAtIndex:[arr count]-1];
}
Georg Schölly's categories are very nice. However, for NSMutableArray, using NSUIntegers for the indices results in a crash when the array is empty. The correct code is:
@implementation NSMutableArray (Reverse)
- (void)reverse {
NSInteger i = 0;
NSInteger j = [self count] - 1;
while (i < j) {
[self exchangeObjectAtIndex:i
withObjectAtIndex:j];
i++;
j--;
}
}
@end
There is a easy way to do it.
NSArray *myArray = @[@"5",@"4",@"3",@"2",@"1"];
NSMutableArray *myNewArray = [[NSMutableArray alloc] init]; //this object is going to be your new array with inverse order.
for(int i=0; i<[myNewArray count]; i++){
[myNewArray insertObject:[myNewArray objectAtIndex:i] atIndex:0];
}
//other way to do it
for(NSString *eachValue in myArray){
[myNewArray insertObject:eachValue atIndex:0];
}
//in both cases your new array will look like this
NSLog(@"myNewArray: %@", myNewArray);
//[@"1",@"2",@"3",@"4",@"5"]
I hope this helps.
NSMutableArray *objMyObject = [NSMutableArray arrayWithArray:[self reverseArray:objArrayToBeReversed]];
// Function reverseArray
-(NSArray *) reverseArray : (NSArray *) myArray {
return [[myArray reverseObjectEnumerator] allObjects];
}
Swift 3 syntax :
let reversedArray = array.reversed()
Reverse array and looping through it:
[[[startArray reverseObjectEnumerator] allObjects] enumerateObjectsUsingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
...
}];
Source: Stackoverflow.com