Reputation: 65
My NSMutableDictionary
contains four NSArray
s with their respective keys. These NSArray
s are of size 2 and contain 2D coordinates. I want to get the most common coordinate among them. For example, If a coordinate is common to three arrays, it would be my first choice. How can I find that any coordinate is common to at least two arrays?
Upvotes: 1
Views: 271
Reputation: 470
Basically you want to find the pair of coordinates that has max occurences. So you can create an array of all coordinates and find its mode.
NSMutableArray *allCoordinates = [NSMutableArray new];
[dictionary enumerateKeysAndObjectsUsingBlock:^(id key, id obj, BOOL *stop) {
if ([key isEqualToString:@"arrayKey1"] || [key isEqualToString:@"arrayKey2"]) {
NSArray *coordinates = (NSArray *)obj;
[coordinates enumerateObjectsUsingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
[allCoordinates addObject:coordinates];
}];
}
}];
Now, U need to write a custom method to find mode of an array of coordinates (with an additional condition of frequency being >=3).
Upvotes: 2
Reputation: 26026
Here is some code that should work, using NSCountedSet
.
In a few words:
I use a NSCountedSet
that will work as a NSSet
keeping also the numbers of occurrences (duplicates times).
Then, I create a NSArray
sorting the values descending according to the number of occurrences.
I wrote explicitly the "count1/count2" comparison in case you want to apply a different value if the number of occurrences is the same.
NSDictionary *allData = @{@"Key1: %@":@[@(0.0), @(0.1)],
@"Key2: %@":@[@(0.1), @(0.1)],
@"Key3: %@":@[@(0.2), @(0.1)],
@"Key4: %@":@[@(0.1), @(0.1)]};
NSCountedSet *countedSet = [[NSCountedSet alloc] initWithArray:[allData allValues]];
for (NSArray *array in countedSet)
NSLog(@"For %@ counted %@ time(s)", array, @([countedSet countForObject:array]));
NSArray *sortedCountedSetArray = [[countedSet allObjects] sortedArrayUsingComparator:^NSComparisonResult(id obj1, id obj2)
{
NSUInteger count1 = [countedSet countForObject:obj1];
NSUInteger count2 = [countedSet countForObject:obj2];
if (count1 < count2)
return NSOrderedDescending;
if (count1 > count2)
return NSOrderedAscending;
else
return NSOrderedSame; //May want to do additionaly thing (for example if coordinate is closer to actual position, etc.)
}];
NSLog(@"sortedCountedSetArray: %@", sortedCountedSetArray);
NSArray *bestOption = [sortedCountedSetArray firstObject]; //Coordinates the most "popular"
NSLog(@"BestOption: %@", bestOption);
The outputs are:
> For (
"0.1",
"0.1"
) counted 2 time(s)
> For (
"0.2",
"0.1"
) counted 1 time(s)
> For (
0,
"0.1"
) counted 1 time(s)
> sortedCountedSetArray: (
(
"0.1",
"0.1"
),
(
"0.2",
"0.1"
),
(
0,
"0.1"
)
)
> BestOption: (
"0.1",
"0.1"
)
Upvotes: 0
Reputation: 2165
This is a working example. I'm assuming that your dictionary looks like coordinatesDict
.
NSDictionary *coordinatesDict = @{@"first": @[@11.58, @40.20], @"second": @[@12.12, @100.12], @"third": @[@11.58, @40.20], @"fourth": @[@13.2, @14.5]};
NSCountedSet *coordinates = [NSCountedSet setWithArray:[coordinatesDict allValues]];
for (NSArray *coordinateArray in coordinates) {
NSUInteger coordinateCount = [coordinates countForObject:coordinateArray];
NSLog(@"Array %@ is included %lu times", coordinateArray, (unsigned long)coordinateCount);
if (coordinateCount >= 2) {
// You found your best coordinate
}
}
Upvotes: 0