2D NSDictionary with NSArray
up vote
-3
down vote
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The data set i'm trying to declare would look something like,
$results[@"Key1"][@"Key2"] = @[int, int, int, int];
How would one declare this combination of NSDictionary
& NSArray
?
objective-c
add a comment |
up vote
-3
down vote
favorite
The data set i'm trying to declare would look something like,
$results[@"Key1"][@"Key2"] = @[int, int, int, int];
How would one declare this combination of NSDictionary
& NSArray
?
objective-c
@@"Key1": @@"key2: @[@(int1),@(int2),@(int3),@(int4) ]
?
– Larme
Nov 10 at 17:47
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
The data set i'm trying to declare would look something like,
$results[@"Key1"][@"Key2"] = @[int, int, int, int];
How would one declare this combination of NSDictionary
& NSArray
?
objective-c
The data set i'm trying to declare would look something like,
$results[@"Key1"][@"Key2"] = @[int, int, int, int];
How would one declare this combination of NSDictionary
& NSArray
?
objective-c
objective-c
asked Nov 10 at 17:43
Simon.
88641640
88641640
@@"Key1": @@"key2: @[@(int1),@(int2),@(int3),@(int4) ]
?
– Larme
Nov 10 at 17:47
add a comment |
@@"Key1": @@"key2: @[@(int1),@(int2),@(int3),@(int4) ]
?
– Larme
Nov 10 at 17:47
@@"Key1": @@"key2: @[@(int1),@(int2),@(int3),@(int4) ]
?– Larme
Nov 10 at 17:47
@@"Key1": @@"key2: @[@(int1),@(int2),@(int3),@(int4) ]
?– Larme
Nov 10 at 17:47
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
So as far as I understand it, you're looking for an array within a nested dictionary - so an array, within a dictionary, within another dictionary.
I would declare it like this:
NSMutableDictionary *firstDictionary = [[NSMutableDictionary alloc] init];
NSMutableDictionary *secondDictionary = [[NSMutableDictionary alloc] init];
NSArray *array = @[@1, @2, @3, @4];
[secondDictionary setObject:array forKey:@"key2"];
[firstDictionary setObject:secondDictionary forKey:@"key1"];
Then you can access the contents of your data structure like this:
NSArray *results = firstDictionary[@"key1"][@"key2"];
NSLog(@"results: %@", results);
And it will output:
results: (
1,
2,
3,
4
)
Hope this helps :)
Edit: just as an aside, if you're trying to create a 2d matrix, where you use 2 coordinates to access your data, then you should probably use a nested array, instead of a nested dictionary.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
So as far as I understand it, you're looking for an array within a nested dictionary - so an array, within a dictionary, within another dictionary.
I would declare it like this:
NSMutableDictionary *firstDictionary = [[NSMutableDictionary alloc] init];
NSMutableDictionary *secondDictionary = [[NSMutableDictionary alloc] init];
NSArray *array = @[@1, @2, @3, @4];
[secondDictionary setObject:array forKey:@"key2"];
[firstDictionary setObject:secondDictionary forKey:@"key1"];
Then you can access the contents of your data structure like this:
NSArray *results = firstDictionary[@"key1"][@"key2"];
NSLog(@"results: %@", results);
And it will output:
results: (
1,
2,
3,
4
)
Hope this helps :)
Edit: just as an aside, if you're trying to create a 2d matrix, where you use 2 coordinates to access your data, then you should probably use a nested array, instead of a nested dictionary.
add a comment |
up vote
0
down vote
So as far as I understand it, you're looking for an array within a nested dictionary - so an array, within a dictionary, within another dictionary.
I would declare it like this:
NSMutableDictionary *firstDictionary = [[NSMutableDictionary alloc] init];
NSMutableDictionary *secondDictionary = [[NSMutableDictionary alloc] init];
NSArray *array = @[@1, @2, @3, @4];
[secondDictionary setObject:array forKey:@"key2"];
[firstDictionary setObject:secondDictionary forKey:@"key1"];
Then you can access the contents of your data structure like this:
NSArray *results = firstDictionary[@"key1"][@"key2"];
NSLog(@"results: %@", results);
And it will output:
results: (
1,
2,
3,
4
)
Hope this helps :)
Edit: just as an aside, if you're trying to create a 2d matrix, where you use 2 coordinates to access your data, then you should probably use a nested array, instead of a nested dictionary.
add a comment |
up vote
0
down vote
up vote
0
down vote
So as far as I understand it, you're looking for an array within a nested dictionary - so an array, within a dictionary, within another dictionary.
I would declare it like this:
NSMutableDictionary *firstDictionary = [[NSMutableDictionary alloc] init];
NSMutableDictionary *secondDictionary = [[NSMutableDictionary alloc] init];
NSArray *array = @[@1, @2, @3, @4];
[secondDictionary setObject:array forKey:@"key2"];
[firstDictionary setObject:secondDictionary forKey:@"key1"];
Then you can access the contents of your data structure like this:
NSArray *results = firstDictionary[@"key1"][@"key2"];
NSLog(@"results: %@", results);
And it will output:
results: (
1,
2,
3,
4
)
Hope this helps :)
Edit: just as an aside, if you're trying to create a 2d matrix, where you use 2 coordinates to access your data, then you should probably use a nested array, instead of a nested dictionary.
So as far as I understand it, you're looking for an array within a nested dictionary - so an array, within a dictionary, within another dictionary.
I would declare it like this:
NSMutableDictionary *firstDictionary = [[NSMutableDictionary alloc] init];
NSMutableDictionary *secondDictionary = [[NSMutableDictionary alloc] init];
NSArray *array = @[@1, @2, @3, @4];
[secondDictionary setObject:array forKey:@"key2"];
[firstDictionary setObject:secondDictionary forKey:@"key1"];
Then you can access the contents of your data structure like this:
NSArray *results = firstDictionary[@"key1"][@"key2"];
NSLog(@"results: %@", results);
And it will output:
results: (
1,
2,
3,
4
)
Hope this helps :)
Edit: just as an aside, if you're trying to create a 2d matrix, where you use 2 coordinates to access your data, then you should probably use a nested array, instead of a nested dictionary.
edited Nov 10 at 20:13
answered Nov 10 at 20:03
simsula
277
277
add a comment |
add a comment |
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@@"Key1": @@"key2: @[@(int1),@(int2),@(int3),@(int4) ]
?– Larme
Nov 10 at 17:47