How to count the total keys in an array based on the value of that key

Question

This might be a basic question, but I am not getting any efficient way to do this. Suppose I have an array of multiple objects containing multiple key value like this. Now, I want the total count of grActual whose value is 16- Jan-13.I can iterate the array, and create a new array with key grAcutal and then I can count. Is there any other way to do that?

    (
    {
    gIActual = "17-Jan-13";
    grActual = "16-Jan-13";
},
    {
    gIActual = "18-Jan-13";
    grActual = "16-Jan-13";
},
    {
    gIActual = "16-Jan-13";
    grActual = "16-Jan-13";      
},
    {
    gIActual = "16-Feb-13";
    grActual = "16-Jan-13";
})

You can use Predicate to search objects like NSArray *filteredarray = [array filteredArrayUsingPredicate:[NSPredicate predicateWithFormat:@"(grActual == %@)", @"16- Jan-13"]]; which will give you number of objects in the array, the total count of the array would be the Count you need. — iphonic
– iphonic, Commented Jul 9, 2016 at 10:21

Droppy · Accepted Answer · 2016-07-09 11:17:20Z

3

I believe this will do it, not that I use NSPredicate very often:

NSArray *filtered = [array filteredArrayUsingPredicate:
    [NSPredicate predicateWithFormat:@"grActual == %@", @"16-Jan-13"]];
NSUInteger count = [filtered count];

Tip: Store data using the correct type, which for 16-Jan-13 is NSDate, not NSString (not to mention Y2K and localization issues...).

edited Jul 9, 2016 at 11:17

answered Jul 9, 2016 at 10:20

Droppy

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4 Comments

iphonic Over a year ago

It can be searched with NSString using [NSPredicate predicateWithFormat:@"(grActual == %@)", @"16-Jan-13"]

Wain Over a year ago

you can't assign in a predicate and the brackets are meaningless in this case so the original was simpler and also correct...

iphonic Over a year ago

@Wain Yes brackets are meaningless, you can assign using == and it works too.

Droppy Over a year ago

@Wain Thanks. As you can see I don't use NSPredicate often enough (I am happy with filtering by block, which can do anything).

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