NSSortDescriptor - Sort descriptor based on another array

Question

I hava a core data application. And I want to fetch one kind of object, User. User have the property userId.

I have another a

Answer 1

As @MartinR said, you cannot do that with sort descriptors.
To achieve a scalable solution in O(N) complexity (if you really need it):

//Not tested
NSArray* permutation = ...;//@[@1,@4,@3,@5] (user ids);
NSMutableDictionary* map = [[NSMutableDictionary alloc] initWithCapacity:[permutation count]];
for (NSUInteger i =0; i< [permutation count]; ++i) {
    map[permutation[i]] = @(i);
}
NSMutableArray* results = ...;//A mutable copy of the results
for (NSUInteger i = 0; i < [permutation count];) {
    id userId = [results[i] valueForKey:@"userId"];
    NSUInteger key = [map[userId] unsignedIntegerValue];//where we like this object to be
    if (key != i) {
        [results exchangeObjectAtIndex:i withObjectAtIndex:key];
    } else {
        ++i;
    }
}

Analysis: (Please correct me if you spot any mistakes)

The mapping stage takes N operation to complete: O(N)
The array copy takes N operations: O(N)
The loop iterate N times, but:
it perform at max N exchanges
and at max N comparisons
hence: O(N) + O(N) + O(N) + O(N) = 4*O(N) = O(N)

@MartinR solution will be based on a comparison based sort limited to a minimum O(N*lg(N)) operations.
however each comparison takes O(N) to get the first index and O(N) to get the second index.
hence: O(N*lg(N))*(O(N) + O(N)) = 2*O(N)*O(N*lg(N)) = O((N^2)*lg(N))

For small enough N there is no real difference (N=10 ==> ~40 operations for my solution and ~200 operations for Martin's solution).
For large N Martin's solutions will not scale well:
N == 1000 ==> ~4000 operations for my solution and ~2*1000*1000*8 =~ 16*10^6