CODEWITHSUNDEEP
c++boostboost-multi-index
UncleZeiv
UncleZeiv

Reputation: 18488

Boost multi_index: retrieve unique values of a non-unique key

I have a boost::multi_index_container whose elements are structs like this:

struct Elem {
    A a;
    B b;
    C c;
};

The main key (in a database sense) is a composite_key of a and b. Other keys exist to perform various types of queries.

I now need to retrieve a set of all different values of c. These values are by all means not unique, but iterating through all entries (albeit ordered), or using std::unique seems quite a waste, considering that the number of different values of c is expected to be << than the total number of entries (say, 10 to 1000).

Am I missing a simple way to obtain this result more efficiently?

Upvotes: 4

Views: 1772

Answers (2)

LTMDUM
LTMDUM

Reputation: 51

The approach I took to solving this problem was to use boost range adaptors as follows

const auto& indexedContainer = container.get<IndexType>();
const auto uniqueIndexRange = indexedContainer 
    | boost::adaptors::transformed([&](auto&& v) {
        return indexedContainer.key_extractor()(v); })
    | boost::adaptors::uniqued;

Upvotes: 1

Emile Cormier
Emile Cormier

Reputation: 29209

I scoured the Boost.MultiIndex documentation and can't seem to find a way to do what you want. I'm interested in knowing if it's doable.

Perhaps the best you can do is maintain a std::map<C, size_t> (or hash map) alongside your multi_index_container and keep them both "synchronized".

The map associates a C value with its occurrence count (frequency). It's essentially a histogram of C values. Each time you add an Elem to your multi_index_container, you increment the corresponding frequency in the histogram. When you remove an Elem from your multi_index_counter, you decrement the corresponding frequency in the histogram. When the frequency reaches zero, you delete that entry from the histogram.

To retrieve the set of distinct C values, you simply iterate through the <key,value> pairs in the histogram and look at the key part of each pair. If you used a std::map, then the distinct C values will come out sorted.

If you're going to examine the set of distinct C values only once (or rarely) then the approach I described above may be overkill. A simpler approach would be to insert all C values into a std::set<C> and then iterate through the set to retrieve the distinct C values.

You said that the set of distinct C's is much smaller then the total number of C's. The std::set<C> approach should therefore waste much less space than copying the C's to a std::vector, sorting the vector, then running std::unique.

Let's compare the time complexity of copying to a set versus copying to a vector, sorting, then running unique. Let N be the total number of C values, and let M be the number of distinct C values. The set approach, by my reckoning, should have a time complexity of O(N*log(M)). Since M is small and does not grow much with higher N's, the complexity effectively becomes O(N). The sorting+unique technique, on the other hand, should have a time complexity of O(N*log(N)).

Upvotes: 1

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