How to Sort One `std::vector` Based on the Order of Another?

Sorting a std::vector by Values of a Different std::vector

This problem involves aligning the ordering of a vector with the order of another vector. For instance, the vector Index contains numbers [3, 1, 2] and the vector Values contains strings "Third, First, Second." Sorting Index in ascending order ([1, 2, 3]) should result in Values being sorted in the corresponding order (["First", "Second", "Third"]).

Solution:

A common approach involves creating a vector order by combining indices from Index and their corresponding elements from Values. This vector can then be sorted using a custom comparator that compares the elements in Index:

typedef vector<int>::const_iterator myiter;
vector<pair<size_t, myiter>> order(Index.size());

size_t n = 0;
for (myiter it = Index.begin(); it != Index.end(); ++it, ++n)
    order[n] = make_pair(n, it);

struct ordering {
    bool operator ()(pair<size_t, myiter> const&amp; a, pair<size_t, myiter> const&amp; b) {
        return *(a.second) < *(b.second);
    }
};

sort(order.begin(), order.end(), ordering());

The resulting order vector now contains the sorted indices. To apply this order to Values, the following function can be used:

template <typename T>
vector<T> sort_from_ref(
    vector<T> const&amp; in,
    vector<pair<size_t, myiter>> const&amp; reference
) {
    vector<T> ret(in.size());

    size_t const size = in.size();
    for (size_t i = 0; i < size; ++i)
        ret[i] = in[reference[i].first];

    return ret;
}

Values = sort_from_ref(Values, order);

This function takes the original vector in and the reference vector containing the sorted indices and returns a copy of in sorted accordingly.

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