根据一个向量内的值从两个向量中删除项目

Question

I have two integer vectors of equal length. Let's say I want to remove all items in the first vector which are NAN. Obviously, I use the remove_if algorithm. Let's say this removes elements that were at indexes 1,2,5. I then want to remove items from the second vector at these indexes.

What's the most canonical C++ way of doing this?

Answer 1

This can be done using Boost by creating a zip_iterator and then iterating over the tuple of iterators from both containers in parallel.

First pass a pair of zip_iterators to std::remove_if , and have the predicate inspect the elements of the first vector for NaN

auto result = std::remove_if(boost::make_zip_iterator(boost::make_tuple(v1.begin(), v2.begin())),
                             boost::make_zip_iterator(boost::make_tuple(v1.end(),   v2.end())),
                             [](boost::tuple<double, int> const& elem) {
                                 return std::isnan(boost::get<0>(elem));
                             });

Then use vector::erase to remove the unneeded elements.

v1.erase(boost::get<0>(result.get_iterator_tuple()), v1.end());
v2.erase(boost::get<1>(result.get_iterator_tuple()), v2.end());

Live demo

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The boilerplate required to create the zipped iterator ranges can be further reduced by using boost::combine and Boost.Range's version of remove_if .

auto result = boost::remove_if(boost::combine(v1, v2),
                               [](boost::tuple<double, int> const& elem) {
                                    return std::isnan(boost::get<0>(elem));
                               });

Live demo

Answer 2

Use a vector<pair<int, int>> to tie the two vector together. Then, perform your remove based on the first element to and get rid of both at the same time.