根据一个向量内的值从两个向量中删除项 [英] Remove items from two vectors depending on the values inside one vector

问题描述

我有两个相等长度的整数向量。让我们说，我想删除第一个向量中的所有项目是NAN。显然，我使用remove_if算法。让我们假设这将删除在索引1,2,5的元素。然后，我要在这些索引中从第二个向量中删除项目。

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.

这是最经典的C ++方法是什么？

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

推荐答案

这可以使用Boost通过创建 zip_iterator ，然后从两个容器中并行迭代 tuple 。

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

首先将一对 zip_iterators 传递给 std :: remove_if ，并使谓语检查NaN的第一个向量的元素

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));
                             });

然后使用 vector :: erase 不需要的元素。

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());

现场演示

创建压缩迭代器范围所需的样板文件可以进一步使用 boost :: combine 和Boost.Range的版本 remove_if 。

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));
                               });

现场演示

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