Question

我想计算成对矢量中相同点的出现次数。

这是我的代码：

vector<Point2f> points_1, points_2;

points_1.push_back(Point2f(1.0, 2.0));
points_1.push_back(Point2f(2.0, 2.0));
points_1.push_back(Point2f(3.0, 2.0));
points_1.push_back(Point2f(1.0, 2.0));
points_1.push_back(Point2f(2.0, 2.0));
points_1.push_back(Point2f(3.0, 2.0));
points_1.push_back(Point2f(1.0, 2.0));

points_2.push_back(Point2f(1.0, 1.0));
points_2.push_back(Point2f(1.0, 1.0));
points_2.push_back(Point2f(1.0, 2.0));
points_2.push_back(Point2f(1.0, 1.0));
points_2.push_back(Point2f(1.0, 1.0));
points_2.push_back(Point2f(1.0, 1.0));
points_2.push_back(Point2f(1.0, 1.0));

vector<pair<Point2f, Point2f>> point_pairs;
for (size_t i = 0; i < points_1.size(); i++) {
    cout << points_1[i] << " " << points_2[i] << endl;
    point_pairs.push_back(make_pair(points_1[i], points_2[i]));
}

结果应为：

[1, 2] [1, 1] - 3 Occurrences
[2, 2] [1, 1] - 2 Occurrences
[3, 2] [1, 2] - 1 Occurrence
[3, 2] [1, 1] - 1 Occurrence

我知道你可以使用带有键的地图（如直方图）跟踪事件，我尝试过这样的事情：

map<pair<Point2f, Point2f>, int> pairToCount;

但我不是C ++ lambda的专家，也不知道如何使用这对TOCount

Answer 1

假设：

std::ostream& operator << (std::ostream& os, const Point2f& pt)
{
    return os << "[" << pt.x << ", " << pt.y << "]";
}

bool operator < (const Point2f& lhs, const Point2f& rhs)
{
    return std::tie(lhs.x, lhs.y) < std::tie(rhs.x, rhs.y);
}

您可以这样做：

const std::vector<std::pair<Point2f, Point2f>> point_pairs /* = */;
map<pair<Point2f, Point2f>, int> counts;

for (const auto& p : point_pairs) {
    ++counts[p];
}

for (const auto& p : counts) {
    const auto& p1 = p.first.first;
    const auto& p2 = p.first.second;
    int count = p.second;
    std::cout << p1 << " " << p2 << " - " << count << " Occurrence(s)\n";
}

Demo

计算对的向量中的出现次数

1 个答案: