C++使用算法过滤类向量

Question

如何使用算法过滤其国家/地区的类向量studentList？意思我只显示来自"America"国家的学生的详细信息.

bool checkCountry (string x, string y) 
{
  return (x == y);
}
vector<Student> studentList;
studentList.push_back(Student("Tom", 'M', "91213242", "America"));
studentList.push_back(Student("Jessilyn", 'F', "98422333", "Europe"));

Answer 1

using std::copy_if;
using std::ostream_iterator;
using std::cout;

enum Region {
    AMERICA,
    EUROPE,
    REST_OF_WORLD
};

bool is_american(const Student& student)
{
    return student.isFrom(AMERICA);
}

copy_if(students.begin(), students.end(),
        ostream_iterator<Student>(cout, "\n"),
        is_american);

在C++ 11中使用lambda,并允许选择的区域:

void show_students_from_region(const Region& region)
{
    copy_if(students.begin(), students.end(),
            ostream_iterator<Student>(cout, "\n"),
            [&](const Student& student) { return student.isFrom(region); });
}

Answer 2

你可以使用filter_iteratorboost.这是一个示例,底层集合是一个普通的数组.

以下是您可以使用的示例未经测试的代码; 我不得不做出某些假设Student(operator<<对输出有效,国家暴露通过std::string country() const)

struct checkCountry
{
  std::string country;
  bool operator()(const Student& x) 
  {
    return (x.country() == country);
  }
};

int main()
{
  std::vector<Student> studentList;
  studentList.push_back(Student("Tom", 'M', "91213242", "America"));
  studentList.push_back(Student("Jessilyn", 'F', "98422333", "Europe"));

  typedef boost::filter_iterator<checkCountry, std::vector<Student>::iterator> FilterIter;
  checkCountry predicate;
  predicate.country = "America";
  FilterIter filter_iter_first(predicate, studentList.begin(), studentList.end());
  FilterIter filter_iter_last(predicate,  studentList.end(),  studentList.end());

  std::copy(filter_iter_first, filter_iter_last, std::ostream_iterator<Student>(std::cout, " "));
}

Answer 3

我认为这就是你要找的东西:

struct country_filter
{
    country_filter(const std::string& a_country): country(a_country) {}
    void operator()(const Student& a_s) const
    {
        if (country == a_s.country)
        {
            std::cout << a_s.name << "\n";
        }
    }
    std::string country;
};

// 
std::for_each(studentList.begin(), studentList.end(), country_filter("Ireland"));

C++ 11:

std::string country = "America";
std::for_each(studentList.begin(), studentList.end(), [&country] (const Student& a_s)
{
    if (a_s.country == country)
    {
        std::cout << a_s.name << "\n";
    }
});

Answer 4

您可以使用实现()运算符的类的对象.这称为仿函数:

struct checkCountry {
  const string& compare;
  checkCountry(const string& compare) : compare(compare) {}
  bool operator()(const string& x) { return x == compare; }
};

vector<Student> studentList;
studentList.push_back(Student("Tom", 'M', "91213242", "America"));
studentList.push_back(Student("Jessilyn", 'F', "98422333", "Europe"));
howMany = std::count_if(studentList.begin(), studentList.end(), checkCountry("America"));

您可以在需要的元谓词任何算法使用仿函数,例如std::count_if,std::find_if等等.