我正在使用多步操作,并希望合并结果。例如:
std::vector<int> A;
std::vector<int> B;
std::vector<int> AB;
我想让AB有A的内容和B的内容按照这个顺序。做这样的事情最有效的方法是什么?
我正在使用多步操作,并希望合并结果。例如:
std::vector<int> A;
std::vector<int> B;
std::vector<int> AB;
我想让AB有A的内容和B的内容按照这个顺序。做这样的事情最有效的方法是什么?
当前回答
还有一个没有提到的简单变体:
copy(A.begin(),A.end(),std::back_inserter(AB));
copy(B.begin(),B.end(),std::back_inserter(AB));
并使用合并算法:
#include <algorithm>
#include <vector>
#include <iterator>
#include <iostream>
#include <sstream>
#include <string>
template<template<typename, typename...> class Container, class T>
std::string toString(const Container<T>& v)
{
std::stringstream ss;
std::copy(v.begin(), v.end(), std::ostream_iterator<T>(ss, ""));
return ss.str();
};
int main()
{
std::vector<int> A(10);
std::vector<int> B(5); //zero filled
std::vector<int> AB(15);
std::for_each(A.begin(), A.end(),
[](int& f)->void
{
f = rand() % 100;
});
std::cout << "before merge: " << toString(A) << "\n";
std::cout << "before merge: " << toString(B) << "\n";
merge(B.begin(),B.end(), begin(A), end(A), AB.begin(), [](int&,int&)->bool {});
std::cout << "after merge: " << toString(AB) << "\n";
return 1;
}
其他回答
如果你的向量排序*,检查set_union从<algorithm>。
set_union(A.begin(), A.end(), B.begin(), B.end(), AB.begin());
链接中有一个更详细的例子。
还有一个没有提到的简单变体:
copy(A.begin(),A.end(),std::back_inserter(AB));
copy(B.begin(),B.end(),std::back_inserter(AB));
并使用合并算法:
#include <algorithm>
#include <vector>
#include <iterator>
#include <iostream>
#include <sstream>
#include <string>
template<template<typename, typename...> class Container, class T>
std::string toString(const Container<T>& v)
{
std::stringstream ss;
std::copy(v.begin(), v.end(), std::ostream_iterator<T>(ss, ""));
return ss.str();
};
int main()
{
std::vector<int> A(10);
std::vector<int> B(5); //zero filled
std::vector<int> AB(15);
std::for_each(A.begin(), A.end(),
[](int& f)->void
{
f = rand() % 100;
});
std::cout << "before merge: " << toString(A) << "\n";
std::cout << "before merge: " << toString(B) << "\n";
merge(B.begin(),B.end(), begin(A), end(A), AB.begin(), [](int&,int&)->bool {});
std::cout << "after merge: " << toString(AB) << "\n";
return 1;
}
对于这个用例,如果您预先知道每个线程产生的结果的数量,那么您可以预先分配AB并将std::span传递给每个线程。这样就不需要进行连接。例子:
std::vector<int> AB(total_number_of_results, 0);
std::size_t chunk_length = …;
std::size_t chunk2_start = chunk_length;
std::size_t chunk3_start = 2 * chunk_length; // If needed
…
// Pass these to the worker threads.
std::span<int> A(AB.data(), chunk_length);
std::span<int> B(AB.data() + chunk2_start, chunk_length);
…
我的回答是基于罗纳德·索萨先生最初的解决方案。除了他最初的解决方案,我还写了一个支持迭代器的向量代理!
对于那些不知道原始解决方案上下文的人来说,简短的描述:joined_vector模板类(即向量代理)将两个向量的两个引用作为构造函数参数,然后将它们视为一个连续的向量。我的实现还支持前向迭代器。
用法:
int main()
{
std::vector<int> a1;
std::vector<int> a2;
joined_vector<std::vector<int>> jv(a1,a2);
for (int i = 0; i < 5; i++)
a1.push_back(i);
for (int i = 5; i <=10; i++)
a2.push_back(i);
for (auto e : jv)
std::cout << e<<"\n";
for (int i = 0; i < jv.size(); i++)
std::cout << jv[i] << "\n";
return 0;
}
实现:
template<typename _vec>
class joined_vector
{
_vec& m_vec1;
_vec& m_vec2;
public:
struct Iterator
{
typedef typename _vec::iterator::value_type type_value;
typedef typename _vec::iterator::value_type* pointer;
typedef typename _vec::iterator::value_type& reference;
typedef std::forward_iterator_tag iterator_category;
typedef std::ptrdiff_t difference_type;
_vec* m_vec1;
_vec* m_vec2;
Iterator(pointer ptr) :m_ptr(ptr)
{
}
Iterator operator++()
{
if (m_vec1->size() > 0 && m_ptr == &(*m_vec1)[m_vec1->size() - 1] && m_vec2->size() != 0)
m_ptr = &(*m_vec2)[0];
else
++m_ptr;
return m_ptr;
}
Iterator operator++(int)
{
pointer curr = m_ptr;
if (m_vec1->size() > 0 && m_ptr == &(*m_vec1)[m_vec1->size() - 1] && m_vec2->size() != 0)
m_ptr = &(*m_vec2)[0];
else
++m_ptr;
return curr;
}
reference operator *()
{
return *m_ptr;
}
pointer operator ->()
{
return m_ptr;
}
friend bool operator == (Iterator& itr1, Iterator& itr2)
{
return itr1.m_ptr == itr2.m_ptr;
}
friend bool operator != (Iterator& itr1, Iterator& itr2)
{
return itr1.m_ptr != itr2.m_ptr;
}
private:
pointer m_ptr;
};
joined_vector(_vec& vec1, _vec& vec2) :m_vec1(vec1), m_vec2(vec2)
{
}
Iterator begin()
{
//checkes if m_vec1 is empty and gets the first elemet's address,
//if it's empty then it get's the first address of the second vector m_vec2
//if both of them are empty then nullptr is returned as the first pointer
Iterator itr_beg((m_vec1.size() != 0) ? &m_vec1[0] : ((m_vec2.size() != 0) ? &m_vec2[0] : nullptr));
itr_beg.m_vec1 = &m_vec1;
itr_beg.m_vec2 = &m_vec2;
return itr_beg;
}
Iterator end()
{
//check if m_vec2 is empty and get the last address of that vector
//if the second vector is empty then the m_vec1's vector/the first vector's last element's address is taken
//if both of them are empty then a null pointer is returned as the end pointer
typename _vec::value_type* p = ((m_vec2.size() != 0) ? &m_vec2[m_vec2.size() - 1] : ((m_vec1.size()) != 0 ? &m_vec1[m_vec1.size() - 1] : nullptr));
Iterator itr_beg(p != nullptr ? p + 1 : nullptr);
itr_beg.m_vec1 = &m_vec1;
itr_beg.m_vec2 = &m_vec2;
return itr_beg;
}
typename _vec::value_type& operator [](int i)
{
if (i < m_vec1.size())
return m_vec1[i];
else
return m_vec2[i - m_vec1.size()];
}
size_t size()
{
return m_vec1.size() + m_vec2.size();
}
};
AB.reserve( A.size() + B.size() ); // preallocate memory
AB.insert( AB.end(), A.begin(), A.end() );
AB.insert( AB.end(), B.begin(), B.end() );