一起压缩几个std :: list迭代器 [英] Zip several std::list iterators together

问题描述

使用boost库，可以使用

Using the boost library it is possible to zip together a known number of iterators using a zip iterator, but what about when the number of iterators to be zipped is not known until runtime?

要稍微扩展一下，我有一个大小相同的列表列表，我需要将每个索引处的所有值组合在一起，然后将它们输入另一个操作中.现在这都是手动的，我觉得应该有更好的方法.

To expand a little bit, I have a list of lists that are all the same size, and I need to group together all the values at each each index and feed them into another operation. Right now this is all manual, and I feel like there should be a better way.

示例:

说我有3个列表:

• [1、2、3、4、5]
• [11、12、13、14、15]
• [21，22，23，24，25]

我需要将这些列表转换为:

I need to transform these lists into:

• [1，11，12]
• [2，12，22]
• [3，13，23]
• [4，14，24]
• ...等等

在运行之前，我不知道输入中有多少个列表.

I do not know how many lists are in the input until runtime.

推荐答案

花费了近1/2个小时之后，我想到了可以进一步改进的dynamic_zip_iterator类，使其看起来像类似于STL的迭代器.到目前为止，它非常具体，因为我已经在其中硬编码了std::list，您可以将其替换为std::vector，或者可以使其更通用:

Alright after spending almost 1/2 hour, I came up with this dynamic_zip_iterator class which can be further improved, to make it look like STL-like iterators. As of now, it's very specific, as I've hardcoded std::list in it which you can replace with std::vector or can make even more generic:

无论如何，看看它:

template<typename T>
struct dynamic_zip_iterator
{
   typedef typename std::list<T>::iterator list_iterator;
   std::list<list_iterator> iterators;
   std::list<std::list<T>> * plists;
   dynamic_zip_iterator(std::list<std::list<T>> & lists, bool isbegin) : plists(&lists) 
   {
         auto it = plists->begin();
         for( ; it != plists->end(); ++it)
         {
           if ( isbegin )
                iterators.push_back(it->begin()); 
           else
                iterators.push_back(it->end()); 
         }
   }
   dynamic_zip_iterator(const dynamic_zip_iterator & zip) : 
          plists(zip.plists),iterators(zip.iterators) {}

   dynamic_zip_iterator operator++()
   { 
     auto it = iterators.begin();
     for( ; it != iterators.end(); ++it)
          ++(*it);
     return *this;
   }
   std::list<T> operator*() 
   { 
     std::list<T> lst;
     auto it = iterators.begin();
     for( ; it != iterators.end(); ++it)
          lst.push_back(*(*it));     
     return lst;
   }
   bool operator!=(dynamic_zip_iterator &zip)
   { 
     auto it1 = iterators.begin();
     auto it2 = zip.iterators.begin();
     return (*it1) != (*it2);
   }
   static dynamic_zip_iterator begin(std::list<std::list<T>> & lists)
   {
      return dynamic_zip_iterator<T>(lists, true);
   }
   static dynamic_zip_iterator end(std::list<std::list<T>> & lists)
   {
      return dynamic_zip_iterator<T>(lists, false);
   }
};

使用它，您的问题将减少到此功能:

Using it your problem reduces to this function:

std::list<std::list<int>> create_lists(std::list<std::list<int>>& lists)
{
  std::list<std::list<int>> results;
  auto begin = dynamic_zip_iterator<int>::begin(lists);
  auto end = dynamic_zip_iterator<int>::end(lists);
  for( ; begin != end ; ++begin)
  {
     results.push_back(*begin);
  }
  return results;    
}

测试代码:

int main() {
        int a[] = {1, 2, 3, 4, 5}, b[] = {11, 12, 13, 14, 15}, c[] = {21, 22, 23, 24, 25};
        std::list<int> l1(a,a+5), l2(b,b+5), l3(c,c+5);
        std::list<std::list<int>> lists;
        lists.push_back(l1); 
        lists.push_back(l2);
        lists.push_back(l3);
        std::list<std::list<int>> newlists = create_lists(lists);
        for(auto lst = newlists.begin(); lst != newlists.end(); ++lst)
        {
                std::cout << "[";
                std::copy(lst->begin(), lst->end(), std::ostream_iterator<int>(std::cout, " "));
                std::cout << "]" << std::endl;
        }
        return 0;
}

输出:

[1 11 21 ]
[2 12 22 ]
[3 13 23 ]
[4 14 24 ]
[5 15 25 ]

在线演示: http://ideone.com/3FJu1

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