如何在保持使用算法的原始排序的同时从未排序的std :: vector中删除重复项？

I have an array of integers that I need to remove duplicates from while maintaining the order of the first occurrence of each integer. I can see doing it like this, but imagine there is a better way that makes use of STL algorithms better? The insertion is out of my control, so I cannot check for duplicates before inserting.

我有一个整数数组,我需要删除重复项,同时保持每个整数第一次出现的顺序。我可以看到这样做,但想象有一种更好的方法可以更好地利用STL算法吗?插入不受我的控制,所以在插入之前我无法检查重复项。

int unsortedRemoveDuplicates(std::vector<int> &numbers) {
    std::set<int> uniqueNumbers;
    std::vector<int>::iterator allItr = numbers.begin();
    std::vector<int>::iterator unique = allItr;
    std::vector<int>::iterator endItr = numbers.end();

    for (; allItr != endItr; ++allItr) {
        const bool isUnique = uniqueNumbers.insert(*allItr).second;

        if (isUnique) {
            *unique = *allItr;
            ++unique;
        }
    }

    const int duplicates = endItr - unique;

    numbers.erase(unique, endItr);
    return duplicates;
}

How can this be done using STL algorithms?

如何使用STL算法完成?

8 个解决方案

#1

The naive way is to use std::set as everyone tells you. It's overkill and has poor cache locality (slow).
The smart* way is to use std::vector appropriately (make sure to see footnote at bottom):

天真的方式是使用std :: set,大家告诉你。它太过分了,缓存局部性很差(慢)。 smart *方法是适当地使用std :: vector(确保在底部看到脚注):

#include <algorithm>
#include <vector>
struct target_less
{
    template<class It>
    bool operator()(It const &a, It const &b) const { return *a < *b; }
};
struct target_equal
{
    template<class It>
    bool operator()(It const &a, It const &b) const { return *a == *b; }
};
template<class It> It uniquify(It begin, It const end)
{
    std::vector<It> v;
    v.reserve(static_cast<size_t>(std::distance(begin, end)));
    for (It i = begin; i != end; ++i)
    { v.push_back(i); }
    std::sort(v.begin(), v.end(), target_less());
    v.erase(std::unique(v.begin(), v.end(), target_equal()), v.end());
    std::sort(v.begin(), v.end());
    size_t j = 0;
    for (It i = begin; i != end && j != v.size(); ++i)
    {
        if (i == v[j])
        {
            using std::iter_swap; iter_swap(i, begin);
            ++j;
            ++begin;
        }
    }
    return begin;
}

Then you can use it like:

然后你就可以使用它:

int main()
{
    std::vector<int> v;
    v.push_back(6);
    v.push_back(5);
    v.push_back(5);
    v.push_back(8);
    v.push_back(5);
    v.push_back(8);
    v.erase(uniquify(v.begin(), v.end()), v.end());
}

*Note: That's the smart way in typical cases, where the number of duplicates isn't too high. For a more thorough performance analysis, see this related answer to a related question.

*注意:这是典型情况下的智能方式,其中重复次数不是太高。有关更全面的性能分析,请参阅相关问题的相关答案。

#2

Sounds like a job for std::copy_if. Define a predicate that keeps track of elements that already have been processed and return false if they have.

听起来像是std :: copy_if的工作。定义一个跟踪已经处理过的元素的谓词,如果有,则返回false。

If you don't have C++11 support, you can use the clumsily named std::remove_copy_if and invert the logic.

如果您没有C ++ 11支持,则可以使用名为std :: remove_copy_if的笨拙并反转逻辑。

This is an untested example:

这是一个未经测试的例子:

template <typename T>
struct NotDuplicate {
  bool operator()(const T& element) {
    return s_.insert(element).second; // true if s_.insert(element);
  }
 private:
  std::set<T> s_;
};

Then

std::vector<int> uniqueNumbers;
NotDuplicate<int> pred;
std::copy_if(numbers.begin(), numbers.end(), 
             std::back_inserter(uniqueNumbers),
             std::ref(pred));

where an std::ref has been used to avoid potential problems with the algorithm internally copying what is a stateful functor, although std::copy_if does not place any requirements on side-effects of the functor being applied.

其中使用std :: ref来避免算法内部复制有状态仿函数的潜在问题,尽管std :: copy_if对正在应用的仿函数的副作用没有任何要求。

#3

int unsortedRemoveDuplicates(std::vector<int>& numbers)
{
    std::set<int> seenNums; //log(n) existence check

    auto itr = begin(numbers);
    while(itr != end(numbers))
    {
        if(seenNums.find(*itr) != end(seenNums)) //seen? erase it
            itr = numbers.erase(itr); //itr now points to next element
        else
        {
            seenNums.insert(*itr);
            itr++;
        }
    }

    return seenNums.size();
}


//3 6 3 8 9 5 6 8
//3 6 8 9 5

#4

Fast and simple, C++11:

快速而简单,C ++ 11:

template<typename T>
size_t RemoveDuplicatesKeepOrder(std::vector<T>& vec)
{
    std::set<T> seen;

    auto newEnd = std::remove_if(vec.begin(), vec.end(), [&seen](const T& value)
    {
        if (seen.find(value) != std::end(seen))
            return true;

        seen.insert(value);
        return false;
    });

    vec.erase(newEnd, vec.end());

    return vec.size();
}

#5

To verify the performance of the proposed solutions, I've tested three of them, listed below. The tests are using random vectors with 1 mln elements and different ratio of duplicates (0%, 1%, 2%, ..., 10%, ..., 90%, 100%).

为了验证所提出的解决方案的性能,我已经测试了其中的三个,如下所示。测试使用具有1毫升元素和不同重复比例的随机载体(0%,1%,2%,...,10%,...,90%,100%)。

• Mehrdad's solution, currently the accepted answer:

  Mehrdad的解决方案,目前是公认的答案:

  void uniquifyWithOrder_sort(const vector<int>&, vector<int>& output)
  {
      using It = vector<int>::iterator;
      struct target_less
      {
          bool operator()(It const &a, It const &b) const { return *a < *b; }
      };
  
      struct target_equal
      {
          bool operator()(It const &a, It const &b) const { return *a == *b; }
      };
  
      auto begin = output.begin();
      auto const end = output.end();
      {
          vector<It> v;
          v.reserve(static_cast<size_t>(distance(begin, end)));
          for (auto i = begin; i != end; ++i)
          {
              v.push_back(i);
          }
          sort(v.begin(), v.end(), target_less());
          v.erase(unique(v.begin(), v.end(), target_equal()), v.end());
          sort(v.begin(), v.end());
          size_t j = 0;
          for (auto i = begin; i != end && j != v.size(); ++i)
          {
              if (i == v[j])
              {
                  using std::iter_swap; iter_swap(i, begin);
                  ++j;
                  ++begin;
              }
          }
      }
  
      output.erase(begin, output.end());
  }
  

• juanchopanza's solution

  void uniquifyWithOrder_set_copy_if(const vector<int>& input, vector<int>& output)
  {
      struct NotADuplicate
      {
          bool operator()(const int& element)
          {
              return _s.insert(element).second;
          }
  
      private:
          set<int> _s;
      };
  
      vector<int> uniqueNumbers;
      NotADuplicate pred;
  
      output.clear();
      output.reserve(input.size());
      copy_if(
          input.begin(),
          input.end(),
          back_inserter(output),
          ref(pred));
  }
  

juanchopanza的解决方案void uniquifyWithOrder_set_copy_if(const vector &input,vector &output) {     struct NotADuplicate     {         bool operator()(const int&element)         {             return _s.insert(element).second;         }     私人的:         set _s;     };     vector uniqueNumbers;     NotADuplicate pred;     output.clear();     output.reserve(input.size());     copy_if(         input.begin()         input.end()         back_inserter(输出),         REF(预解码值)); }

• Leviathan's solution

  void uniquifyWithOrder_set_remove_if(const vector<int>& input, vector<int>& output)
  {
      set<int> seen;
  
      auto newEnd = remove_if(output.begin(), output.end(), [&seen](const int& value)
      {
          if (seen.find(value) != end(seen))
              return true;
  
          seen.insert(value);
          return false;
      });
  
      output.erase(newEnd, output.end());
  }
  

Leviathan的解决方案void uniquifyWithOrder_set_remove_if(const vector &input,vector &output) {     set seen;     auto newEnd = remove_if(output.begin(),output.end(),[&seen](const int&value)     {         if(seen.find(value)!= end(see))             返回true;         seen.insert(值);         返回虚假;     });     output.erase(newEnd,output.end()); }

They are slightly modified for simplicity, and to allow comparing in-place solutions with not in-place ones. The full code used to test is available here.

为简单起见,对它们进行了略微修改,并允许将就地解决方案与非就地解决方案进行比较。用于测试的完整代码可在此处获得。

The results suggest that if you know you'll have at least 1% duplicates the remove_if solution with std::set is the best one. Otherwise, you should go with the sort solution:

结果表明,如果您知道至少有1%重复,则使用std :: set的remove_if解决方案是最好的。否则,您应该使用排序解决方案:

// Intel(R) Core(TM) i7-2600 CPU @ 3.40 GHz 3.40 GHz
// 16 GB RAM, Windows 7, 64 bit
//
// cl 19
// /GS /GL /W3 /Gy /Zc:wchar_t /Zi /Gm- /O2 /Zc:inline /fp:precise /D "NDEBUG" /D "_CONSOLE" /D "_UNICODE" /D "UNICODE" /WX- /Zc:forScope /Gd /Oi /MD /EHsc /nologo /Ot 
//
// 1000 random vectors with 1 000 000 elements each.
// 11 tests: with 0%, 10%, 20%, ..., 90%, 100% duplicates in vectors.

// Ratio: 0
// set_copy_if   : Time : 618.162 ms +- 18.7261 ms
// set_remove_if : Time : 650.453 ms +- 10.0107 ms
// sort          : Time : 212.366 ms +- 5.27977 ms
// Ratio : 0.1
// set_copy_if   : Time : 34.1907 ms +- 1.51335 ms
// set_remove_if : Time : 24.2709 ms +- 0.517165 ms
// sort          : Time : 43.735 ms +- 1.44966 ms
// Ratio : 0.2
// set_copy_if   : Time : 29.5399 ms +- 1.32403 ms
// set_remove_if : Time : 20.4138 ms +- 0.759438 ms
// sort          : Time : 36.4204 ms +- 1.60568 ms
// Ratio : 0.3
// set_copy_if   : Time : 32.0227 ms +- 1.25661 ms
// set_remove_if : Time : 22.3386 ms +- 0.950855 ms
// sort          : Time : 38.1551 ms +- 1.12852 ms
// Ratio : 0.4
// set_copy_if   : Time : 30.2714 ms +- 1.28494 ms
// set_remove_if : Time : 20.8338 ms +- 1.06292 ms
// sort          : Time : 35.282 ms +- 2.12884 ms
// Ratio : 0.5
// set_copy_if   : Time : 24.3247 ms +- 1.21664 ms
// set_remove_if : Time : 16.1621 ms +- 1.27802 ms
// sort          : Time : 27.3166 ms +- 2.12964 ms
// Ratio : 0.6
// set_copy_if   : Time : 27.3268 ms +- 1.06058 ms
// set_remove_if : Time : 18.4379 ms +- 1.1438 ms
// sort          : Time : 30.6846 ms +- 2.52412 ms
// Ratio : 0.7
// set_copy_if   : Time : 30.3871 ms +- 0.887492 ms
// set_remove_if : Time : 20.6315 ms +- 0.899802 ms
// sort          : Time : 33.7643 ms +- 2.2336 ms
// Ratio : 0.8
// set_copy_if   : Time : 33.3077 ms +- 0.746272 ms
// set_remove_if : Time : 22.9459 ms +- 0.921515 ms
// sort          : Time : 37.119 ms +- 2.20924 ms
// Ratio : 0.9
// set_copy_if   : Time : 36.0888 ms +- 0.763978 ms
// set_remove_if : Time : 24.7002 ms +- 0.465711 ms
// sort          : Time : 40.8233 ms +- 2.59826 ms
// Ratio : 1
// set_copy_if   : Time : 21.5609 ms +- 1.48986 ms
// set_remove_if : Time : 14.2934 ms +- 0.535431 ms
// sort          : Time : 24.2485 ms +- 0.710269 ms

---

// Ratio: 0
// set_copy_if   : Time: 666.962 ms +- 23.7445 ms
// set_remove_if : Time: 736.088 ms +- 39.8122 ms
// sort          : Time: 223.796 ms +- 5.27345 ms
// Ratio: 0.01
// set_copy_if   : Time: 60.4075 ms +- 3.4673 ms
// set_remove_if : Time: 43.3095 ms +- 1.31252 ms
// sort          : Time: 70.7511 ms +- 2.27826 ms
// Ratio: 0.02
// set_copy_if   : Time: 50.2605 ms +- 2.70371 ms
// set_remove_if : Time: 36.2877 ms +- 1.14266 ms
// sort          : Time: 62.9786 ms +- 2.69163 ms
// Ratio: 0.03
// set_copy_if   : Time: 46.9797 ms +- 2.43009 ms
// set_remove_if : Time: 34.0161 ms +- 0.839472 ms
// sort          : Time: 59.5666 ms +- 1.34078 ms
// Ratio: 0.04
// set_copy_if   : Time: 44.3423 ms +- 2.271 ms
// set_remove_if : Time: 32.2404 ms +- 1.02162 ms
// sort          : Time: 57.0583 ms +- 2.9226 ms
// Ratio: 0.05
// set_copy_if   : Time: 41.758 ms +- 2.57589 ms
// set_remove_if : Time: 29.9927 ms +- 0.935529 ms
// sort          : Time: 54.1474 ms +- 1.63311 ms
// Ratio: 0.06
// set_copy_if   : Time: 40.289 ms +- 1.85715 ms
// set_remove_if : Time: 29.2604 ms +- 0.593869 ms
// sort          : Time: 57.5436 ms +- 5.52807 ms
// Ratio: 0.07
// set_copy_if   : Time: 40.5035 ms +- 1.80952 ms
// set_remove_if : Time: 29.1187 ms +- 0.63127 ms
// sort          : Time: 53.622 ms +- 1.91357 ms
// Ratio: 0.08
// set_copy_if   : Time: 38.8139 ms +- 1.9811 ms
// set_remove_if : Time: 27.9989 ms +- 0.600543 ms
// sort          : Time: 50.5743 ms +- 1.35296 ms
// Ratio: 0.09
// set_copy_if   : Time: 39.0751 ms +- 1.71393 ms
// set_remove_if : Time: 28.2332 ms +- 0.607895 ms
// sort          : Time: 51.2829 ms +- 1.21077 ms
// Ratio: 0.1
// set_copy_if   : Time: 35.6847 ms +- 1.81495 ms
// set_remove_if : Time: 25.204 ms +- 0.538245 ms
// sort          : Time: 46.4127 ms +- 2.66714 ms

#6

Here is what WilliamKF is searching for. It uses the erase statement. This code is good for lists but isn t good for vectors. For vectors you should not use the erase statement.

这是WilliamKF正在寻找的东西。它使用erase语句。这段代码适用于列表,但不适用于矢量。对于向量,您不应使用erase语句。

//makes uniques in one shot without sorting !! 
template<class listtype> inline
void uniques(listtype* In)
    {

    listtype::iterator it = In->begin();
    listtype::iterator it2= In->begin();

    int tmpsize = In->size();

        while(it!=In->end())
        {
        it2 = it;
        it2++;
        while((it2)!=In->end())
            {
            if ((*it)==(*it2))
                In->erase(it2++);
            else
                ++it2;
            }
        it++;

        }
    }

What I have tryed for vectors without using sort is that:

我没有使用sort尝试的矢量是:

//makes vectors as fast as possible unique
template<typename T> inline
void vectoruniques(std::vector<T>* In)
    {

    int tmpsize = In->size();

        for (std::vector<T>::iterator it = In->begin();it<In->end()-1;it++)
        {
            T tmp = *it;
            for (std::vector<T>::iterator it2 = it+1;it2<In->end();it2++)
            {
                if (*it2!=*it)
                    tmp = *it2;
                else
                    *it2 = tmp;
            }
        }
        std::vector<T>::iterator it = std::unique(In->begin(),In->end());
        int newsize = std::distance(In->begin(),it);
            In->resize(newsize);
    }

Somehow it looks like this would work. I tested it a bit maybe can somebody tell if this really works ! This solution doesn t need any greater operator. I mean why use the greater operator for seaching unique elements ? Usage for Vectors:

不知怎的,它看起来会起作用。我测试了一下也许有人可以告诉我这是否真的有效!该解决方案不需要任何更大的操作员。我的意思是为什么使用更大的运算符来搜索唯一元素?矢量用法:

int myints[] = {21,10,20,20,20,30,21,31,20,20,2}; 
std::vector<int> abc(myints , myints+11);
vectoruniques(&abc);

#7

Here's something that handles POD and non-POD types with move support. Uses default operator== or a custom equality predicate. Does not require sorting/operator<, key generation, or a separate set. No idea if this is more efficient than the other methods described above.

这里有处理POD和非POD类型的东西,有移动支持。使用默认运算符==或自定义等式谓词。不需要排序/运算符<,密钥生成或单独的集合。不知道这是否比上述其他方法更有效。

template <typename Cnt, typename _Pr = std::equal_to<typename Cnt::value_type>>
void remove_duplicates( Cnt& cnt, _Pr cmp = _Pr() )
{
    Cnt result;
    result.reserve( std::size( cnt ) );  // or cnt.size() if compiler doesn't support std::size()

    std::copy_if( 
        std::make_move_iterator( std::begin( cnt ) )
        , std::make_move_iterator( std::end( cnt ) )
        , std::back_inserter( result )
        , [&]( const typename Cnt::value_type& what ) 
        { 
            return std::find_if( 
                std::begin( result )
                , std::end( result )
                , [&]( const typename Cnt::value_type& existing ) { return cmp( what, existing ); }
            ) == std::end( result );
        }
    );  // copy_if

    cnt = std::move( result );  // place result in cnt param
}   // remove_duplicates

Usage/tests:

{
    std::vector<int> ints{ 0,1,1,2,3,4 };
    remove_duplicates( ints );
    assert( ints.size() == 5 );
}

{
    struct data 
    { 
        std::string foo; 
        bool operator==( const data& rhs ) const { return this->foo == rhs.foo; }
    };

    std::vector<data>
        mydata{ { "hello" }, {"hello"}, {"world"} }
        , mydata2 = mydata
        ;

    // use operator==
    remove_duplicates( mydata );
    assert( mydata.size() == 2 );

    // use custom predicate
    remove_duplicates( mydata2, []( const data& left, const data& right ) { return left.foo == right.foo; } );
    assert( mydata2.size() == 2 );

}

#8

Here is a c++11 generic version that works with iterators and doesn't allocate additional storage. It may have the disadvantage of being O(n^2) but is likely faster for smaller input sizes.

这是一个与迭代器一起使用的c ++ 11通用版本,不分配额外的存储空间。它可能具有O(n ^ 2)的缺点,但对于较小的输入尺寸可能更快。

template<typename Iter>
Iter removeDuplicates(Iter begin,Iter end)
{
    auto it = begin;
    while(it != end)
    {
        auto next = std::next(it);
        if(next == end)
        {
            break;
        }
        end = std::remove(next,end,*it);
        it = next;
    }

    return end;
}

....

std::erase(removeDuplicates(vec.begin(),vec.end()),vec.end());

Sample Code: http://cpp.sh/5kg5n

示例代码:http://cpp.sh/5kg5n