将vector <int>转换为字符串

Question

我有一个vector<int>有整数的容器(例如{1,2,3,4}),我想转换为表格的字符串

"1,2,3,4"

在C++中最干净的方法是什么？在Python中,我就是这样做的:

>>> array = [1,2,3,4]
>>> ",".join(map(str,array))
'1,2,3,4'

Answer 1

绝对不如Python优雅,但没有什么比C++中的Python更优雅.

你可以使用stringstream......

#include <sstream>
//...

std::stringstream ss;
for(size_t i = 0; i < v.size(); ++i)
{
  if(i != 0)
    ss << ",";
  ss << v[i];
}
std::string s = ss.str();

你也可以使用std::for_each.

Answer 2

使用std :: copy和std :: ostream_iterator,我们可以得到像python一样优雅的东西.

#include <iostream>
#include <sstream>

int main()
{
     int  array[] = {1,2,3,4};
     std::for_each(std::begin(array), std::end(array),
                   [&std::cout, sep=' '](int x) mutable {
                       out << sep << x; sep=',';
                   });
}

看到这个问题,我写了一个小班.这不会打印尾随的逗号.此外,如果我们假设C++ 14将继续为我们提供基于范围的等效算法,如下所示:

namespace std {
   // I am assuming something like this in the C++14 standard
   // I have no idea if this is correct but it should be trivial to write if it  does not appear.
   template<typename C, typename I>
   void copy(C const& container, I outputIter) {copy(begin(container), end(container), outputIter);}
}
using POI = PrefexOutputIterator;   
int main()
{
     int  array[] = {1,2,3,4};
     std::copy(array, POI(std::cout, ","));
  // ",".join(map(str,array))               // closer
}

Answer 3

你可以使用std :: accumulate.请考虑以下示例

if (v.empty() 
    return std::string();
std::string s = std::accumulate(v.begin()+1, v.end(), std::to_string(v[0]),
                     [](const std::string& a, int b){
                           return a + ',' + std::to_string(b);
                     });

Answer 4

另一种选择是使用std::copy和ostream_iterator类:

#include <iterator>  // ostream_iterator
#include <sstream>   // ostringstream
#include <algorithm> // copy

std::ostringstream stream;
std::copy(array.begin(), array.end(), std::ostream_iterator<>(stream));
std::string s=stream.str();
s.erase(s.length()-1);

也不如Python好.为此,我创建了一个join函数:

template <class T, class A>
T join(const A &begin, const A &end, const T &t)
{
  T result;
  for (A it=begin;
       it!=end;
       it++)
  {
    if (!result.empty())
      result.append(t);
    result.append(*it);
  }
  return result;
}

然后像这样使用它:

std::string s=join(array.begin(), array.end(), std::string(","));

你可能会问为什么我传入了迭代器.好吧,实际上我想要反转数组,所以我像这样使用它:

std::string s=join(array.rbegin(), array.rend(), std::string(","));

理想情况下,我想模板化它可以推断出char类型,并使用字符串流,但我还是无法解决这个问题.

Answer 5

使用Boost和C++ 11,可以像这样实现:

auto array = {1,2,3,4};
join(array | transformed(tostr), ",");

好吧,差不多.这是完整的例子:

#include <array>
#include <iostream>

#include <boost/algorithm/string/join.hpp>
#include <boost/range/adaptor/transformed.hpp>

int main() {
    using boost::algorithm::join;
    using boost::adaptors::transformed;
    auto tostr = static_cast<std::string(*)(int)>(std::to_string);

    auto array = {1,2,3,4};
    std::cout << join(array | transformed(tostr), ",") << std::endl;

    return 0;
}

感谢执政官.

您可以处理任何值类型,如下所示:

template<class Container>
std::string join(Container const & container, std::string delimiter) {
  using boost::algorithm::join;
  using boost::adaptors::transformed;
  using value_type = typename Container::value_type;

  auto tostr = static_cast<std::string(*)(value_type)>(std::to_string);
  return join(container | transformed(tostr), delimiter);
};

Answer 6

这只是试图解决1800 INFORMATION对他缺乏通用性的第二个解决方案的评论所给出的谜语,而不是试图回答这个问题:

template <class Str, class It>
Str join(It begin, const It end, const Str &sep)
{
  typedef typename Str::value_type     char_type;
  typedef typename Str::traits_type    traits_type;
  typedef typename Str::allocator_type allocator_type;
  typedef std::basic_ostringstream<char_type,traits_type,allocator_type>
                                       ostringstream_type;
  ostringstream_type result;

  if(begin!=end)
    result << *begin++;
  while(begin!=end) {
    result << sep;
    result << *begin++;
  }
  return result.str();
}

Works On My Machine(TM).

Answer 7

很多模板/想法.我的不是通用的或有效的,但我只是遇到了同样的问题,并希望把它作为简短而甜蜜的东西.它赢得最短的线数... :)

std::stringstream joinedValues;
for (auto value: array)
{
    joinedValues << value << ",";
}
//Strip off the trailing comma
std::string result = joinedValues.str().substr(0,joinedValues.str().size()-1);

Answer 8

string s;
for (auto i : v)
    s += (s.empty() ? "" : ",") + to_string(i);