我有一个vector<int>容器，它有整数（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'
答案 0 :(得分：86)
绝对不如Python那么优雅，但没有什么比C ++中的Python更优雅。
您可以使用stringstream ...
std::stringstream ss;
for(size_t i = 0; i < v.size(); ++i)
{
if(i != 0)
ss << ",";
ss << v[i];
}
std::string s = ss.str();
您也可以使用std::for_each代替。
答案 1 :(得分：43)
使用std :: for_each和lambda，你可以做一些有趣的事情。
#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=',';
});
}
请参阅this question我写的一个小班。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算法等价物：
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
}
答案 2 :(得分：17)
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，但我还是想不出来。
答案 3 :(得分：17)
您可以使用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);
});
答案 4 :(得分：12)
使用Boost和C ++ 11，可以这样实现：
empKeyauto array = {1,2,3,4};
join(array | transformed(tostr), ",");
归功于Praetorian。
您可以处理任何类型的值：
#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;
}
答案 5 :(得分：11)
这只是试图解决1800 INFORMATION's remark关于他缺乏通用性的第二个解决方案所给出的谜语，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）。
答案 6 :(得分：7)
很多模板/想法。我不是一般的或有效的，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。它以最短的线数获胜...... :)
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);
答案 7 :(得分：4)
如果你想std::cout << join(myVector, ",") << std::endl;，你可以这样做：
template <typename C, typename T> class MyJoiner
{
C &c;
T &s;
MyJoiner(C &&container, T&& sep) : c(std::forward<C>(container)), s(std::forward<T>(sep)) {}
public:
template<typename C, typename T> friend std::ostream& operator<<(std::ostream &o, MyJoiner<C, T> const &mj);
template<typename C, typename T> friend MyJoiner<C, T> join(C &&container, T&& sep);
};
template<typename C, typename T> std::ostream& operator<<(std::ostream &o, MyJoiner<C, T> const &mj)
{
auto i = mj.c.begin();
if (i != mj.c.end())
{
o << *i++;
while (i != mj.c.end())
{
o << mj.s << *i++;
}
}
return o;
}
template<typename C, typename T> MyJoiner<C, T> join(C &&container, T&& sep)
{
return MyJoiner<C, T>(std::forward<C>(container), std::forward<T>(sep));
}
注意，此解决方案直接连接到输出流而不是创建辅助缓冲区，并且可以使用具有运算符＆lt;＆lt;到了ostream。
如果boost::algorithm::join()代替vector<char*>而vector<string>失败，这也适用。
答案 8 :(得分：2)
我喜欢1800的回答。但是我会将第一次迭代移出循环，因为if语句的结果仅在第一次迭代后更改一次
template <class T, class A>
T join(const A &begin, const A &end, const T &t)
{
T result;
A it = begin;
if (it != end)
{
result.append(*it);
++it;
}
for( ;
it!=end;
++it)
{
result.append(t);
result.append(*it);
}
return result;
}
如果您愿意，这当然可以减少到更少的陈述：
template <class T, class A>
T join(const A &begin, const A &end, const T &t)
{
T result;
A it = begin;
if (it != end)
result.append(*it++);
for( ; it!=end; ++it)
result.append(t).append(*it);
return result;
}
答案 9 :(得分：2)
有一些有趣的尝试为这个问题提供了一个优雅的解决方案。我有一个想法，使用模板化的流来有效地回答OP的原始困境。虽然这是一个老帖子，但我希望未来的用户偶然发现这个问题会让我的解决方案变得有益。
首先，一些答案（包括已接受的答案）不会促进可重用性。由于C ++没有提供一种优雅的方式来连接标准库中的字符串（我已经看到），因此创建一个灵活且可重用的字符串变得非常重要。这是我的镜头：
// Replace with your namespace //
namespace my {
// Templated join which can be used on any combination of streams, iterators and base types //
template <typename TStream, typename TIter, typename TSeperator>
TStream& join(TStream& stream, TIter begin, TIter end, TSeperator seperator) {
// A flag which, when true, has next iteration prepend our seperator to the stream //
bool sep = false;
// Begin iterating through our list //
for (TIter i = begin; i != end; ++i) {
// If we need to prepend a seperator, do it //
if (sep) stream << seperator;
// Stream the next value held by our iterator //
stream << *i;
// Flag that next loops needs a seperator //
sep = true;
}
// As a convenience, we return a reference to the passed stream //
return stream;
}
}
现在使用它，您可以简单地执行以下操作：
// Load some data //
std::vector<int> params;
params.push_back(1);
params.push_back(2);
params.push_back(3);
params.push_back(4);
// Store and print our results to standard out //
std::stringstream param_stream;
std::cout << my::join(param_stream, params.begin(), params.end(), ",").str() << std::endl;
// A quick and dirty way to print directly to standard out //
my::join(std::cout, params.begin(), params.end(), ",") << std::endl;
注意使用流如何使这个解决方案非常灵活，因为我们可以将结果存储在字符串流中以便以后回收它，或者我们可以直接写入标准输出，文件，甚至是作为实现的网络连接流。要打印的类型必须是可迭代的并且与源流兼容。 STL提供与各种类型兼容的各种流。所以你真的可以带着它到镇上去。在我的脑海中，你的向量可以是int，float，double，string，unsigned int，SomeObject *等等。
答案 10 :(得分：2)
string s;
for (auto i : v)
s += (s.empty() ? "" : ",") + to_string(i);
答案 11 :(得分：1)
我已经创建了一个帮助程序头文件来添加扩展联接支持。
只需将以下代码添加到常规头文件中，并在需要时包含它。
用法示例：
/* An example for a mapping function. */
ostream&
map_numbers(ostream& os, const void* payload, generic_primitive data)
{
static string names[] = {"Zero", "One", "Two", "Three", "Four"};
os << names[data.as_int];
const string* post = reinterpret_cast<const string*>(payload);
if (post) {
os << " " << *post;
}
return os;
}
int main() {
int arr[] = {0,1,2,3,4};
vector<int> vec(arr, arr + 5);
cout << vec << endl; /* Outputs: '0 1 2 3 4' */
cout << join(vec.begin(), vec.end()) << endl; /* Outputs: '0 1 2 3 4' */
cout << join(vec.begin(), vec.begin() + 2) << endl; /* Outputs: '0 1 2' */
cout << join(vec.begin(), vec.end(), ", ") << endl; /* Outputs: '0, 1, 2, 3, 4' */
cout << join(vec.begin(), vec.end(), ", ", map_numbers) << endl; /* Outputs: 'Zero, One, Two, Three, Four' */
string post = "Mississippi";
cout << join(vec.begin() + 1, vec.end(), ", ", map_numbers, &post) << endl; /* Outputs: 'One Mississippi, Two mississippi, Three mississippi, Four mississippi' */
return 0;
}
场景背后的代码：
#include <iostream>
#include <vector>
#include <list>
#include <set>
#include <unordered_set>
using namespace std;
#define GENERIC_PRIMITIVE_CLASS_BUILDER(T) generic_primitive(const T& v) { value.as_##T = v; }
#define GENERIC_PRIMITIVE_TYPE_BUILDER(T) T as_##T;
typedef void* ptr;
/** A union that could contain a primitive or void*,
* used for generic function pointers.
* TODO: add more primitive types as needed.
*/
struct generic_primitive {
GENERIC_PRIMITIVE_CLASS_BUILDER(int);
GENERIC_PRIMITIVE_CLASS_BUILDER(ptr);
union {
GENERIC_PRIMITIVE_TYPE_BUILDER(int);
GENERIC_PRIMITIVE_TYPE_BUILDER(ptr);
};
};
typedef ostream& (*mapping_funct_t)(ostream&, const void*, generic_primitive);
template<typename T>
class Join {
public:
Join(const T& begin, const T& end,
const string& separator = " ",
mapping_funct_t mapping = 0,
const void* payload = 0):
m_begin(begin),
m_end(end),
m_separator(separator),
m_mapping(mapping),
m_payload(payload) {}
ostream&
apply(ostream& os) const
{
T begin = m_begin;
T end = m_end;
if (begin != end)
if (m_mapping) {
m_mapping(os, m_payload, *begin++);
} else {
os << *begin++;
}
while (begin != end) {
os << m_separator;
if (m_mapping) {
m_mapping(os, m_payload, *begin++);
} else {
os << *begin++;
}
}
return os;
}
private:
const T& m_begin;
const T& m_end;
const string m_separator;
const mapping_funct_t m_mapping;
const void* m_payload;
};
template <typename T>
Join<T>
join(const T& begin, const T& end,
const string& separator = " ",
ostream& (*mapping)(ostream&, const void*, generic_primitive) = 0,
const void* payload = 0)
{
return Join<T>(begin, end, separator, mapping, payload);
}
template<typename T>
ostream&
operator<<(ostream& os, const vector<T>& vec) {
return join(vec.begin(), vec.end()).apply(os);
}
template<typename T>
ostream&
operator<<(ostream& os, const list<T>& lst) {
return join(lst.begin(), lst.end()).apply(os);
}
template<typename T>
ostream&
operator<<(ostream& os, const set<T>& s) {
return join(s.begin(), s.end()).apply(os);
}
template<typename T>
ostream&
operator<<(ostream& os, const Join<T>& vec) {
return vec.apply(os);
}
答案 12 :(得分：1)
扩展@sbi在generic solution上的尝试，该尝试不限于std::vector<int>或特定的返回字符串类型。下面显示的代码可以像这样使用：
std::vector<int> vec{ 1, 2, 3 };
// Call modern range-based overload.
auto str = join( vec, "," );
auto wideStr = join( vec, L"," );
// Call old-school iterator-based overload.
auto str = join( vec.begin(), vec.end(), "," );
auto wideStr = join( vec.begin(), vec.end(), L"," );
在原始代码中，如果分隔符是字符串文字，则模板参数推导无法产生正确的返回字符串类型（如上述示例中所示）。在这种情况下，函数主体中的Str::value_type之类的typedef不正确。该代码假定Str始终是std::basic_string之类的类型，因此对于字符串文字而言显然失败。
为解决此问题，下面的代码尝试从分隔符参数中仅推断出 character 类型，并使用该类型产生默认的返回字符串类型。这是通过boost::range_value实现的，它从给定的 range 类型中提取元素类型。
#include <string>
#include <sstream>
#include <boost/range.hpp>
template< class Sep, class Str = std::basic_string< typename boost::range_value< Sep >::type >, class InputIt >
Str join( InputIt first, const InputIt last, const Sep& sep )
{
using char_type = typename Str::value_type;
using traits_type = typename Str::traits_type;
using allocator_type = typename Str::allocator_type;
using ostringstream_type = std::basic_ostringstream< char_type, traits_type, allocator_type >;
ostringstream_type result;
if( first != last )
{
result << *first++;
}
while( first != last )
{
result << sep << *first++;
}
return result.str();
}
现在，我们可以轻松提供基于范围的重载，将其简单地转发到基于迭代器的重载：
template <class Sep, class Str = std::basic_string< typename boost::range_value<Sep>::type >, class InputRange>
Str join( const InputRange &input, const Sep &sep )
{
// Include the standard begin() and end() in the overload set for ADL. This makes the
// function work for standard types (including arrays), aswell as any custom types
// that have begin() and end() member functions or overloads of the standalone functions.
using std::begin; using std::end;
// Call iterator-based overload.
return join( begin(input), end(input), sep );
}
答案 13 :(得分：1)
以下是将vector中的元素转换为string的简单实用方法：
std::string join(const std::vector<int>& numbers, const std::string& delimiter = ",") {
std::ostringstream result;
for (const auto number : numbers) {
if (result.tellp() > 0) { // not first round
result << delimiter;
}
result << number;
}
return result.str();
}
#include <sstream>需要ostringstream。
答案 14 :(得分：1)
Here's a generic C++11 solution that will let you do
int main() {
vector<int> v {1,2,3};
cout << join(v, ", ") << endl;
string s = join(v, '+').str();
}
The code is:
template<typename Iterable, typename Sep>
class Joiner {
const Iterable& i_;
const Sep& s_;
public:
Joiner(const Iterable& i, const Sep& s) : i_(i), s_(s) {}
std::string str() const {std::stringstream ss; ss << *this; return ss.str();}
template<typename I, typename S> friend std::ostream& operator<< (std::ostream& os, const Joiner<I,S>& j);
};
template<typename I, typename S>
std::ostream& operator<< (std::ostream& os, const Joiner<I,S>& j) {
auto elem = j.i_.begin();
if (elem != j.i_.end()) {
os << *elem;
++elem;
while (elem != j.i_.end()) {
os << j.s_ << *elem;
++elem;
}
}
return os;
}
template<typename I, typename S>
inline Joiner<I,S> join(const I& i, const S& s) {return Joiner<I,S>(i, s);}
答案 15 :(得分：0)
我写了以下代码。它基于C＃string.join。它适用于std :: string和std :: wstring以及许多类型的向量。 （评论中的例子）
这样称呼：
std::vector<int> vVectorOfIds = {1, 2, 3, 4, 5};
std::wstring wstrStringForSQLIn = Join(vVectorOfIds, L',');
代码：
// Generic Join template (mimics string.Join() from C#)
// Written by RandomGuy (stackoverflow) 09-01-2017
// Based on Brian R. Bondy anwser here:
// http://stackoverflow.com/questions/1430757/c-vector-to-string
// Works with char, wchar_t, std::string and std::wstring delimiters
// Also works with a different types of vectors like ints, floats, longs
template<typename T, typename D>
auto Join(const std::vector<T> &vToMerge, const D &delimiter)
{
// We use std::conditional to get the correct type for the stringstream (char or wchar_t)
// stringstream = basic_stringstream<char>, wstringstream = basic_stringstream<wchar_t>
using strType =
std::conditional<
std::is_same<D, std::string>::value,
char,
std::conditional<
std::is_same<D, char>::value,
char,
wchar_t
>::type
>::type;
std::basic_stringstream<strType> ss;
for (size_t i = 0; i < vToMerge.size(); ++i)
{
if (i != 0)
ss << delimiter;
ss << vToMerge[i];
}
return ss.str();
}
答案 16 :(得分：0)
我从@ sbi的答案开始，但大部分时间最终将结果字符串汇总到一个流，因此创建了下面的解决方案，可以通过管道连接到流，而无需在内存中创建完整字符串的开销
使用如下：
#include "string_join.h"
#include <iostream>
#include <vector>
int main()
{
std::vector<int> v = { 1, 2, 3, 4 };
// String version
std::string str = join(v, std::string(", "));
std::cout << str << std::endl;
// Directly piped to stream version
std::cout << join(v, std::string(", ")) << std::endl;
}
其中string_join.h是：
#pragma once
#include <iterator>
#include <sstream>
template<typename Str, typename It>
class joined_strings
{
private:
const It begin, end;
Str sep;
public:
typedef typename Str::value_type char_type;
typedef typename Str::traits_type traits_type;
typedef typename Str::allocator_type allocator_type;
private:
typedef std::basic_ostringstream<char_type, traits_type, allocator_type>
ostringstream_type;
public:
joined_strings(It begin, const It end, const Str &sep)
: begin(begin), end(end), sep(sep)
{
}
operator Str() const
{
ostringstream_type result;
result << *this;
return result.str();
}
template<typename ostream_type>
friend ostream_type& operator<<(
ostream_type &ostr, const joined_strings<Str, It> &joined)
{
It it = joined.begin;
if(it!=joined.end)
ostr << *it;
for(++it; it!=joined.end; ++it)
ostr << joined.sep << *it;
return ostr;
}
};
template<typename Str, typename It>
inline joined_strings<Str, It> join(It begin, const It end, const Str &sep)
{
return joined_strings<Str, It>(begin, end, sep);
}
template<typename Str, typename Container>
inline joined_strings<Str, typename Container::const_iterator> join(
Container container, const Str &sep)
{
return join(container.cbegin(), container.cend(), sep);
}
答案 17 :(得分：0)
我使用类似的东西
namespace std
{
// for strings join
string to_string( string value )
{
return value;
}
} // namespace std
namespace // anonymous
{
template< typename T >
std::string join( const std::vector<T>& values, char delimiter )
{
std::string result;
for( typename std::vector<T>::size_type idx = 0; idx < values.size(); ++idx )
{
if( idx != 0 )
result += delimiter;
result += std::to_string( values[idx] );
}
return result;
}
} // namespace anonymous
答案 18 :(得分：0)
正如@capone所做的那样，
std::string join(const std::vector<std::string> &str_list ,
const std::string &delim=" ")
{
if(str_list.size() == 0) return "" ;
return std::accumulate( str_list.cbegin() + 1,
str_list.cend(),
str_list.at(0) ,
[&delim](const std::string &a , const std::string &b)
{
return a + delim + b ;
} ) ;
}
template <typename ST , typename TT>
std::vector<TT> map(TT (*op)(ST) , const vector<ST> &ori_vec)
{
vector<TT> rst ;
std::transform(ori_vec.cbegin() ,
ori_vec.cend() , back_inserter(rst) ,
[&op](const ST& val){ return op(val) ;} ) ;
return rst ;
}
然后我们可以这样打电话：
int main(int argc , char *argv[])
{
vector<int> int_vec = {1,2,3,4} ;
vector<string> str_vec = map<int,string>(to_string, int_vec) ;
cout << join(str_vec) << endl ;
return 0 ;
}
就像python：
>>> " ".join( map(str, [1,2,3,4]) )
答案 19 :(得分：0)
这是将整数向量转换为字符串的简单方法。
#include <bits/stdc++.h>
using namespace std;
int main()
{
vector<int> A = {1, 2, 3, 4};
string s = "";
for (int i = 0; i < A.size(); i++)
{
s = s + to_string(A[i]) + ",";
}
s = s.substr(0, s.length() - 1); //Remove last character
cout << s;
}
答案 20 :(得分：0)
我使用template function来加入vector项，并通过仅迭代{{1}中的第一个倒数第二个项来删除了不必要的if语句}}，然后加入vector循环之后的最后一项。这也消除了需要额外的代码来删除连接字符串末尾的额外分隔符的需求。因此，没有for语句会减慢迭代速度，也没有多余的分隔符需要整理。
这会产生优美的函数调用，以加入if，vector或string等中的integer。
我写了两个版本：一个返回一个字符串；另一个返回一个字符串。另一个直接写入流。
double#include <iostream>
#include <sstream>
#include <string>
#include <vector>
using namespace std;
// Return a string of joined vector items.
template<typename T>
string join(const vector<T>& v, const string& sep)
{
ostringstream oss;
const auto LAST = v.end() - 1;
// Iterate through the first to penultimate items appending the separator.
for (typename vector<T>::const_iterator p = v.begin(); p != LAST; ++p)
{
oss << *p << sep;
}
// Join the last item without a separator.
oss << *LAST;
return oss.str();
}
// Write joined vector items directly to a stream.
template<typename T>
void join(const vector<T>& v, const string& sep, ostream& os)
{
const auto LAST = v.end() - 1;
// Iterate through the first to penultimate items appending the separator.
for (typename vector<T>::const_iterator p = v.begin(); p != LAST; ++p)
{
os << *p << sep;
}
// Join the last item without a separator.
os << *LAST;
}
int main()
{
vector<string> strings
{
"Joined",
"from",
"beginning",
"to",
"end"
};
vector<int> integers{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
vector<double> doubles{ 1.2, 3.4, 5.6, 7.8, 9.0 };
cout << join(strings, "... ") << endl << endl;
cout << join(integers, ", ") << endl << endl;
cout << join(doubles, "; ") << endl << endl;
join(strings, "... ", cout);
cout << endl << endl;
join(integers, ", ", cout);
cout << endl << endl;
join(doubles, "; ", cout);
cout << endl << endl;
return 0;
}
答案 21 :(得分：0)
#include <iostream>
#include <vector>
int main()
{
std::vector<int> v{{1,2,3,4}};
std::string str;
// ----->
if (! v.empty())
{
str = std::to_string(*v.begin());
for (auto it = std::next(v.begin()); it != v.end(); ++it)
str.append("," + std::to_string(*it));
}
// <-----
std::cout << str << "\n";
}
答案 22 :(得分：0)
为什么这里的答案如此复杂可笑
string vec2str( vector<int> v){
string s="";
for (auto e: v){
s+=to_string(e);
s+=',';
}
s.pop_back();
return s;
}
答案 23 :(得分：0)
使用数字库 (#include <numeric>) 中的 std::accumulate 解决此问题的另一种方法：
std::vector<int> v{1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
auto comma_fold = [](std::string a, int b) { return std::move(a) + ',' + std::to_string(b); };
std::string s = std::accumulate(std::next(v.begin()), v.end(),
std::to_string(v[0]), // start with first element
comma_fold);
std::cout << s << std::endl; // 1,2,3,4,5,6,7,8,9,10