69

如何使用 C++/STL 执行以下等效操作？我想std::vector用一系列值 [min, max) 填充 a。

# Python
>>> x = range(0, 10)
>>> x
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

我想我可以使用std::generate_n并提供一个仿函数来生成序列，但我想知道是否有使用 STL 的更简洁的方法？

4

10 回答 10

75

在 C++11 中，有std::iota：

#include <vector>
#include <numeric> //std::iota

std::vector<int> x(10);
std::iota(std::begin(x), std::end(x), 0); //0 is the starting number
于 2012-10-31T06:17:52.453 回答
24

有boost::irange：

std::vector<int> x;
boost::push_back(x, boost::irange(0, 10));
于 2012-10-31T06:21:39.063 回答
5

我最终编写了一些实用程序函数来做到这一点。您可以按如下方式使用它们：

auto x = range(10); // [0, ..., 9]
auto y = range(2, 20); // [2, ..., 19]
auto z = range(10, 2, -2); // [10, 8, 6, 4]

编码：

#include <vector>
#include <stdexcept>

template <typename IntType>
std::vector<IntType> range(IntType start, IntType stop, IntType step)
{
  if (step == IntType(0))
  {
    throw std::invalid_argument("step for range must be non-zero");
  }

  std::vector<IntType> result;
  IntType i = start;
  while ((step > 0) ? (i < stop) : (i > stop))
  {
    result.push_back(i);
    i += step;
  }

  return result;
}

template <typename IntType>
std::vector<IntType> range(IntType start, IntType stop)
{
  return range(start, stop, IntType(1));
}

template <typename IntType>
std::vector<IntType> range(IntType stop)
{
  return range(IntType(0), stop, IntType(1));
}
于 2015-05-18T20:55:26.743 回答
4

多年来，我一直将这个库用于这个确切的目的：

https://github.com/klmr/cpp11-range

工作得很好，代理被优化了。

for (auto i : range(1, 5))
    cout << i << "\n";

for (auto u : range(0u))
    if (u == 3u) 
        break;
    else         
        cout << u << "\n";

for (auto c : range('a', 'd'))
    cout << c << "\n";

for (auto i : range(100).step(-3))
    if (i < 90) 
        break;
    else        
        cout << i << "\n";

for (auto i : indices({"foo", "bar"}))
    cout << i << '\n';
于 2017-08-01T22:49:51.477 回答
3

有boost::irange，但不提供浮点，负步，不能直接初始化stl容器。

我的RO库里也numeric_range有

在 RO 中，初始化一个向量：

vector<int> V=range(10);

从文档页面剪切-n-粘贴示例（scc-c++ 代码段评估器）：

// [0,N)  open-ended range. Only range from 1-arg  range() is open-ended.
scc 'range(5)'
{0, 1, 2, 3, 4}

// [0,N]  closed range
scc 'range(1,5)'
{1, 2, 3, 4, 5}

// floating point 
scc 'range(1,5,0.5)'
{1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5}

// negative step
scc 'range(10,0,-1.5)'
{10, 8.5, 7, 5.5, 4, 2.5, 1}

// any arithmetic type
scc "range('a','z')"
a b c d e f g h i j k l m n o p q r s t u v w x y z

// no need for verbose iota. (vint - vector<int>)
scc 'vint V = range(5);   V' 
{0, 1, 2, 3, 4}

// is lazy
scc 'auto NR = range(1,999999999999999999l);  *find(NR.begin(), NR.end(), 5)'
5

//  Classic pipe. Alogorithms are from std:: 
scc 'vint{3,1,2,3} | sort | unique | reverse'
{3, 2, 1}

//  Assign 42 to 2..5
scc 'vint V=range(0,9);   range(V/2, V/5) = 42;  V'
{0, 1, 42, 42, 42, 5, 6, 7, 8, 9}

//  Find (brute force algorithm) maximum of  `cos(x)` in interval: `8 < x < 9`:
scc 'range(8, 9, 0.01) * cos  || max'
-0.1455

//  Integrate sin(x) from 0 to pi
scc 'auto d=0.001;  (range(0,pi,d) * sin || add) * d'
2

//  Total length of strings in vector of strings
scc 'vstr V{"aaa", "bb", "cccc"};  V * size ||  add'
9

//  Assign to c-string, then append `"XYZ"` and then remove `"bc"` substring :
scc 'char s[99];  range(s) = "abc";  (range(s) << "XYZ") - "bc"'
aXYZ


// Hide phone number:
scc "str S=\"John Q Public  (650)1234567\";  S|isdigit='X';  S"
John Q Public  (XXX)XXXXXXX
于 2012-12-15T06:34:15.847 回答
3

对于那些不能使用 C++11 或库的人：

vector<int> x(10,0); // 0 is the starting number, 10 is the range size
transform(x.begin(),x.end(),++x.begin(),bind2nd(plus<int>(),1)); // 1 is the increment
于 2014-04-30T22:12:11.193 回答
2

类似于下面的 range() 函数将有所帮助：

#include <algorithm>
#include <iostream>
#include <numeric>
#include <vector>
using namespace std;

// define range function (only once)
template <typename T>
vector <T> range(T N1, T N2) {
    vector<T> numbers(N2-N1);
    iota(numbers.begin(), numbers.end(), N1);
    return numbers;
}


vector <int> arr = range(0, 10);
vector <int> arr2 = range(5, 8);

for (auto n : arr) { cout << n << " "; }    cout << endl;
// output:    0 1 2 3 4 5 6 7 8 9

for (auto n : arr2) { cout << n << " "; }   cout << endl;
// output:    5 6 7
于 2017-09-16T10:47:53.173 回答
1

我不知道有一种方法可以像在 python 中那样做，但另一种选择显然是 for 循环遍历它：

for (int i = range1; i < range2; ++i) {
    x.push_back(i);
}

如果你有 c++11，克里斯的答案会更好

于 2012-10-31T06:19:25.083 回答
1

如果你不能使用 C++11，你可以使用它std::partial_sum来生成 1 到 10 的数字。如果你需要 0 到 9 的数字，你可以使用 transform 减 1：

std::vector<int> my_data( 10, 1 );
std::partial_sum( my_data.begin(), my_data.end(), my_data.begin() );
std::transform(my_data.begin(), my_data.end(), my_data.begin(), bind2nd(std::minus<int>(), 1));
于 2014-08-29T15:45:22.700 回答
1

前段时间我写了下面的_range类，它的行为类似于 Python range（把它放到“range.h”中）：

#pragma once
#include <vector>
#include <cassert>

template < typename T = size_t >
class _range 
{
        const T kFrom, kEnd, kStep;

    public:

        ///////////////////////////////////////////////////////////
        // Constructor 
        ///////////////////////////////////////////////////////////
        //
        // INPUT:
        //      from - Starting number of the sequence.
        //      end - Generate numbers up to, but not including this number.
        //      step -  Difference between each number in the sequence.     
        //
        // REMARKS:
        //      Parameters must be all positive or all negative
        //
        _range( const T from, const T end, const T step = 1 ) 
            : kFrom( from ), kEnd( end ), kStep( step ) 
        {
            assert( kStep != 0 );
            assert( ( kFrom >= 0 && kEnd > 0 && kStep > 0 ) || ( kFrom < 0 && kEnd < 0 && kStep < 0 ) );
        }

        // Default from==0, step==1
        _range( const T end ) 
            : kFrom( 0 ), kEnd( end ), kStep( 1 ) 
        {
            assert( kEnd > 0 );
        }

    public:

        class _range_iter 
        {
            T fVal;
            const T kStep;
        public:
            _range_iter( const T v, const T step ) : fVal( v ), kStep( step ) {}
            operator T  () const            { return fVal; }
            operator const T & ()           { return fVal; }
            const T operator * () const     { return fVal; }
            const _range_iter & operator ++ ()  { fVal += kStep; return * this; }


            bool operator == ( const _range_iter & ri ) const
            {
                return ! operator != ( ri );
            }

            bool operator != ( const _range_iter & ri ) const
            {   
                // This is a tricky part - when working with iterators
                // it checks only once for != which must be a hit to stop;
                // However, this does not work if increasing kStart by N times kSteps skips over kEnd
                return fVal < 0 ? fVal > ri.fVal : fVal < ri.fVal;  
            }                                               
        };                                                  

        const _range_iter begin()   { return _range_iter( kFrom, kStep ); }
        const _range_iter end()     { return _range_iter( kEnd, kStep ); }

    public:

        // Conversion to any vector< T >
        operator std::vector< T > ( void ) 
        {
            std::vector< T > retRange;
            for( T i = kFrom; i < kEnd; i += kStep )
                retRange.push_back( i );
            return retRange;    // use move semantics here
        }
};


// A helper to use pure range meaning _range< size_t >
typedef _range<>    range;

一些测试代码如下所示：

#include "range.h" 
#include <iterator>
#include <fstream>

using namespace std;

void RangeTest( void )
{
    ofstream ostr( "RangeTest.txt" );
    if( ostr.is_open() == false )
        return;

    // 1:
    ostr << "1st test:" << endl;

    vector< float > v = _range< float >( 256 );
    copy( v.begin(), v.end(), ostream_iterator< float >( ostr, ", " ) );

    // 2:
    ostr << endl << "2nd test:" << endl;

    vector< size_t >    v_size_t( range( 0, 100, 13 ) );
    for( auto a : v_size_t )
        ostr << a << ", ";

    // 3:
    ostr << endl << "3rd test:" << endl;

    auto vvv = range( 123 );    // 0..122 inclusive, with step 1
    for( auto a : vvv )
        ostr << a << ", ";

    // 4:
    ostr << endl << "4th test:" << endl;

    // Can be used in the nested loops as well
    for( auto i : _range< float >( 0, 256, 16.5 ) ) 
    {
        for( auto j : _range< int >( -2, -16, -3 ) ) 
        {
            ostr << j << ", ";
        }
        ostr << endl << i << endl;
    }

}
于 2018-03-12T13:16:18.807 回答