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？
问问题
39765 次
10 回答
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
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