python linspace in c++
python linspace in c++
im 试图编写类似版本的 Python numpy.linspace 函数。
double linspace(int a, int b, int c){
double line[c];
double delta =b-a/(c-1);
for (int i=0; i<c; ++i){
line[i]=0 + (i*delta);
}
return line;
A 和 B 是数组中的第一个和最后一个组件,C 指定数组中元素的数量。但是当我编译这个脚本时,它返回:
linspace.cpp: In function ‘double linspace(int, int, int)’:
linspace.cpp:11:9: error: cannot convert ‘double*’ to ‘double’ in return
return line;
^
有人碰巧知道如何解决这个问题吗?
像这样的事情怎么样:
#include <iostream>
#include <vector>
template<typename T>
std::vector<double> linspace(T start_in, T end_in, int num_in)
{
std::vector<double> linspaced;
double start = static_cast<double>(start_in);
double end = static_cast<double>(end_in);
double num = static_cast<double>(num_in);
if (num == 0) { return linspaced; }
if (num == 1)
{
linspaced.push_back(start);
return linspaced;
}
double delta = (end - start) / (num - 1);
for(int i=0; i < num-1; ++i)
{
linspaced.push_back(start + delta * i);
}
linspaced.push_back(end); // I want to ensure that start and end
// are exactly the same as the input
return linspaced;
}
void print_vector(std::vector<double> vec)
{
std::cout << "size: " << vec.size() << std::endl;
for (double d : vec)
std::cout << d << " ";
std::cout << std::endl;
}
int main()
{
std::vector<double> vec_1 = linspace(1, 10, 3);
print_vector(vec_1);
std::vector<double> vec_2 = linspace(6.0, 23.4, 5);
print_vector(vec_2);
std::vector<double> vec_3 = linspace(0.0, 2.0, 1);
print_vector(vec_3);
std::vector<double> vec_4 = linspace(0.0, 2.0, 0);
print_vector(vec_4);
return 0;
}
C++结果:
size: 3
1 5.5 10
size: 5
6 10.35 14.7 19.05 23.4
size: 1
0
size: 0
数字结果:
In [14]: np.linspace(1, 10, 3)
Out[14]: array([ 1. , 5.5, 10. ])
In [15]: np.linspace(6, 23.4, 5)
Out[15]: array([ 6. , 10.35, 14.7 , 19.05, 23.4 ])
In [16]: np.linspace(0.0, 2.0, 1)
Out[16]: array([ 0.])
In [17]: np.linspace(0.0, 2.0, 0)
Out[17]: array([], dtype=float64)
你正在尝试做的事情是行不通的。 首先,您将在堆栈上分配内存 linspace
double line[c];
您要么在调用之前重新分配内存并将其传入,要么动态分配内存并返回它(并记住稍后释放它)。
要动态分配,您可以执行以下操作:
double * line = new double[c];
同样,这将需要在稍后完成的某个时候释放,否则您将出现内存泄漏。
delete line[];
此外,double line[c];
会创建一个双精度数组,并line
指向此。 所以线是一个double *
. 您将函数的返回类型指定为 double
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