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使用 STL 将向量元素乘以标量值

[英]Multiply vector elements by a scalar value using STL

Hi I want to (multiply,add,etc) vector by scalar value for example myv1 * 3 , I know I can do a function with a forloop , but is there a way of doing this using STL function?嗨,我想(乘、加等)向量按标量值,例如myv1 * 3 ,我知道我可以用 forloop 做一个函数,但是有没有办法使用 STL 函数来做到这一点? Something like the {Algorithm.h :: transform function }? {Algorithm.h :: 变换函数}之类的东西?

Yes, using std::transform :是的,使用std::transform

std::transform(myv1.begin(), myv1.end(), myv1.begin(),
               std::bind(std::multiplies<T>(), std::placeholders::_1, 3));

Before C++17 you could use std::bind1st() , which was deprecated in C++11.在 C++17 之前,您可以使用在 C++11 中已弃用的std::bind1st()

std::transform(myv1.begin(), myv1.end(), myv1.begin(),
               std::bind1st(std::multiplies<T>(), 3));

For the placeholders;对于占位符;

#include <functional> 

If you can use a valarray instead of a vector , it has builtin operators for doing a scalar multiplication.如果您可以使用valarray而不是vector ,它具有用于执行标量乘法的内置运算符。

v *= 3;

If you have to use a vector , you can indeed use transform to do the job:如果您必须使用vector ,您确实可以使用transform来完成这项工作:

transform(v.begin(), v.end(), v.begin(), _1 * 3);

(assuming you have something similar to Boost.Lambda that allows you to easily create anonymous function objects like _1 * 3 :-P) (假设您有类似于Boost.Lambda的东西,它允许您轻松创建匿名函数对象,例如_1 * 3 :-P)

Modern C++ solution for your question.您的问题的现代 C++ 解决方案。

#include <algorithm>
#include <vector>

std::vector<double> myarray;
double myconstant{3.3};
std::transform(myarray.begin(), myarray.end(), myarray.begin(), [&myconstant](auto& c){return c*myconstant;});

I think for_each is very apt when you want to traverse a vector and manipulate each element according to some pattern, in this case a simple lambda would suffice:我认为当你想遍历一个向量并根据某种模式操作每个元素时, for_each非常合适,在这种情况下,一个简单的 lambda 就足够了:

std::for_each(myv1.begin(), mtv1.end(), [](int &el){el *= 3; });

note that any variable you want to capture for the lambda function to use (say that you eg wanted to multiply with some predetermined scalar), goes into the bracket as a reference.请注意,您想要捕获以供 lambda 函数使用的任何变量(例如,您想要与某个预定的标量相乘),都会作为参考放入括号中。

If you had to store the results in a new vector , then you could use the std::transform() from the <algorithm> header:如果您必须将结果存储在一个新的 vector中,那么您可以使用<algorithm>标头中的std::transform()

#include <algorithm>
#include <vector>

int main() {
    const double scale = 2;
    std::vector<double> vec_input{1, 2, 3};
    std::vector<double> vec_output(3); // a vector of 3 elements, Initialized to zero
    // ~~~
    std::transform(vec_input.begin(), vec_input.end(), vec_output.begin(),
                   [&scale](double element) { return element *= scale; });
    // ~~~
    return 0;
}

So, what we are saying here is,所以,我们在这里要说的是,

  • take the values ( element s) of vec_input starting from the beginning ( vec_input.begin() ) to the end ( vec_input.begin() ),vec_input从开头 ( vec_input.begin() ) 到结尾 ( vec_input.begin() ) 的值 ( element s),
    • essentially, with the first two arguments, you specify a range of elements ( [beginning, end) ) to transform, range本质上,使用前两个参数,您指定要转换的元素范围( [beginning, end) ),范围
  • pass each element to the last argument, lambda expression,将每个element传递给最后一个参数 lambda 表达式,
  • take the output of lambda expression and put it in the vec_output starting from the beginning ( vec_output.begin() ).获取 lambda 表达式的输出并将其放入vec_output从头开始​​( vec_output.begin() )。
    • the third argument is to specify the beginning of the destination vector.第三个参数是指定目标向量的开始。

The lambda expression拉姆达表达式

  • captures the value of scale factor ( [&scale] ) from outside by reference,通过引用从外部捕获比例因子 ( [&scale] ) 的值,
  • takes as its input a vector element of type double (passed to it by std::transform() )将 double 类型的向量元素作为输入(由std::transform()传递给它)
  • in the body of the function, it returns the final result,在函数体中,它返回最终结果,
    • which, as I mentioned above, will be consequently stored in the vec_input .正如我上面提到的,它将因此存储在vec_input中。

Final note: Although unnecessary, you could pass lambda expression per below:最后说明:虽然没有必要,但您可以通过以下 lambda 表达式:

[&scale](double element) -> double { return element *= scale; }

It explicitly states that the output of the lambda expression is a double.它明确指出 lambda 表达式的输出是双精度数。 However, we can omit that, because the compiler, in this case, can deduce the return type by itself.但是,我们可以省略它,因为在这种情况下,编译器可以自己推断返回类型。

I know this not STL as you want, but it is something you can adapt as different needs arise.我知道这不是你想要的 STL,但它是你可以适应不同需求的东西。

Below is a template you can use to calculate;以下是您可以用来计算的模板; 'func' would be the function you want to do: multiply, add, and so on; 'func' 将是您想要执行的函数:乘法、加法等; 'parm' is the second parameter to the 'func'. 'parm' 是 'func' 的第二个参数。 You can easily extend this to take different func's with more parms of varied types.您可以轻松地扩展它以使用更多不同类型的参数来获取不同的函数。

template<typename _ITStart, typename _ITEnd, typename _Func , typename _Value >
_ITStart xform(_ITStart its, _ITEnd ite, _Func func, _Value parm)
{
    while (its != ite) { *its = func(*its, parm); its++; }
    return its;
}
...

int mul(int a, int b) { return a*b; }

vector< int > v;

xform(v.begin(), v.end(), mul, 3); /* will multiply each element of v by 3 */

Also, this is not a 'safe' function, you must do type/value-checking etc. before you use it.此外,这不是一个“安全”功能,您必须在使用它之前进行类型/值检查等。

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