I have a functor and the overloaded operator() of it returns a double. In an easy way I can write:
int main(){
auto f=[](double t){return 1.0/(pow(7.0*t,2.0)+1.0);};
std::vector<double> nodes(n);
bestpolchevnod(f,nodes); //calculate nodes = f_i(cheby_point(i))
ChebyPoly c_p = ChebyPoly(nodes);//call the constructor with the calculated nodes
std::cout << c_p(0.6) << std::endl; //evaluate at c_p(0.6) as an approx of f(0.6)
};
Now it is possible to go through a set of values by using for_each() like:
std::vector<double> xpoints={0.4,0.5,0.6,0.7};
std::for_each(xpoints.begin(), xpoints.end(), ChebyPoly(nodes));
Is there a sharp/short way to save the calculated values of this algorithm directly for example in a vector? I know there are ways to do it otherwise. But I wonder if there is something similar like
std::vector<double> resvec(xpoints.size());
resvec.push_back(std::for_each(xpoints.begin(), xpoints.end(), ChebyPoly(nodes))); // wrong
std::for_each is the wrong algorithm, you want std::transform
std::vector<double> resvec(xpoints.size());
std::transform(xpoints.begin(), xpoints.end(), resvec.begin(), ChebyPoly(nodes));
Or without zero-initialising the elements of resvec
std::vector<double> resvec;
resvec.reserve(xpoints.size());
std::transform(xpoints.begin(), xpoints.end(), std::back_inserter(resvec), ChebyPoly(nodes));
Or without allocating a result, instead having a lazy view
auto result = xpoints | std::ranges::views::transform(ChebyPoly(nodes));
Related
Suppose I have the following function:
void sum(const std::vector<int*>& input) {
return ... ; // the sum
}
I store a vector of int pointers somewhere
...
std::vector<std::unique_ptr<int>> my_ints;
Is there a way to pass my_ints to sum() without any extra allocations such as an intermediate vector of the unique_ptrs converted to a vector of raw pointers?
Obviously, I could refacor sum() to take a vector of unique ptrs instead. Or overload it. But I'm hoping to find a way where I don't have to, and let the user decide whether or not to use a vector of unique_ptrs or raw pointers.
Not like you want, but you should think about sum() differently. It looks like an algorithm that operates on a range, so you should make it more like this:
template <typename It>
ValueType sum(It begin, It end) {
// ... iterate and calculate sum
return sum;
}
Then suddenly, you can start to use ranges to do cool things!
std::vector<std::unique_ptr<int>> my_ints;
auto range = my_ints | ranges::views::transform
(
[](auto smart_ptr) {
return smart_ptr.get();
}
);
This is a range that will transform as you use it! Then you could enter it into your sum() like this:
auto my_sum = sum(std::begin(range), std::end(range));
Also look up std::accumulate(), which does what you want here, I would say.
No, there is absolutely no way to pass those pointer values to that sum method without changing the method.
I try to implement that summing up all elements of a vector<vector<int>> in a non-loop ways.
I have checked some relevant questions before, How to sum up elements of a C++ vector?.
So I try to use std::accumulate to implement it but I find it is hard for me to overload a Binary Operator in std::accumulate and implement it.
So I am confused about how to implement it with std::accumulate or is there a better way?
If not mind could anyone help me?
Thanks in advance.
You need to use std::accumulate twice, once for the outer vector with a binary operator that knows how to sum the inner vector using an additional call to std::accumulate:
int sum = std::accumulate(
vec.begin(), vec.end(), // iterators for the outer vector
0, // initial value for summation - 0
[](int init, const std::vector<int>& intvec){ // binaryOp that sums a single vector<int>
return std::accumulate(
intvec.begin(), intvec.end(), // iterators for the inner vector
init); // current sum
// use the default binaryOp here
}
);
In this case, I do not suggest using std::accumulate as it would greatly impair readability. Moreover, this function use loops internally, so you would not save anything. Just compare the following loop-based solution with the other answers that use std::accumulate:
int result = 0 ;
for (auto const & subvector : your_vector)
for (int element : subvector)
result += element;
Does using a combination of iterators, STL functions, and lambda functions makes your code easier to understand and faster? For me, the answer is clear. Loops are not evil, especially for such simple application.
According to https://en.cppreference.com/w/cpp/algorithm/accumulate , looks like BinaryOp has the current sum on the left hand, and the next range element on the right. So you should run std::accumulate on the right hand side argument, and then just sum it with left hand side argument and return the result. If you use C++14 or later,
auto binary_op = [&](auto cur_sum, const auto& el){
auto rhs_sum = std::accumulate(el.begin(), el.end(), 0);
return cur_sum + rhs_sum;
};
I didn't try to compile the code though :). If i messed up the order of arguments, just replace them.
Edit: wrong terminology - you don't overload BinaryOp, you just pass it.
Signature of std::accumulate is:
T accumulate( InputIt first, InputIt last, T init,
BinaryOperation op );
Note that the return value is deduced from the init parameter (it is not necessarily the value_type of InputIt).
The binary operation is:
Ret binary_op(const Type1 &a, const Type2 &b);
where... (from cppreference)...
The type Type1 must be such that an object of type T can be implicitly converted to Type1. The type Type2 must be such that an object of type InputIt can be dereferenced and then implicitly converted to Type2. The type Ret must be such that an object of type T can be assigned a value of type Ret.
However, when T is the value_type of InputIt, the above is simpler and you have:
using value_type = std::iterator_traits<InputIt>::value_type;
T binary_op(T,value_type&).
Your final result is supposed to be an int, hence T is int. You need two calls two std::accumulate, one for the outer vector (where value_type == std::vector<int>) and one for the inner vectors (where value_type == int):
#include <iostream>
#include <numeric>
#include <iterator>
#include <vector>
template <typename IT, typename T>
T accumulate2d(IT outer_begin, IT outer_end,const T& init){
using value_type = typename std::iterator_traits<IT>::value_type;
return std::accumulate( outer_begin,outer_end,init,
[](T accu,const value_type& inner){
return std::accumulate( inner.begin(),inner.end(),accu);
});
}
int main() {
std::vector<std::vector<int>> x{ {1,2} , {1,2,3} };
std::cout << accumulate2d(x.begin(),x.end(),0);
}
Solutions based on nesting std::accumulate may be difficult to understand.
By using a 1D array of intermediate sums, the solution can be more straightforward (but possibly less efficient).
int main()
{
// create a unary operator for 'std::transform'
auto accumulate = []( vector<int> const & v ) -> int
{
return std::accumulate(v.begin(),v.end(),int{});
};
vector<vector<int>> data = {{1,2,3},{4,5},{6,7,8,9}}; // 2D array
vector<int> temp; // 1D array of intermediate sums
transform( data.begin(), data.end(), back_inserter(temp), accumulate );
int result = accumulate(temp);
cerr<<"result="<<result<<"\n";
}
The call to transform accumulates each of the inner arrays to initialize the 1D temp array.
To avoid loops, you'll have to specifically add each element:
std::vector<int> database = {1, 2, 3, 4};
int sum = 0;
int index = 0;
// Start the accumulation
sum = database[index++];
sum = database[index++];
sum = database[index++];
sum = database[index++];
There is no guarantee that std::accumulate will be non-loop (no loops). If you need to avoid loops, then don't use it.
IMHO, there is nothing wrong with using loops: for, while or do-while. Processors that have specialized instructions for summing arrays use loops. Loops are a convenient method for conserving code space. However, there may be times when loops want to be unrolled (for performance reasons). You can have a loop with expanded or unrolled content in it.
With range-v3 (and soon with C++20), you might do
const std::vector<std::vector<int>> v{{1, 2}, {3, 4, 5, 6}};
auto flat = v | ranges::view::join;
std::cout << std::accumulate(begin(flat), end(flat), 0);
Demo
All
suppose I got vector with data in cm, and would like to construct another vector but in mm (or mm with a shift, or ..., so it's not quite simple).
What would be good way to accomplish such task?
I wrote some code doing iterator adapter
struct scaling_iterator_adaptor {
...
};
vector v_mm{ scaling_iterator_adaptor{v_cm.begin()}, scaling_iterator_adaptor{v_cm.end()} };
Is there a better way to do such task? Conceptually different way?
If it is not essential to construct it with all the data contained already, you can use standard algorithms:
std::vector<double> v_cm{1, 3.14, 4.2};
std::vector<double> v_mm(v_cm.size());
std::transform(v_cm.cbegin(), v_cm.cend(), v_mm.begin(), [](double x){ return x * 10; });
You can use std::back_inserter if you don't want to prefill the target with zeroes.
I want to create a program that uses a vector to sort it for testing reasons. So I want to calculate the CPU time by a benchmark that sorts the vector a certain amount of times. So the original vector needs to remain constant, and then use another vector so that it can be sorted.
So what I have done is...
#include <iostream>
#include <vector>
#include <random>
#include <chrono>
using namespace std;
typedef vector<int> intv;
int main(){
intv vi;
// Stuff to create my vector with certain characteristics...
intv vii=vi;
cout << "Size: \n";
cin >> tt ;
for(i=0; i<tt; ++i){
tb=sort(t,vii);
m=m+tb;
vii=vi;
}
m=m/tt;
cout << "BS" << m << "\n";
}
So I pass the vector by reference, and make a copy for each sorting so that I can sort it again. How can I do this a better way? Is it better to pass it by value, and in that case, Could someone provide me a minimum example of the best way to do this?
sort is a basic bubble sorting function:
double sort(int t, intv &vii){
vii.reserve(t);
bool swapped=true;
int a;
auto t0 =chrono::high_resolution_clock::now();
while (swapped==true){
for (int i=1; i<t; ++i){
swapped=false;
if (vii[i-1]>vii[i]){
a=vii[i];
vii[i]=vii[i-1];
vii[i-1]=a;
swapped=true;
}
}
t=t-1;
}
auto t1 = chrono::high_resolution_clock::now();
double T = chrono::duration_cast<chrono::nanoseconds>(t1-t0).count();
return T;
}
Once you have sorted, you have to do something that is equivalent to:
vii=vi;
I think assigning vi to vii will be the most efficient method of copying the contents of vi to vii. You can try:
size_t index = 0;
for ( auto const& val : vi )
{
vii[index++] = val;
}
However, I will be really surprised if the second method is more efficient than the first.
Nothing wrong with sorting in-place, and making a copy of the vector. The code you have should work, though it is not clear from where your parameter t is coming.
Note that the statement vii.reserve(t) is not doing anything useful in your sort routine: either t is less than or equal to the size of vii, in which case the reserve call does nothing, or it is greater than the size of vii, in which case you are accessing values outside the range of the vector. Better to check t against the vector size and throw an error or similar if it is too big.
Passing by value is straightforward: just declare your sort routine as double sort(int t, intv vii). When the function is called, vii will be copied from whichever vector you pass in as the second argument.
From a design point of view though, it is better to make a copy and then pass a reference. Sorting should change the thing being sorted; passing by value in the context of your code would mean that nothing would be able to inspect the sorted result.
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? Something like the {Algorithm.h :: transform function }?
Yes, using 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.
std::transform(myv1.begin(), myv1.end(), myv1.begin(),
std::bind1st(std::multiplies<T>(), 3));
For the placeholders;
#include <functional>
If you can use a valarray instead of a vector, it has builtin operators for doing a scalar multiplication.
v *= 3;
If you have to use a vector, you can indeed use transform to do the job:
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)
Modern C++ solution for your question.
#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:
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 e.g. wanted to multiply with some predetermined scalar), goes into the bracket as a reference.
If you had to store the results in a new vector, then you could use the std::transform() from the <algorithm> header:
#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,
take the values (elements) of vec_input starting from the beginning (vec_input.begin()) to the end (vec_input.begin()),
essentially, with the first two arguments, you specify a range of elements ([beginning, end)) to transform,
range
pass each element to the last argument, lambda expression,
take the output of lambda expression and put it in the vec_output starting from the beginning (vec_output.begin()).
the third argument is to specify the beginning of the destination vector.
The lambda expression
captures the value of scale factor ([&scale]) from outside by reference,
takes as its input a vector element of type double (passed to it by std::transform())
in the body of the function, it returns the final result,
which, as I mentioned above, will be consequently stored in the vec_input.
Final note: Although unnecessary, you could pass lambda expression per below:
[&scale](double element) -> double { return element *= scale; }
It explicitly states that the output of the lambda expression is a double. However, we can omit that, because the compiler, in this case, can deduce the return type by itself.
I know this not STL as you want, but it is something you can adapt as different needs arise.
Below is a template you can use to calculate; 'func' would be the function you want to do: multiply, add, and so on; 'parm' is the second parameter to the 'func'. You can easily extend this to take different func's with more parms of varied types.
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.