I have a vector of pairs of int, and I want to add all the first elements of each pair. I have written following code
#include <iostream>
#include <numeric>
#include <vector>
#include <utility>
#define PII pair<int,int>
using namespace std;
int main() {
vector<pair<int,int>> v;
v.push_back(PII(1,2));
v.push_back(PII(3,4));
v.push_back(PII(5,6));
cout<<accumulate(v.begin(),v.end(),0,[](auto &a, auto &b){return a.first+b.first;});
return 0;
}
Here it is giving errors http://ideone.com/Kf2i7d.
The required answer is 1+3+5 = 9. I can't understand the errors it is giving.
In this call of the algorithm
cout<<accumulate(v.begin(),v.end(),0,[](auto &a, auto &b){return a.first+b.first;});
its third parameter is initialized by 0 and hence has deduced type int.
It corresponds to the accumulator of the algorithm that accumulates the values that are suplied by the second parameter of the lambda expression.
So you have to write
cout<<accumulate(v.begin(),v.end(),0,[](auto &a, auto &b){return a + b.first;});
As for me I would initialize it with integer literal of type long long int. For example
cout<<accumulate(v.begin(),v.end(),0ll,[](auto &a, auto &b){return a +b.first;});
std::accumulate iterates over each element and calls the supplied function with the current element and the current value of the accumulator.
The accumulator has type int, not pair<int, int> so you need to fix your lambda function to accept the right argument types.
Related
I have a vector of complex numbers and I need to sort them by their argument. Sadly, the numbers have type complex<int>, so function arg(c) returns an integer in range [-3,3] instead of a float and the numbers can't be sorted properly.
I've tried also
typedef complex<int> ci;
typedef complex<double> cd;
vector<ci> a;
sort(a.begin(), a.end(), [](ci v, ci u) { return arg(cd(v)) < arg(cd(u)); });
but it does not work either (compilation error: no matching function for call to ‘std::complex<double>::complex(ci&)).
Can I sort these numbers without changing their type?
You get the error because there is no converting constructor from std::complex<int> to std::complex<double> you have to construct the std::complex<double> by passing real and imaginary parts to the constructor:
#include <vector>
#include <complex>
#include <algorithm>
int main() {
std::vector<std::complex<int>> a;
std::sort(a.begin(), a.end(), [](const auto& v,const auto& u) {
return std::arg(std::complex<double>(v.real(),v.imag())) < std::arg(std::complex<double>(u.real(),u.imag()));
});
}
Note that you can also use atan2 directly without constructing the std::complex<double> as mentioned by user17732522.
Last but not least, reconsider if you really need int with std::complex. From cppreference:
T - the type of the real and imaginary components. The behavior is unspecified (and may fail to compile) if T is not float, double, or long double and undefined if T is not NumericType.
Basically this means that you need to check the implementation you are using whether it supports std::complex<int> at all.
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
I want to find both the minimum and maximum of a long vector. The following code works, but I need to traverse the vector twice.
I could use an old fashioned for loop, but I wonder if there is an elegant (c++11, std) way of doing it.
#include <vector>
#include <algorithm>
using namespace std;
int main(int argc, char** argv) {
vector<double> C;
// code to insert values in C not shown here
const double cLower = *min_element(C.begin(), C.end());
const double cUpper = *max_element(C.begin(), C.end());
// code using cLower and cUpper
}
You mean like std::minmax_element?
auto mm = std::minmax_element(std::begin(c), std::end(c));
const double cLower = *mm.first;
const double cUpper = *mm.second;
Note this assumes the range is not empty (as does your existing solution), else the iterator dereferences are Undefined Behaviour.
Also note this is not quite the same as your solution, as max_element returns the first largest element, and minmax_element returns the last largest element. E.g.
1 2 1 2
^ ^
A B
Where A is found by your solution, and B is found by mine. (This is for reasons of stability; Alex Stepanov got the definition of max wrong in the original STL.)
I wrote a function for vector comparison.
#include <vector>
#include <iostream>
#include <algorithm>
using std::vector;
using std::cout;
bool mycomp(const vector<int>& vi_a, const vector<int>& vi_b){
for(auto x:vi_a) cout << x;
cout << '\n';
return true;
}
int main(){
vector<int> vi1{2,9,8};
vector<int> vi2{3,5,6};
vector<int> vi = min(vi1, vi2, mycomp);
for(auto x:vi) cout << x;
cout << '\n';
}
Output:
356
356
What's weird is that the output is 356, instead of 298. It seems that the two vectors are switched when calling mycomp.
p.s. I'd rather not use lambda here, because mycomp contains more than one line of code, which is more readable this way.
If you take a look at the cppreference documentation for std::min() you will see in the possible implementation that the second element is used as the first element in the comparator.
template<class T, class Compare>
const T& min(const T& a, const T& b, Compare comp)
{
return (comp(b, a)) ? b : a;
}
And when you call min() with the arguments vi2 and vi1, the second argument is passed to the comparator (which in your case is vi2), upon which the comparator returns true (implying that vi2 is the minimum of the two). While doing this the comparator also prints out 356 because that is the first element passed to the comparator according to the possible implementation above.
After that happens you take the presumably smaller of the two vectors, which is the first one passed to the comparator according to the possible implementation above (which is vi2). And you print that out. Therefore you get 356 again.
Note that the reason b is passed as the first argument to std::min is because in the case where the two compare equal the algorithm is required to return a. At the same time it is required to call the comparator only once. Combining these two requirements gives the possible implementation above.
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.