How to send a portion of a QVector to a function?
QVector<int> a;
a.append(1);
a.append(2);
a.append(3);
a.append(4);
a.append(5);
Some printing function should print "2 3 4" taking the subset of the vector as an argument.
In R this would be possible using a[2:4].
Is this at all possible?
Note: In the std::vector, it is advised to use the insert function to create a new variable. This is a different insert though than QVector has, and thus I cannot find a recommended method.
I must write at least 30 characters so I can tell you, you should try:
a.mid(1,3);
You could always write a function to do this operation for you, e.g.
QVector<int> sub_vector(const QVector<int>& vec, size_t from, size_t to)
{
QVector<int> subvec;
for (size_t i = from; i <= to; ++i)
subvec.append(vec[i]);
return subvec;
}
Yes it is possible, but you must pass a pair of iterators (begin and end of the range you want, you can use std::pair to pass only one argument or use a clearer method that take two QVector::iterator arguments and that way it's clearer that you meant that function to take a range) or if it's simpler to you (or the elements you want are not in continuous order in original QVector) construct another QVector that contains only the selected elements (kind of the solution proposed by john).
Related
I am trying to store a vector of objects and sort them by a string member possessed by each object. It doesn't need to be sorted alphabetically, it only needs to group every object with an identical string together in the vector.
IE reading through the vector and outputting the strings from beginning to end should return something like:
string_bulletSprite
string_bulletSprite
string_bulletSprite
string_playerSprite
string_enemySprite
string_enemySprite
But should NEVER return something like:
string_bulletSprite
string_playerSprite
string_bulletSprite
[etc.]
Currently I am using std:sort and a custom comparison function:
std::vector<GameObject*> worldVector;
[...]
std::sort(worldVector.begin(), worldVector.end(), compString);
And the comparison function used in the std::sort looks like this:
bool compString(GameObject* a, GameObject* b)
{
return a->getSpriteNameAndPath() < b->getSpriteNameAndPath();
}
getSpriteNameAndPath() is a simple accessor which returns a normal string.
This seems to work fine. I've stress tested this a fair bit and it seems to always group things together the way I wanted.
My question is, is this the ideal or most logical/efficient way of accomplishing the stated goal? I get the impression Sort isn't quite meant to be used this way and I'm wondering if there's a better way to do this if all I want to do is group but don't care about doing so in alphabetic order.
Or is this fine?
If you have lots of equivalent elements in your range, then std::sort is less efficient than manually sorting the elements.
You can do this by shifting the minimum elements to the beginning of the range, and then repeating this process on the remaining non-minimum elements
// given some range v
auto b = std::begin(v); // keeps track of remaining elements
while (b != std::end(v)) // while there's elements to be arranged
{
auto min = *std::min_element(b, std::end(v)); // find the minimum
// move elements matching that to the front
// and simultaneously update the remaining range
b = std::partition(b, std::end(v),
[=](auto const & i) {
return i == min;
});
}
Of course, a custom comparator can be passed to min_element, and the lambda in partition can be modified if equivalence is defined some other way.
Note that if you have very few equivalent elements, this method is much less efficient than using std::sort.
Here's a demo with a range of ints.
I hope I understood your question correctly, if so, I will give you a little example of std::map which is great for grouping things by keys, which will most probably be a std::string.
Please take a look:
class Sprite
{
public:
Sprite(/* args */)
{
}
~Sprite()
{
}
};
int main(int argc, char ** argv){
std::map <std::string, std::map<std::string, Sprite>> sprites;
std::map <std::string, Sprite> spaceships;
spaceships.insert(std::make_pair("executor", Sprite()));
spaceships.insert(std::make_pair("millennium Falcon", Sprite()));
spaceships.insert(std::make_pair("death star", Sprite()));
sprites.insert(std::make_pair("spaceships",spaceships));
std::cout << sprites["spaceships"]["executor"].~member_variable_or_function~() << std::endl;
return 0;
}
Seems like Functor or Lambda is the way to go for this particular program, but I realized some time after posting that I could just create an ID for the images and sort those instead of strings. Thanks for the help though, everyone!
In my C++ code,
vector <string> strVector = GetStringVector();
vector <int> intVector = GetIntVector();
So I combined these two vectors into a single one,
void combineVectors(vector<string>& strVector, vector <int>& intVector, vector < pair <string, int>>& pairVector)
{
for (int i = 0; i < strVector.size() || i < intVector.size(); ++i )
{
pairVector.push_back(pair<string, int> (strVector.at(i), intVector.at(i)));
}
}
Now this function is called like this,
vector <string> strVector = GetStringVector();
vector <int> intVector = GetIntVector();
vector < pair <string, int>> pairVector
combineVectors(strVector, intVector, pairVector);
//rest of the implementation
The combineVectors function uses a loop to add the elements of other 2 vectors to the vector pair. I doubt this is a efficient way as this function gets called hundrands of times passing different data. This might cause a performance issue because everytime it goes through the loop.
My goal is to copy both the vectors in "one go" to the vector pair. i.e., without using a loop. Am not sure whether that's even possible.
Is there a better way of achieving this without compromising the performance?
You have clarified that the arrays will always be of equal size. That's a prerequisite condition.
So, your situation is as follows. You have vector A over here, and vector B over there. You have no guarantees whether the actual memory that vector A uses and the actual memory that vector B uses are next to each other. They could be anywhere.
Now you're combining the two vectors into a third vector, C. Again, no guarantees where vector C's memory is.
So, you have really very little to work with, in terms of optimizations. You have no additional guarantees whatsoever. This is pretty much fundamental: you have two chunks of bytes, and those two chunks need to be copied somewhere else. That's it. That's what has to be done, that's what it all comes down to, and there is no other way to get it done, other than doing exactly that.
But there is one thing that can be done to make things a little bit faster. A vector will typically allocate memory for its values in incremental steps, reserving some extra space, initially, and as values get added to the vector, one by one, and eventually reach the vector's reserved size, the vector has to now grab a new larger block of memory, copy everything in the vector to the larger memory block, then delete the older block, and only then add the next value to the vector. Then the cycle begins again.
But you know, in advance, how many values you are about to add to the vector, so you simply instruct the vector to reserve() enough size in advance, so it doesn't have to repeatedly grow itself, as you add values to it. Before your existing for loop, simply:
pairVector.reserve(pairVector.size()+strVector.size());
Now, the for loop will proceed and insert new values into pairVector which is guaranteed to have enough space.
A couple of other things are possible. Since you have stated that both vectors will always have the same size, you only need to check the size of one of them:
for (int i = 0; i < strVector.size(); ++i )
Next step: at() performs bounds checking. This loop ensures that i will never be out of bounds, so at()'s bound checking is also some overhead you can get rid of safely:
pairVector.push_back(pair<string, int> (strVector[i], intVector[i]));
Next: with a modern C++ compiler, the compiler should be able to optimize away, automatically, several redundant temporaries, and temporary copies here. It's possible you may need to help the compiler, a little bit, and use emplace_back() instead of push_back() (assuming C++11, or later):
pairVector.emplace_back(strVector[i], intVector[i]);
Going back to the loop condition, strVector.size() gets evaluated on each iteration of the loop. It's very likely that a modern C++ compiler will optimize it away, but just in case you can also help your compiler check the vector's size() only once:
int i=strVector.size();
for (int i = 0; i < n; ++i )
This is really a stretch, but it might eke out a few extra quantums of execution time. And that pretty much all obvious optimizations here. Realistically, the most to be gained here is by using reserve(). The other optimizations might help things a little bit more, but it all boils down to moving a certain number of bytes from one area in memory to another area. There aren't really special ways of doing that, that's faster than other ways.
We can use std:generate() to achieve this:
#include <bits/stdc++.h>
using namespace std;
vector <string> strVector{ "hello", "world" };
vector <int> intVector{ 2, 3 };
pair<string, int> f()
{
static int i = -1;
++i;
return make_pair(strVector[i], intVector[i]);
}
int main() {
int min_Size = min(strVector.size(), intVector.size());
vector< pair<string,int> > pairVector(min_Size);
generate(pairVector.begin(), pairVector.end(), f);
for( int i = 0 ; i < 2 ; i++ )
cout << pairVector[i].first <<" " << pairVector[i].second << endl;
}
I'll try and summarize what you want with some possible answers depending on your situation. You say you want a new vector that is essentially a zipped version of two other vectors which contain two heterogeneous types. Where you can access the two types as some sort of pair?
If you want to make this more efficient, you need to think about what you are using the new vector for? I can see three scenarios with what you are doing.
The new vector is a copy of your data so you can do stuff with it without affecting the original vectors. (ei you still need the original two vectors)
The new vector is now the storage mechanism for your data. (ei you
no longer need the original two vectors)
You are simply coupling the vectors together to make use and representation easier. (ei where they are stored doesn't actually matter)
1) Not much you can do aside from copying the data into your new vector. Explained more in Sam Varshavchik's answer.
3) You do something like Shakil's answer or here or some type of customized iterator.
2) Here you make some optimisations here where you do zero coping of the data with the use of a wrapper class. Note: A wrapper class works if you don't need to use the actual std::vector < std::pair > class. You can make a class where you move the data into it and create access operators for it. If you can do this, it also allows you to decompose the wrapper back into the original two vectors without copying. Something like this might suffice.
class StringIntContainer {
public:
StringIntContaint(std::vector<std::string>& _string_vec, std::vector<int>& _int_vec)
: string_vec_(std::move(_string_vec)), int_vec_(std::move(_int_vec))
{
assert(string_vec_.size() == int_vec_.size());
}
std::pair<std::string, int> operator[] (std::size_t _i) const
{
return std::make_pair(string_vec_[_i], int_vec_[_i]);
}
/* You may want methods that return reference to data so you can edit it*/
std::pair<std::vector<std::string>, std::vector<int>> Decompose()
{
return std::make_pair(std::move(string_vec_), std::move(int_vec_[_i])));
}
private:
std::vector<std::string> _string_vec_;
std::vector<int> int_vec_;
};
I would like to create a vector (arma::uvec) of integers - I do not ex ante know the size of the vector. I could not find approptiate function in Armadillo documentation, but moreover I was not successfull with creating the vector by a loop. I think the issue is initializing the vector or in keeping track of its length.
arma::uvec foo(arma::vec x){
arma::uvec vect;
int nn=x.size();
vect(0)=1;
int ind=0;
for (int i=0; i<nn; i++){
if ((x(i)>0)){
ind=ind+1;
vect(ind)=i;
}
}
return vect;
}
The error message is: Error: Mat::operator(): index out of bounds.
I would not want to assign 1 to the first element of the vector, but could live with that if necessary.
PS: I would really like to know how to obtain the vector of unknown length by appending, so that I could use it even in more general cases.
Repeatedly appending elements to a vector is a really bad idea from a performance point of view, as it can cause repeated memory reallocations and copies.
There are two main solutions to that.
Set the size of the vector to the theoretical maximum length of your operation (nn in this case), and then use a loop to set some of the values in the vector. You will need to keep a separate counter for the number of set elements in the vector so far. After the loop, take a subvector of the vector, using the .head() function. The advantage here is that there will be only one copy.
An alternative solution is to use two loops, to reduce memory usage. In the first loop work out the final length of the vector. Then set the size of the vector to the final length. In the second loop set the elements in the vector. Obviously using two loops is less efficient than one loop, but it's likely that this is still going to be much faster than appending.
If you still want to be a lazy coder and inefficiently append elements, use the .insert_rows() function.
As a sidenote, your foo(arma::vec x) is already making an unnecessary copy the input vector. Arguments in C++ are by default passed by value, which basically means C++ will make a copy of x before running your function. To avoid this unnecessary copy, change your function to foo(const arma::vec& x), which means take a constant reference to x. The & is critical here.
In addition to mtall's answer, which i agree with,
for a case in which performance wasn't needed i used this:
void uvec_push(arma::uvec & v, unsigned int value) {
arma::uvec av(1);
av.at(0) = value;
v.insert_rows(v.n_rows, av.row(0));
}
Suppose Foo is any class.
Foo f[5];
std::vector<Foo*> v;
I can insert the elements into vector of pointers using a for loop statement:
for (size_t i = 0; i < 5; i++)
v.push_back(&f[i]);
Is it possible to insert them using std::vector::insert() function and why not? I have tried several times it failed something like this:
v.insert(v.end(), &f[0], &f[5]); // error
If you mean, with a single call to insert, then no - that can copy a range, performing type conversions if needed, but can't apply arbitrary transformations like taking the address of each element.
You could use std::transform:
std::transform(std::begin(f), std::end(f),
std::back_inserter(v),
[](Foo & f) {return &f;});
although that's probably less clear than a simple loop, especially if you use new-style syntax
for (Foo & foo : f) {
v.push_back(&foo);
}
Yes you can use insert also. But there are few differences between these two operations:-
push_back puts a new element at the end of the vector and insert allows you to select position. This impacts the performance. insert forces to move all elements after the selected position of a new element. You simply have to make a place for it. This is why insert might often be less efficient than push_back.
Since there is no .resize() member function in C++ std::map I was wondering, how one can get a std::map with at most n elements.
The obvious solution is to create a loop from 0 to n and use the nth iterator as the first parameter for std::erase().
I was wondering if there is any solution that does not need the loop (at least not in my user code) and is more "the STL way to go".
You can use std::advance( iter, numberofsteps ) for that.
Universal solution for almost any container, such as std::list, std::map, boost::multi_index. You must check the size of your map only.
template<class It>
It myadvance(It it, size_t n) {
std::advance(it, n);
return it;
}
template<class Cont>
void resize_container(Cont & cont, size_t n) {
cont.erase(myadvance(cont.begin(), std::min(n, cont.size())),
cont.end());
}
The correct way for this is to use std::advance. But here is a funny (slow) way allowing to 'use resize on map'. More generally, this kind of trick can be used for other things working on vector but not on map.
map<K,V> m; //your map
vector< pair<K,V> > v(m.begin(), m.end());
v.resize(n);
m = map<K,V>(v.begin(),v.end());
Why would you want to resize a map?
The elements in a map aren't stored in any order - the first 'n' doesn't really mean anything
edit:
Interestingly std::map does have an order, not sure how useful this concept is.
Are the entries in the same sort order as the keys?
What does that mean? If you have Names keyed by SSN does that mean the names are stored in SSN numeric order?
A std::map is not a list. There are no "first n" elements.
BTW: Iterators become invalid if the container is changed.
If you really need a smaller map you could iterate though it and add all elements up to the n-th into a new map.