I am relative new at C++ and I have little problem. I have vector and in that vector are vectors with 3 integers.
Inner vector represents like one person. 3 integers inside that inner vector represents distance from start, velocity and original index (because in input integers aren't sorted and in output I need to print original index not index in this sorted vector).
Now I have given some points representing distance from start and I need to find which person will be first at that point so I have been thinking that my first step would be that I would find closest person to the given point so basically I need to find lower_bound/upper_bound.
How can I use lower_bound if I want to find the lower_bound of first item in inner vectors? Or should I use struct/class instead of inner vectors?
You would use the version of std::lower_bound which takes a custom comparator (the versions marked "(2)" at the link); and you would write a comparator of vectors which compares vectors by their first item (or whatever other way you like).
Howerver:
As #doctorlove points out, std::lower_bound doesn't compare the vectors to each other, it compares them to a given value (be it a vector or a scalar). So it's possible you actually want to do something else.
It's usually not a good idea to keep fixed-length sequences of elements in std::vector's. Have you considered std::array?
It's very likely that your "vectors with 3 integers" actually stand for something else, e.g. points in a 3-dimensional geometric space; in which case, yes, they should be in some sort of class.
I am not sure that your inner things should be std::vector-s of 3 elements.
I believe that they should std::array-s of 3 elements (because you know that the size is 3 and won't change).
So you probably want to have
typedef std::array<double,3> element_ty;
then use std::vector<element_ty> and for the rest (your lower_bound point) do like in einpoklum's answer.
BTW, you probably want to use std::min_element with an explicit compare.
Maybe you want something like:
std::vector<element_ty> vec;
auto minit =
std::min_element(vec.begin(), vec.end(),
[](const element_ty& x, const element_ty&y) {
return x[0] < y[0]));
Related
I have a vector < vector <int> > like so:
v = {{1,2,3}, {4,2,1}, {3,1,1}....}}
All v's elements like v[0], v[1], v[2]... have the same size. There may be duplicate elements.
What I am trying to do is to find and delete vectors (like v[2]) that are "majorized" by another vector (like v[1]), i.e. all elements of v[1] are greater than/equal to the respective elements(in order of indices) in v[2].
A naive way of doing this would be to loop thorough v and compare each vector with another vector and further compare each element with another vector's element.
But I feel there must a better way to do this without getting O(n^3) in the number of elements of all the vectors in v.
If multiple vectors are equal, I need only one of them (i.e delete all except one). A random choice would be sufficient.
Any thoughts or ideas are appreciated!
This is called the maxima of a point set. For two and three dimensions, this can be solved in O(n log n) time. For more than three dimensions, this can be solved in O(n(log n)^(d − 3) log log n) time. For random points, a linear expected time algorithm is available.
Suppose I have a vector A = {1,1,1,0,0};
Is there any inbuilt function in vector header to find all the indices of vector where A is repeated?
suppose for 1, returning, { 0,1,2 }
for 0, {3,4}
If not, is there any time efficient way to do so?
If not, is there any time efficient way to do so?
Sort your vector and use std::equal_range to find iterators range, then convert them to indexes. If you cannot sort the vector, create vector of indexes, sort it and copy the range from it to result.
If the vector is or can be sorted, then you can use std::equal_range.
http://en.cppreference.com/w/cpp/algorithm/equal_range
I have a matrix consisting of a vector of which each element representing the rows is composed of a vector representing the columns of the matrix. I would like to sort the rows according to the 1st column.
Each element inside this matrix is a double, although the first column contains a number that serves as an identifier (but is not unique).
My goal is to have something like the aggregate functions available in SQL, such as count() and sum() when I group by the first column.
For instance, if I have:
ID VALUE
1 10
2 20
1 30
2 40
3 60
I would like to get:
ID COUNT MEAN
1 2 20
2 2 30
3 1 60
However, I am stuck in the very first step: how do I sort the rows according to the value of the first element of each row?
I found a clue on this topic, and changed adapted the comparator to:
bool compareFunction (double i,double j)
{
return (i<j);
}
But the compiler was not very happy about that (making a reference to the stl_algo.h file):
error: cannot convert 'std::vector<double>' to 'double' in argument passing
I was therefore wondering if there is a way to sort such a vector of vectors when it contains doubles.
Answer (imho): use a different datastructure. What you are trying to do is setup a multimap. Oh hey look:
http://www.cplusplus.com/reference/map/multimap/
stl::multimap - how do i get groups of data?
It'll be faster for large numbers of elements. And is actually a map rather than a vector of vector of double.
Either that, or skip the sorting all together, and count by key using std::map, std::unordered_map, or (if you know the number of keys and/or the keys are offset by 1 with no breaks) std::vector.
To expand, sorting your list to get means will be slow. Sorting (using std::sort) is O(nlogn), and will be O(nlogn) every time you compute this mean. And it is an unessisary step: your stuff is grouped by key reguardless of order. std::map and std::multimap will "sort as you go" which will be just a little faster than sorting every time, but you won't have to sort the whole thing to get the list. Then you can just iterate the multimap to get the mean, O(n) each mean calculation. (It is still O(nlg(n)) to add all the elements to the multimap)
But if you know the key output is going to be 1,2,3...n-1,n, than sorting is a complete waste of time. Just make a counter for each key (since you know what the keys can be) and add to the key while iterating the array.
BUT WAIT THERE IS MORE
If the keys are actually setup the way you are thinking, than the best way from the get go is to forget the table structure, and make build it like this:
Index VALUE
0 10,30
1 20,40
2 60
Count is now constant time for each row. Mean for each row is O(n). Getting a list is constant time for each row. EVERYBODY WINS.
You need to create a comparator function comparing vector<double>:
struct VecComp {
bool operator()(const vector<double>& _a, const vector<double>& _b) {
//compare first elements
}
}
Then you can use std::sort on your structure with the new comparator function:
std::sort(myMat.begin(), myMat.end(), VecComp());
If you are using c++11 features you can also utilize lambda functions here:
std::sort(myMat.begin(), myMat.end(), [](const vector<double>& a, const vector<double>& b) {
//compare the first elements
}
);
You need to write your own comparator functor to pass into your vector declaration:
struct comp {
bool operator() (const std::vector<double>& i,
const std::vector<double>& j) {
return i[0] < j[0];
}
Have you tried just this?:
std::sort(vecOfVecs.begin(), vecOfVecs.end());
That should work as std::vector has operator< which provides lexicographical sorting, which is (a little more specific than) what you want.
I have an std::vector of floats that I want to not contain duplicates but the math that populates the vector isn't 100% precise. The vector has values that differ by a few hundredths but should be treated as the same point. For example here's some values in one of them:
...
X: -43.094505
X: -43.094501
X: -43.094498
...
What would be the best/most efficient way to remove duplicates from a vector like this.
First sort your vector using std::sort. Then use std::unique with a custom predicate to remove the duplicates.
std::unique(v.begin(), v.end(),
[](double l, double r) { return std::abs(l - r) < 0.01; });
// treats any numbers that differ by less than 0.01 as equal
Live demo
Sorting is always a good first step. Use std::sort().
Remove not sufficiently unique elements: std::unique().
Last step, call resize() and maybe also shrink_to_fit().
If you want to preserve the order, do the previous 3 steps on a copy (omit shrinking though).
Then use std::remove_if with a lambda, checking for existence of the element in the copy (binary search) (don't forget to remove it if found), and only retain elements if found in the copy.
I say std::sort() it, then go through it one by one and remove the values within certain margin.
You can have a separate write iterator to the same vector and one resize operation at the end - instead of calling erase() for each removed element or having another destination copy for increased performance and smaller memory usage.
If your vector cannot contain duplicates, it may be more appropriate to use an std::set. You can then use a custom comparison object to consider small changes as being inconsequential.
Hi you could comprare like this
bool isAlmostEquals(const double &f1, const double &f2)
{
double allowedDif = xxxx;
return (abs(f1 - f2) <= allowedDif);
}
but it depends of your compare range and the double precision is not on your side
if your vector is sorted you could use std::unique with the function as predicate
I would do the following:
Create a set<double>
go through your vector in a loop or using a functor
Round each element and insert into the set
Then you can swap your vector with an empty vector
Copy all elements from the set to the empty vector
The complexity of this approach will be n * log(n) but it's simpler and can be done in a few lines of code. The memory consumption will double from just storing the vector. In addition set consumes slightly more memory per each element than vector. However, you will destroy it after using.
std::vector<double> v;
v.push_back(-43.094505);
v.push_back(-43.094501);
v.push_back(-43.094498);
v.push_back(-45.093435);
std::set<double> s;
std::vector<double>::const_iterator it = v.begin();
for(;it != v.end(); ++it)
s.insert(floor(*it));
v.swap(std::vector<double>());
v.resize(s.size());
std::copy(s.begin(), s.end(), v.begin());
The problem with most answers so far is that you have an unusual "equality". If A and B are similar but not identical, you want to treat them as equal. Basically, A and A+epsilon still compare as equal, but A+2*epsilon does not (for some unspecified epsilon). Or, depending on your algorithm, A*(1+epsilon) does and A*(1+2*epsilon) does not.
That does mean that A+epsilon compares equal to A+2*epsilon. Thus A = B and B = C does not imply A = C. This breaks common assumptions in <algorithm>.
You can still sort the values, that is a sane thing to do. But you have to consider what to do with a long range of similar values in the result. If the range is long enough, the difference between the first and last can still be large. There's no simple answer.
If I have a map with keys as strings. How can I know which string is in which place in the map? For example, 'cats' is the first entry and 'dogs' are the second entry. How will I know that dogs are the second entry in the map? Should I go through the map and search/check the order or is there any easier way?
You can find() the element, and then compute the distance between the element and the map's begin(). Since this is a map, the complexity of distance will be O(n) with the number of elements in the map. I doubt you'd see much performance improvement versus simply looping the map and counting, but I'd prefer to not write such hand-written loops.
But this smells like an XY Problem. Why would you need the position of an element in a map?
You can't do that anyway better, and here's two reasons for that:
a) std::map has a tree inside that, so initially leaves of tree don't have linear order, only a partial order (of course you can compare two keys, but you can't do that for a general binary tree - that's the reason for complication).
b) std::map has bidirectional iterator, so you can't have effective iterator arithmetics. You could do that like this:
typedef map<string, int> MyMapT;
MyMapT data;
// ...
auto it = data.find("dog");
if (it == data.end()) {
// don't have such a key
}
auto position = distance(data.begin(), data.end());
Unfortunately, that's not the best way to do that, because distance function works in O(n) time for bidirectional iterators. Again, that's because of tree structure inside the std::map.