排序n维点并跟踪原始索引

Question

I have a set of n- dimension point store in vector< vector<double> >

ex A[0][1].............[N], and A[0][0] = X, A[0][1] = Y, A[0][2] = Z

and I want to sort the vector of all of the dimension

ex sort X, Y, Z ,.........N in ascending order


ex              A[0] = (1,5,3), A[1] = (3,2,1) A[2] = (2,8,4) after sorting
index:            0               1             2
                A[0] = (1,5,3), A[1] = (2,8,4) A[2] = (3,2,1)
original index :  0               2             1

I find that sort(vector.begin(), vector.end()) can sort it but how can I record the original index with a additional vector?

Is there a algorithm or C++ feature can solve it?

Thx in advance.

Answer 1

You need to somehow keep the information about the index. I can see two ways of doing this :

1-Since you represent your point by a vector, you can had another dimension that will represent the orginal index :

//Adding index info, inportant that your index be the last dimension , otherwise the sort is incorrect for(auto point = vector.begin(),unsigned int index=0;point != vector.end(); ++point,++index){ point->push_back(index) };

Then sort the same way you are doing now :

sort(vector.begin(), vector.end())

and you access the original index with A[0][n]

The cool thing about that is that it allow you to keep track of the index in a very convenient way, but you need to be able to modify the representation of your point.

2- The other way is to have an external table of indices and sort that using custom comp. operator :

you start by creating a vector of indices

std::vector indices(vector.size());

for(unsigned int index =0;index != indicies.size(); ++point){ indicies[index] = index ; };

//and sort...

std::sort(indices.begin(),indices.end(),[&](unsigned int i,unsigned int j) { return vector[i] < vector[j]})

now you need an additional level of indirection to go trough your points in the sorted order : A[indices[0]],A[indices[1]],... So the original position of A[indices[x]] is simply x

The main to remember between those two ways of doing it is that in the first you move your data around and not in the second , depending on what you are doing on your points, one might be better that the order