I have a unsymmetrical vector in 2D.
vector< vector<int> > Test
where Test =
2 4 6 5 7
6 5 7 9 10
5 9 10
9 10
I am reading the row 1 and if any element of this is present in other rows then delete it. for eaxmple.. After reading row 1, i have to remove 6, 5, and 7 from other rows.
However, It is not working
Here is the code i am trying
Test[i].erase(Test[i].begin()+j);
where i = row and j is col.
My code is :
for (i =0; i < Test.size();i++)
{
for (j=0; j < Test[i].size();j++)
{
// removed repeated element
if (i >0)
{
Test[i].erase(Test[i].begin() +j);
}
}
}
What exactly do you think is Test[i].begin()+j
? Is ist a set of elements you want to erase? I don't think so. It should be just an iterator, that points to a single element , but you want to delete all elements , that are already in your datastructure.
If I understood, what you want to do, try:
for(int j = 0; j < Test.size(); j++){ //iterate over other rows
if(j == i)
continue;
for(int k = 0; k < Test[j].size(); k++){ //iterate over elements of the rows
int elementToRemove = (Test[j])[k];
vector<int>::iterator it = Test[i].begin();
while (it != Test[i].end()) { //iterate over row i
if((*it) == elementToRemove){ //erase the element if it matches the actual
it = Test[i].erase(it);
}else{
it++;
}
}
}
}
You could execute the code for every possible i. Maybe start form i = 0 to n
. If I refer to your code, that you added put the code above in between your
for (i =0; i < Test.size();i++){
//my code here...
}
Edit: Used iterator now to delete. The first version was not correct.
Edit2: Changed the index of the first loop and added continue statement.
Maybe it is nor very nice but it works
#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
int main()
{
std::vector<std::vector<int>> v =
{
{ 2, 4, 6, 5, 7 },
{ 6, 5, 7, 9, 10 },
{ 5, 9, 10 },
{ 9, 10 }
};
for ( const auto &row : v )
{
for ( int x : row ) std::cout << x << ' ';
std::cout << std::endl;
}
std::cout << std::endl;
if ( !v.empty() )
{
for ( auto it = std::next( v.begin() ); it != v.end(); ++it )
{
auto is_present = [&]( int x )
{
return std::find_if( v.begin(), it,
[x]( const std::vector<int> &v1 )
{
return std::find( v1.begin(), v1.end(), x ) != v1.end();
} ) != it;
};
it->erase( std::remove_if( it->begin(), it->end(), is_present ),
it->end() );
}
}
for ( const auto &row : v )
{
for ( int x : row ) std::cout << x << ' ';
std::cout << std::endl;
}
return 0;
}
The output is
2 4 6 5 7
6 5 7 9 10
5 9 10
9 10
2 4 6 5 7
9 10
You can place the values encountered in each row in a set, and then query every element in a new row for existence in that set. Such a function would look like this :
void RemoveRowDuplicates(vector<vector<int>> &v)
{
std::set<int> vals;
for(auto &vec : v)
{
vec.erase(remove_if(vec.begin(), vec.end(), [&](int k){
return vals.find(k) != vals.end();
}), vec.end());
vals.insert(vec.begin(), vec.end());
}
}
This works for me:
int i = 0;
for ( int j = i+1; j < Test.size(); ++j )
{
for ( int k = 0; k < Test[i].size(); ++k )
{
std::vector<int>::iterator iter = Test[j].begin();
std::vector<int>::iterator end = Test[j].end();
for ( ; iter != end; )
{
if ( *iter == Test[i][k] )
{
iter = Test[j].erase(iter);
}
else
{
++iter;
}
}
}
}
Consider the following 2D vector
myVector=
1 2 3 4 5 -6
6 7 8 -9
8 -1 -2 1 0
Suppose we want to delete the element myVector[row][column] for appropriate row and column indices.
Look at the following code:
void delete_element(vector<int>& temp, col)
{
temp.erase(temp.begin()+col);
}
int main()
{
//Assume that the vector 'myVector' is already present.
cin>>row>>column;
delete_element(myVector[row],column);
}
What we basically do is,we get the row and column of the element to be deleted. Now as this 2D vector is a vector of vectors, we pass the vector(the row containing the element to be deleted) and the column as parameters to a function. Note that the row-vector is passed as a reference ('&' in vector parameter). Now the problem becomes quite as simple as deleting an element from a 1D vector.
Hope this helps!
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