How to get frequency of std:vectors in C++? - c++

I have 5 vectors. I want to check how many times these vectors exist. I used the following code to compare if 2 vectors are equal, but now I have more than 2 vectors. I want to compare all these 5 vectors together and count how many times each vector exists.
How can I do it?
The output should be:
(0,0,1,2,3,0,0,0) = 2 time(s)
(0,0,1,2,3,4,0,0) = 1 time(s)
(0,0,2,4,3,0,0,0) = 1 time(s)
(0,0,6,2,3,5,6,0) = 1 time(s)
Here is my code:
#include <stdio.h>
#include <iostream>
#include <vector>
using namespace std;
void checkVec(vector<int> v){
vector<int> v0;
if(v0 == v){
cout << "Equal\n";
}
else{
cout << "Not Equal\n";
}
}
int main(){
vector<int> v1={0,0,1,2,3,0,0,0};
vector<int> v2={0,0,1,2,3,4,0,0};
vector<int> v3={0,0,2,4,3,0,0,0};
vector<int> v4={0,0,1,2,3,0,0,0};
vector<int> v5={0,0,6,2,3,5,6,0};
checkVec(v1);
return 0;
}

You can use std::map counting the number of occurences of each vector:
#include <map>
#include <vector>
#include <iostream>
using vec = std::vector<int>;
int main(){
vec v1={0,0,1,2,3,0,0,0};
vec v2={0,0,1,2,3,4,0,0};
vec v3={0,0,2,4,3,0,0,0};
vec v4={0,0,1,2,3,0,0,0};
vec v5={0,0,6,2,3,5,6,0};
std::map<vec,std::size_t> counter;
// Initializer list creates copies by default
// But you should not create vX variables anyway.
for(const auto& v: {v1,v2,v3,v4,v5}){
++counter[v];
}
std::cout<<"V1 is present " <<counter[v1]<<" times.\n";
return 0;
}
V1 is present 2 times.

Well, this is a contribution for Quimby answer, but if you know how many vectors you will get at compile time, use std::array to contain all that vector. If you know it at runtime instead, use std::vector as shown below
#include <map>
#include <vector>
#include <iostream>
#include <cstddef>
int main(){
std::vector<std::vector<int>> allVector
{
std::vector<int>{0,0,1,2,3,0,0,0},
std::vector<int>{0,0,1,2,3,4,0,0},
std::vector<int>{0,0,2,4,3,0,0,0},
std::vector<int>{0,0,1,2,3,0,0,0},
std::vector<int>{0,0,6,2,3,5,6,0},
};
std::map<std::vector<int>, std::size_t> counter;
for(const auto& v : allVector)
{
++counter[v];
}
// print out the array and it's frequency
for(const auto& pr : counter)
{
std::cout << '(';
for(std::size_t i {0}; i < pr.first.size(); ++i)
{
std::cout << pr.first[i];
if(i != pr.first.size() - 1)
std::cout << ' ';
}
std::cout << ") = " << pr.second << ", ";
}
return 0;
}

Related

How to limit boost::combine to the minimum of two ranges

Finding the following as the source of a segfault just cost me about 4h of work:
#include <boost/range/combine.hpp>
#include <boost/foreach.hpp>
#include <iostream>
#include <vector>
#include <list>
int main(int, const char*[])
{
std::vector<int> v;
std::list<char> l;
for (int i = 0; i < 5; ++i)
{
v.push_back(i);
l.push_back(static_cast<char>(i) + 'a');
}
v.push_back(5);
int ti;
char tc;
BOOST_FOREACH(boost::tie(ti, tc), boost::combine(v, l))
{
std::cout << '(' << ti << ',' << tc << ')' << '\n';
}
return 0;
}
If you execute this example, you will note that combine happily iterates as long as the longer range has values. I do not see this in the documentation.
Is there a way to limit the iteration to the shorter of the two ranges?

c++ return all of the elements that appear n times in a vector as a vector

Goal: Return all elements in vector A that appear N times and put results in vector B.
Expected Result:
--Begin With---
Vector A=(10,20,30,30,40,50,100,50,20,100,10,10,200,300)
Do some code to return the name of elements that appear in Vector A
when N=3
Result should be Vector B=(10) //because only 10 is in vector A N=3 times.
My Attempt:
I got the counts of all the elements placed into another vector but I don't have the part that can give back all of the elements that appear N times. I'm very flexible with how it can be done if it means a speed increase.
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <iterator> // std::back_inserter
#include <algorithm> // std::copy
int main()
{
std::vector<int> v{ 1, 2, 3, 4, 4, 3, 7, 8, 9, 10 };
std::vector<std::pair<int, int> > shows;
int target1;
int num_items1;
int size = static_cast<int>(v.size());
for(int x=0; x<size; x++)
{
target1 = v[x];
num_items1 = std::count(v.begin(), v.end(), target1);
shows.push_back(std::make_pair(target1, num_items1));
std::cout << "number: " << target1 << " count: " << num_items1 << '\n';
}
}
ACCEPTED SOLUTION TO QUESTION
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <iterator> // std::back_inserter
#include <algorithm> // std::copy
#include <set>
#include <map>
using namespace std;
int main()
{
std::vector<int> v{ 1, 2, 3, 4, 4, 3, 7, 8, 9, 10 };
std::vector<int> shows;
std::map<int, int> freqOfv;
for(const auto& i: v)
{
freqOfv[i]++;
}
std::set<int> s(v.begin(), v.end());
int N = 2; //Can be read from stdin as well...
for ( auto it = s.begin(); it != s.end(); it++ )
{
if(freqOfv[*it] ==N)
{
shows.push_back(*it);
}
}
for (std::vector<int>::const_iterator i = shows.begin(); i != shows.end(); ++i)
{
std::cout << *i << ' ';
}
return 0;
}
As suggested in the comments, std::map will simplify the code:
int main()
{
std::vector<int> v{ 1, 2, 3, 4, 4, 3, 7, 8, 9, 10 };
std::map<int, int> freqOfv;
for(const auto& i: v)
freqOfv[i]++;
int N = 2; //Can be read from stdin as well...
for(const auto& i: freqOfv)
{
if(N == i.second)
std::cout << "The value " << i.first << " occurs " << N << " times." << std::endl;
}
}
This produces the following output:
The value 3 occurs 2 times.
The value 4 occurs 2 times.
Of course, you need #include <map> at the beginning to use maps in your code.

Remove all elements from array greater than n

I'm beginner in programming. Something is giving me trouble to code. Suppose, I've an array.
int Array[] = {3,6,9,5,10,21,3,25,14,12,32,41,3,24,15,26,7,8,11,4};
I want to remove all elements which are greater than 9. How can I do this?
You can do this if you use vector. First initialize vector with your array. Then use remove_if() function. Hope this will help.
#include <algorithm>
#include <vector>
int main()
{
int Array[] = {3,6,9,5,10,21,3,25,14,12,32,41,3,24,15,26,7,8,11,4};
vector<int> V(Array, Array+20);
vector<int> :: iterator it;
it = remove_if(V.begin(), V.end(), bind2nd(greater<int>(), 9));
V.erase (it, V.end()); // This is your required vector if you wish to use vector
}
You cannot remove items from an array, since they are fixed in size.
If you used std::vector, then the solution would look like this:
#include <vector>
#include <algorithm>
#include <iostream>
#include <iterator>
using namespace std;
int main()
{
std::vector<int> Array = {3,6,9,5,10,21,3,25,14,12,32,41,3,24,15,26,7,8,11,4};
Array.erase(remove_if(Array.begin(), Array.end(), [](int n) { return n > 9; }),
Array.end());
copy(Array.begin(), Array.end(), ostream_iterator<int>(cout, " "));
}
Live example: http://ideone.com/UjdJ5h
If you want to stick with your array, but mark the items that are greater than 10, you can use the same algorithm std::remove_if.
#include <algorithm>
#include <iostream>
#include <iterator>
using namespace std;
int main()
{
int Array[] = {3,6,9,5,10,21,3,25,14,12,32,41,3,24,15,26,7,8,11,4};
int *overwrite_start = remove_if(std::begin(Array), std::end(Array), [](int n){ return n>9; });
fill(overwrite_start, std::end(Array), -1);
copy(std::begin(Array), std::end(Array), ostream_iterator<int>(cout, " "));
}
The above will move the "erased" items to the end of the array, and mark them with -1.
Live example: http://ideone.com/7rwaXy
Note the usage in both examples of the STL algorithm functions. The second example with the array uses the same remove_if algorithm function. The remove_if returns the start of the "erased" data, as remove_if doesn't actually remove, but moves the data to the end of the sequence.
i am try swap concept without using vector
int Array[] = {3,6,9,5,10,21,3,25,14,12,32,41,3,24,15,26,7,8,11,4};
int n;
int arr_len = sizeof(Array)/sizeof(int);
void print_array_value() {
int i;
cout << "\n";
for (i = 0; i < arr_len; i++) {
cout << Array[i] << ", ";
}
cout << " : " << arr_len << "\n";
}
void swap_array_value(int start) {
int i;
for ( ; (start+1) < arr_len; start++) {
Array[start] = Array[start+1];
}
}
void remove_array_value() {
int i;
for (i = 0; i < arr_len; i++) {
if (Array[i] > n) {
swap_array_value(i);
arr_len--;
i--;
}
}
}
void main () {
clrscr();
cout << "Enter the N value : ";
cin >> n;
print_array_value();
remove_array_value();
print_array_value();
cout << "Array Length : " << arr_len;
getch();
}

swap array values in c++

I want to shift left array values if my v=4 is in a[n],remove 4 from a[n] and at the end index add 0,how i can do this?
#include <iostream>
using namespace std;
const int n=5;
int main()
{
int a[n]={1,5,4,6,8}, v=4;
int b[n];
cout << "Enter a Value" << endl;
cout<<v<<endl;
for(int i=0; i<n; i++){
cout<<a[i];
}
cout<<endl;
for(int j=0; j<n; j++){
b[j]=a[j];
if(a[j]==v)
b[j]=a[++j];
cout<<b[j];
}
return 0;
}
#include <vector> // needed for vector
#include <algorithm> // needed for find
#include <iostream> // needed for cout, cin
using namespace std;
// Vectors are just like dynamic arrays, you can resize vectors on the fly
vector<int> a { 1,5,4,6,8 }; // Prepare required vector
int v;
cout << "enter value"; // Read from user
cin >> v;
auto itr = find( a.begin(), a.end(), v); // Search entire vector for 'v'
if( itr != a.end() ) // If value entered by user is found in vector
{
a.erase(itr); // Delete the element and shift everything after element
// Toward beginning of vector. This reduces vector size by 1
a.push_back(0); // Add 0 in the end. This increases vector size by 1
}
for( int i : a ) // Iterate through all element of a (i holds element)
cout << i; // Print i
cout << '\n'; // Line end
a few helpful links:
vector , find , iterator , erase , push_back
You could use std::rotate. I suggest that you use std::vector instead of C arrays and take full advantage of the STL algorithms. Nevertheless, below I'm illustrating two versions one with C arrays and one with std::vector:
Version with C array:
#include <iostream>
#include <algorithm>
int main()
{
int const n = 5;
int a[n] = {1,5,4,6,8};
std::cout << "Enter a Value" << std::endl;
int v;
std::cin >> v;
for(auto i : a) std::cout << i<< " ";
std::cout << std::endl;
auto it = std::find(std::begin(a), std::end(a), v);
if(it != std::end(a)) {
std::rotate(it + 1, it, std::end(a));
a[n - 1] = 0;
}
for(auto i : a) std::cout << i<< " ";
std::cout << std::endl;
return 0;
}
Version with vector:
#include <iostream>
#include <vector>
#include <algorithm>
int main()
{
std::vector<int> a{1,5,4,6,8};
std::cout << "Enter a Value" << std::endl;
int v;
std::cin >> v;
for(auto i : a) std::cout << i<< " ";
std::cout << std::endl;
auto it = std::find(std::begin(a), std::end(a), v);
if(it != std::end(a)) {
std::rotate(it + 1, it, std::end(a));
a.back() = 0;
}
for(auto i : a) std::cout << i<< " ";
std::cout << std::endl;
return 0;
}
Here's an example using std::array
#include <array>
#include <algorithm>
// defines our array.
std::array<int, 5> a = {{ 1, 2, 3, 4, 5 }};
// find the position of the element with the value 4.
auto where = std::find(a.begin(), a.end(), 4);
// if it wasn't found, give up
if (where == a.end())
return 0;
// move every element past "where" down one.
std::move(where + 1, a.end(), where);
// fill the very last element of the array with zero
a[ a.size() - 1] = 0;
// loop over our array, printing it to stdout
for (int i : a)
std::cout << i << " ";
std::cout << "\n";
Why would anyone use these awkward algorithms? Well, there are a few reasons. Firstly, they are container-independant. This will work with arrays and vectors and deques, no problem. Secondly, they can be easily used to work with a whole range of elements at once, not just single items, and can copy between containers and so on. They're also type-independant... you acn have an array of strings, or an vector of ints, or other more complex things, and the algorithms will still work just fine.
They're quite powerful, once you've got over their initial user-unfriendliness.
You can always use either std::array or std::vector or whatever without using the standard library algorithms, of course.
std::array<int, 5> a = {{ 1, 2, 3, 4, 5 }};
size_t where = 0;
int to_remove = 4;
// scan through until we find our value.
while (a[where] != to_remove && where < a.size())
where++;
// if we didn't find it, give up
if (where == a.size())
return 0;
// shuffle down the values
for (size_t i = where; i < a.size() - 1; i++)
a[i] = a[i + 1];
// set the last element to zero
a[ a.size() - 1] = 0;
As a final example, you can use memmove (as suggested by BLUEPIXY) to do the shuffling-down operation in one function call:
#include <cstring>
if (where < a.size() - 1)
memmove(&a[where], &a[where + 1], a.size() - where);

String vector's sort compare and delete

In C++ I have
vector < vector <string> > Kblist;
Inside Kblist, there are many clause, and numbers of clauses=kblist.size(); and every clause in side Kblist is a string-type vector, and every word in sentence is split inside Kblist[i].
What is the fastest way to find the sentence in same words like one in "I love you" and the other in "you love i" and delete these two sentence from Kblist, My code might be work run, but I think it is too slow because of to many circulate. So I wonder is there any better solution which is fast like using sort, clause1==clause2 or other approach.
for (int a=0; a<KBlist.size(); a++){
for (int b=a+1; b<KBlist.size(); b++){
int checksize=0;
if (KBlist[a].size()==KBlist[b].size()) {
for (int c=0; c<KBlist[a].size(); c++){
for (int d=0; d<KBlist[b].size(); d++){
if (KBlist[a][b]==KBlist[c][d]&&KBlist[a][b+1]==KBlist[c][d]) {
checksize=checksize+1;
break;
}
}
}
if (checksize==c.size()) {
inset=1;
break;
}
}
}
}
}while (duplicate==0);
You could iterate over each std::vector and use the algorithms of the standard-library.
There is std::find
// find example
#include <iostream> // std::cout
#include <algorithm> // std::find
#include <vector> // std::vector
int main () {
int myints[] = { 10, 20, 30 ,40 };
int * p;
// pointer to array element:
p = std::find (myints,myints+4,30);
++p;
std::cout << "The element following 30 is " << *p << '\n';
std::vector<int> myvector (myints,myints+4);
std::vector<int>::iterator it;
// iterator to vector element:
it = find (myvector.begin(), myvector.end(), 30);
++it;
std::cout << "The element following 30 is " << *it << '\n';
return 0;
}
there is std::find_if
// find_if example
#include <iostream> // std::cout
#include <algorithm> // std::find_if
#include <vector> // std::vector
bool IsOdd (int i) {
return ((i%2)==1);
}
int main () {
std::vector<int> myvector;
myvector.push_back(10);
myvector.push_back(25);
myvector.push_back(40);
myvector.push_back(55);
std::vector<int>::iterator it = std::find_if (myvector.begin(), myvector.end(), IsOdd);
std::cout << "The first odd value is " << *it << '\n';
return 0;
}
As you are working with std::string this shouldn't be a large problem.
In your scenario it is probably better to use std::multiset< vector <string> > with comparator that compares std::vector<string> in a way you need it to. This will give you sorted container with duplicated values next to each other and cheap insert/erase.