Determine the third quartile from a collection of integers in C++?
-
11-06-2021 - |
문제
I'm reading Accelerated C++. At the moment I'm at the end of chapter 3 and here's the exercise that I'm trying to do:
"Write a program to compute and print the quartiles of a set of integers."
I found the first and the second quartiles, but I have no idea how to find the third. Here's my code:
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
int main(){
cout<<"Enter numbers:";
int x;
vector<int>integers;
while(cin>>x)
integers.push_back(x);
typedef vector<int>::size_type vec_sz;
vec_sz size = integers.size();
sort(integers.begin(), integers.end());
vec_sz mid = size/2;
vec_sz q1 = mid/2;
double median;
median = size % 2 == 0 ? ((double)integers[mid] + (double)integers[mid-1]) / 2
: integers[mid];
double quartOne = ((double)integers[q1] + (double)integers[q1-1])/2;
cout<<"The First Quartile is: "<<quartOne<<endl;
cout<<"The Second Quartile is: "<<median<<endl;
return 0;
}
해결책
One way would be to sort the collection and then take the 3 dividing items:
vector<int> v = ...;
sort(v.begin(), v.end());
int q12 = v[v.size()*1/4];
int q23 = v[v.size()*2/4];
int q34 = v[v.size()*3/4];
This is O(nlogn) in the number of data items.
Another way would be to perform a binary search of the data for the three divisions seperately. ie propose an initial q12, check if it is correct by making a pass of the data, if it is incorrect adjust it up or down by half, and repeat. Do likewise for q23 and q34.
This is technically O(n) because a 32-bit int has a fixed range and can be binary searched in 32 passes max.
다른 팁
This solutions implements the second method described in wikipedia for computing quartiles. It provides the correct values both for vectors with odd and even lengths.
#include <vector>
#include <tuple>
#include <iostream>
using namespace std;
double median(vector<double>::const_iterator begin,
vector<double>::const_iterator end) {
int len = end - begin;
auto it = begin + len / 2;
double m = *it;
if ((len % 2) == 0) m = (m + *(--it)) / 2;
return m;
}
tuple<double, double, double> quartiles(const vector<double>& v) {
auto it_second_half = v.cbegin() + v.size() / 2;
auto it_first_half = it_second_half;
if ((v.size() % 2) == 0) --it_first_half;
double q1 = median(v.begin(), it_first_half);
double q2 = median(v.begin(), v.end());
double q3 = median(it_second_half, v.end());
return make_tuple(q1, q2, q3);
}
int main() {
vector<double> v = {2, 2, 3, 4, 4, 5, 5, 10};
auto q = quartiles(v);
cout << get<0>(q) << "," << get<1>(q) << "," << get<2>(q) << endl;
return 0;
}
It is designed for real numbers, but it is easily adaptable for integer values (just round the values to their nearest integer).