finding quartiles - c++

Question

I've written a program where the user can enter any number of values into a vector and it's supposed to return the quartiles, but I keep getting a "vector subscript out of range" error :
#include "stdafx.h"
#include <iostream>
#include <string>
#include <algorithm>
#include <iomanip>
#include <ios>
#include <vector>
int main () {
using namespace std;
cout << "Enter a list of numbers: ";
vector<double> quantile;
double x;
//invariant: homework contains all the homework grades so far
while (cin >> x)
quantile.push_back(x);
//check that the student entered some homework grades
//typedef vector<double>::size_type vec_sz;
int size = quantile.size();
if (size == 0) {
cout << endl << "You must enter your numbers . "
"Please try again." << endl;
return 1;
}
sort(quantile.begin(), quantile.end());
int mid = size/2;
double median;
median = size % 2 == 0 ? (quantile[mid] + quantile[mid-1])/2 : quantile[mid];
vector<double> first;
vector<double> third;
for (int i = 0; i!=mid; ++i)
{
first[i] = quantile[i];
}
for (int i = mid; i!= size; ++i)
{
third[i] = quantile[i];
}
double fst;
double trd;
int side_length = 0;
if (size % 2 == 0)
{
side_length = size/2;
}
else {
side_length = (size-1)/2;
}
fst = (size/2) % 2 == 0 ? (first[side_length/2]/2 + first[(side_length-1)/2])/2 : first[side_length/2];
trd = (size/2) % 2 == 0 ? (third[side_length/2]/2 + third[(side_length-1)/2])/2 : third[side_length/2];
streamsize prec = cout.precision();
cout << "The quartiles are" << setprecision(3) << "1st"
<< fst << "2nd" << median << "3rd" << trd << setprecision(prec) << endl;
return 0;
}

Answer 1

Instead of doing std::sort(quantile.begin(), quantile.end()) a somewhat cheaper way would be
auto const Q1 = quantile.size() / 4;
auto const Q2 = quantile.size() / 2;
auto const Q3 = Q1 + Q2;
std::nth_element(quantile.begin(), quantile.begin() + Q1, quantile.end());
std::nth_element(quantile.begin() + Q1 + 1, quantile.begin() + Q2, quantile.end());
std::nth_element(quantile.begin() + Q2 + 1, quantile.begin() + Q3, quantile.end());
This would not sort the complete array, but only do a "between groups" sort of the 4 quartile. This saves on the "within groups" sort that a full std::sort would do.
If your quantile array is not large, it's a small optimization. But the scaling behavior of std::nth_element is O(N) however, rather than O(N log N) of a std::sort.

Answer 2

Here is Quantile function which is MATLAB's equivalent with linear interpolation:
#include <algorithm>
#include <cmath>
#include <vector>
template<typename T>
static inline double Lerp(T v0, T v1, T t)
{
return (1 - t)*v0 + t*v1;
}
template<typename T>
static inline std::vector<T> Quantile(const std::vector<T>& inData, const std::vector<T>& probs)
{
if (inData.empty())
{
return std::vector<T>();
}
if (1 == inData.size())
{
return std::vector<T>(1, inData[0]);
}
std::vector<T> data = inData;
std::sort(data.begin(), data.end());
std::vector<T> quantiles;
for (size_t i = 0; i < probs.size(); ++i)
{
T poi = Lerp<T>(-0.5, data.size() - 0.5, probs[i]);
size_t left = std::max(int64_t(std::floor(poi)), int64_t(0));
size_t right = std::min(int64_t(std::ceil(poi)), int64_t(data.size() - 1));
T datLeft = data.at(left);
T datRight = data.at(right);
T quantile = Lerp<T>(datLeft, datRight, poi - left);
quantiles.push_back(quantile);
}
return quantiles;
}
Find quartiles:
std::vector<double> in = { 1,2,3,4,5,6,7,8,9,10,11 };
auto quartiles = Quantile<double>(in, { 0.25, 0.5, 0.75 });

Answer 3

This C++ template function calculates quartile for you. It assumes x to be sorted.
#include <assert.h>
template <typename T1, typename T2> typename T1::value_type quant(const T1 &x, T2 q)
{
assert(q >= 0.0 && q <= 1.0);
const auto n = x.size();
const auto id = (n - 1) * q;
const auto lo = floor(id);
const auto hi = ceil(id);
const auto qs = x[lo];
const auto h = (id - lo);
return (1.0 - h) * qs + h * x[hi];
}
To use it do:
std::vector<float> x{1,1,2,2,3,4,5,6};
std::cout << quant(x, 0.25) << std::endl;
std::cout << quant(x, 0.50) << std::endl;
std::cout << quant(x, 0.75) << std::endl;

Answer 4

You need to preallocate first and third vectors before you set the contents.
vector<double> first(mid);
vector<double> third(size-mid);
or use push_back instead of assignments to first[i] and third[i]

Answer 5

If only one element in vector, this instruction is out of range:
quantile[mid-1]
"i" starting from mid so third[0] is out of range
for (int i = mid; i!= size; ++i)
{
third[i] = quantile[i];
}

Answer 6

Here is one error:
vector<double> first;
vector<double> third;
for (int i = 0; i!=mid; ++i)
{
first[i] = quantile[i];
}
The vector first doesn't have any contents, but you try to access the contents. Same problem with third and its loop. Do you mean to use push_back instead?

Answer 7

Implementation for weighted quantiles
This implements a weight functionality for the quantile function with linear interpolation in between the gridpoints.
#include <vector>
#include <numeric>
#include <algorithm>
#include <iostream>
#include <assert.h>
// https://stackoverflow.com/a/12399290/7128154
template <typename T>
std::vector<size_t> sorted_index(const std::vector<T> &v) {
std::vector<size_t> idx(v.size());
iota(idx.begin(), idx.end(), 0);
stable_sort(idx.begin(), idx.end(),
[&v](size_t i1, size_t i2) {return v[i1] < v[i2];});
return idx;
}
// https://stackoverflow.com/a/1267878/7128154
template< typename order_iterator, typename value_iterator >
void reorder( order_iterator order_begin, order_iterator order_end, value_iterator v ) {
typedef typename std::iterator_traits< value_iterator >::value_type value_t;
typedef typename std::iterator_traits< order_iterator >::value_type index_t;
typedef typename std::iterator_traits< order_iterator >::difference_type diff_t;
diff_t remaining = order_end - 1 - order_begin;
for ( index_t s = index_t(), d; remaining > 0; ++ s ) {
for ( d = order_begin[s]; d > s; d = order_begin[d] ) ;
if ( d == s ) {
-- remaining;
value_t temp = v[s];
while ( d = order_begin[d], d != s ) {
swap( temp, v[d] );
-- remaining;
}
v[s] = temp;
}
}
}
// https://stackoverflow.com/a/1267878/7128154
template< typename order_iterator, typename value_iterator >
void reorder_destructive( order_iterator order_begin, order_iterator order_end, value_iterator v ) {
typedef typename std::iterator_traits< value_iterator >::value_type value_t;
typedef typename std::iterator_traits< order_iterator >::value_type index_t;
typedef typename std::iterator_traits< order_iterator >::difference_type diff_t;
diff_t remaining = order_end - 1 - order_begin;
for ( index_t s = index_t(); remaining > 0; ++ s ) {
index_t d = order_begin[s];
if ( d == (diff_t) -1 ) continue;
-- remaining;
value_t temp = v[s];
for ( index_t d2; d != s; d = d2 ) {
std::swap( temp, v[d] );
std::swap( order_begin[d], d2 = (diff_t) -1 );
-- remaining;
}
v[s] = temp;
}
}
// https://stackoverflow.com/a/29677616/7128154
// https://stackoverflow.com/a/37708864/7128154
template <typename T>
double quantile(double q, std::vector<T> values, std::vector<double> weights = std::vector<double>())
{
assert( 0. <= q && q <= 1. && "expecting quantile in range [0; 1]");
if (weights.empty())
{
weights = std::vector<double>(values.size(), 1.);
}
else
{
assert (values.size() == weights.size() && "values and weights missfit in quantiles");
std::vector<size_t> inds = sorted_index(values);
reorder_destructive(inds.begin(), inds.end(), weights.begin());
}
stable_sort(values.begin(), values.end());
// values and weights are sorted now
std::vector<double> quantiles (weights.size());
quantiles[0] = weights[0];
for (int ii = 1; ii < quantiles.size(); ii++)
{
quantiles[ii] = quantiles[ii-1] + weights[ii];
}
double norm = std::accumulate(weights.begin(), weights.end(), 0.0);
int ind = 0;
double qCurrent = 0;
for (; ind < quantiles.size(); ind++)
{
qCurrent = (quantiles[ind] - weights[ind] / 2. ) / norm;
quantiles[ind] = qCurrent;
if (qCurrent > q)
{
if (ind == 0) {return values[0];}
double rat = (q - quantiles[ind-1]) / (quantiles[ind] - quantiles[ind-1]);
return values[ind-1] + (values[ind] - values[ind-1]) * rat;
}
}
return values[values.size()-1];
}
template <typename T>
double quantile(double q, std::vector<T> values, std::vector<int> weights)
{
std::vector<double> weights_double (weights.begin(), weights.end());
return quantile(q, values, weights_double);
}
int main()
{
std::vector<int> vals {5, 15, 25, 35, 45, 55, 65, 75, 85, 95};
std::cout << "quantile(0, vals)=" << quantile(0, vals) << std::endl;
std::cout << "quantile(.73, vals)=" << quantile(.73, vals) << std::endl;
std::vector<int> vals2 {1, 2, 3};
std::vector<double> ws2 {1, 2, 3};
std::cout << "quantile(.13, vals2, ws2)=" << quantile(.13, vals2, ws2) << std::endl;
}
Output
quantile(0, vals)=5
quantile(.73, vals)=73
quantile(.13, vals2, ws2)=1.18667
About weighted quantiles
While in the unweighted case, the input values form an equidistant distribution
values: [1, 2, 3] -> positions: [1/6, 3/6, 5/6]
in the weighted case, the distance is modified.
values: [1, 2, 3], weights: [1, 2, 1] -> positions: [1/8, 4/8, 7/8]
This is not equivalent to a repetition of values for the unweighted case
values: [1, 2, 2, 3] -> positions: [1/8, 3/8, 5/8, 7/8],
as a plateau is formed in between the repeated values. This means:
quantile(q=3/8, values=[1, 2, 2, 3]) = 2
but
quantile(q=3/8, values=[1, 2, 3], weights=[1, 2, 1]) = 1.67