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Views: 7867 | Replies: 9
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Join Date: Oct 2004
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Newbie Poster
Need help calculating Median when array is even Please help I am writing a program that calculates the mean and median of a sequence of integers. My program works fine as long as the array is odd. Can someone please help me with a statement to calculate the median when the array is even. I have included my code for the program.
#include <iostream>
float mean (const int* A, size_t size); //calculates mean of data in A
float median (int* A, size_t size); // calculates median of data in A
void swap (int& x, int& y); // interchanges values of x and y
void sort (int*A, size_t size); // sorts the data in A
void bubblesort (int, int);
int main()
{
int n;
size_t size;
int A [100]; size = 0;
std::cout << "Enter data to calculate the mean and median: ";
while((std::cin >> n) && (size < 100))
{
A[size] = n;
++size;
}
for (size_t i = 0; i < size; i++)
std::cout << A[i] << ' ';
std::cout << '\n';
std::cout << "The mean of the data entered is: " << mean(A,size) << '\n';
std::cout << "The median of the data entered is: " << median(A,size)<< '\n';
for (size_t i = 0; i < size; i++)
std::cout << A[i] << ' ';
std::cout << '\n';
return 0;
}
float mean(const int* A, size_t size)
{
float total = 0;
for (size_t i = 0; i < size; i++)
total += A[i];
return total / size;
}
float median(int* A, size_t size)
{
sort(A,size);
return A[size/2];
}
void swap(int& x, int& y)
{
int hold = x;
x = y;
y = hold;
}
void sort(int* A, size_t size)
// bubble sort
{
for (size_t pass = 0; pass < size - 1; pass ++)
for (size_t j = 0; j < size - 1; j++)
if (A[j] > A[j + 1])
swap(A[j], A[j+1]);
}
#include <iostream>
float mean (const int* A, size_t size); //calculates mean of data in A
float median (int* A, size_t size); // calculates median of data in A
void swap (int& x, int& y); // interchanges values of x and y
void sort (int*A, size_t size); // sorts the data in A
void bubblesort (int, int);
int main()
{
int n;
size_t size;
int A [100]; size = 0;
std::cout << "Enter data to calculate the mean and median: ";
while((std::cin >> n) && (size < 100))
{
A[size] = n;
++size;
}
for (size_t i = 0; i < size; i++)
std::cout << A[i] << ' ';
std::cout << '\n';
std::cout << "The mean of the data entered is: " << mean(A,size) << '\n';
std::cout << "The median of the data entered is: " << median(A,size)<< '\n';
for (size_t i = 0; i < size; i++)
std::cout << A[i] << ' ';
std::cout << '\n';
return 0;
}
float mean(const int* A, size_t size)
{
float total = 0;
for (size_t i = 0; i < size; i++)
total += A[i];
return total / size;
}
float median(int* A, size_t size)
{
sort(A,size);
return A[size/2];
}
void swap(int& x, int& y)
{
int hold = x;
x = y;
y = hold;
}
void sort(int* A, size_t size)
// bubble sort
{
for (size_t pass = 0; pass < size - 1; pass ++)
for (size_t j = 0; j < size - 1; j++)
if (A[j] > A[j + 1])
swap(A[j], A[j+1]);
}
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Join Date: Sep 2004
Posts: 6,009
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>Can someone please help me with a statement to calculate the median when the array is even.
Take the median of the two middle items: a[size / 2] and a[size / 2 + 1].
Take the median of the two middle items: a[size / 2] and a[size / 2 + 1].
Member of: Beautiful Code Club.
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Join Date: Apr 2004
Posts: 3,422
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>>Can someone please help me with a statement to calculate the median when the array is even.
>Take the median of the two middle items: a[size / 2] and a[size / 2 + 1].
Wouldn't that be the mean of a[size / 2] and a[size / 2 - 1]?
>Take the median of the two middle items: a[size / 2] and a[size / 2 + 1].
Wouldn't that be the mean of a[size / 2] and a[size / 2 - 1]?
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Join Date: Sep 2004
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Originally Posted by Dave Sinkula
>>Can someone please help me with a statement to calculate the median when the array is even.
>Take the median of the two middle items: a[size / 2] and a[size / 2 + 1].
Wouldn't that be the mean of a[size / 2] and a[size / 2 - 1]?
Member of: Beautiful Code Club.
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Join Date: May 2004
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Newbie Poster
((n/2)+(n+1)/2)/2
n is the nmber of elements in the array
i e
if yr array was x[n];
median = ((x[n]/2)+(x[n]+1)/2)/2;
n is the nmber of elements in the array
i e
if yr array was x[n];
median = ((x[n]/2)+(x[n]+1)/2)/2;
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Join Date: May 2004
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Newbie Poster
Re: Need help calculating Median when array is even •
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Originally Posted by matika
((n/2)+(n+1)/2)/2
n is the nmber of elements in the array
i e
if yr array was x[n];
median = ((x[n]/2)+(x[n]+1)/2)/2;
sorry again it will b ((x[n]/2)+(x[n+1])/2)/2
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Join Date: Apr 2004
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Originally Posted by matika
sorry again it will b ((x[n]/2)+(x[n+1])/2)/2
#include <iostream>
float median(int* x, int n)
{
return ((x[n]/2)+(x[n+1])/2)/2;
}
int main()
{
int i, A[] = {1,2,3,4,5,6,7,8,9,10}, size = sizeof A / sizeof *A;
float result = median(A, size);
for ( i = 0; i < size; i++ )
{
std::cout << A[i] << ' ';
}
std::cout << "\nThe median of the data entered is: " << result << '\n';
return 0;
}
/* my output
1 2 3 4 5 6 7 8 9 10
The median of the data entered is: 1.38446e+06
*/ •
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Join Date: May 2004
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Newbie Poster
float median(int* x, int n)
{
return ((x[n/2])+(x[(n/2)+1]))/2;
}
{
return ((x[n/2])+(x[(n/2)+1]))/2;
}
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Join Date: May 2004
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Newbie Poster
tell me when it works
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Join Date: Apr 2004
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Uh... look up.
Oh nevermind.
Oh nevermind.
float median(int* x, int n)
{
return ( x [ n / 2 ] + x [ n / 2 - 1] ) / 2.0F;
} |
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