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troubled code - finding the mean, median, mode, standard deviation, variance of an array

here is my code ( it sux)
i got the mean, im working on the median, i have no clue as to what the standard deviation or variance means, but one small step at a time i guess...if you can help with any of these it would be much appreciated

#include <stdio.h>
#define MAX 20

void print_inventory( double inventory[], int numitems); //print out the data
int inputinventory( double inventory[], int maxnum); //input the data function
void print_avg( double inventory[], int numitems); //print the average function

void swap( double *a, double *b);
void bubble_sort( double data[], int num_elts );

//void median( double elements[], int num_elts) //find the median fuction

int main( void )
{
      double inventory[MAX];
      int num_items;

      printf( "Please enter the number of items in stock.");
      printf( "enter -1 when you are done.\n");

      num_items = input_inventory( inventory, MAX );

      print_inventory(inventory, num_items);
      print_avg( inventory, num_items);

      bubble_sort( inventory, num_items );
      print_inventory(inventory, num_items);

      //void median( double elements[], int num_elts)

      return 0;
}

int input_inventory( double inventory[], int maxnum )
{
      int index = 0;

      scanf( "%lf", &inventory[ index ] );
                   printf("current input %lf:\n", inventory[index]);
      while (index < maxnum-1 && inventory[ index ] != -1 ) {
       index++;
              scanf( "%lf", &inventory[index] );
              printf("current input%lf:\n", inventory[index]);
      }
      if (index == maxnum - 1){
            printf("No room for more items.\n" );
            return (index+1);
      }
      else return ( index );
}

void print_inventory( double inventory[], int numitems)
{
      int index;


      for (index = 0; index < numitems; index++) {
      printf("Item number %d:\t\t", index+1);
      printf("Number on hand %lf\n", inventory[index] );
      }

}

void print_avg( double inventory[], int numitems)
{
      double sum=0;
      int index;

      for(index = 0; index < numitems ; index++){
            sum += inventory[index];
      }
      printf("\nthe average of all items is %lf\n", sum / index);
}

void bubble_sort( double data[], int num_elts )
{
      int i, j;

      for ( i = 0; i < num_elts; i++ )
                  for ( j = num_elts-1; j > i; j--)
                        if ( data[j] < data[j-1] )
                              swap( &data[j], &data[j-1] );
}

void swap( double *a, double *b)
{
      int temp;

      temp = *a;
      *a = *b;
      *b = temp;
}

/* troubled code */
void median( double inventory[], int num_elts)
{
      int i;
      for( i = 0; i < num_elts; i++;){
            if(num_elts % 2 == 0)
                  printf("number of elements is even: %d", num_elts);
            else
                  printf("number of elements is odd: %d", num_elts);
      }

}
/*
here i was just trying to determine wether the number of elements was even or odd so i know whether i have to divide by two or not at all to find the median*/

Sjef Bosman

Probably you meant:

void median( double inventory[], int num_elts)
{
int i;

if(num_elts % 2 == 0)
printf("number of elements is even: %d", num_elts);
else
printf("number of elements is odd: %d", num_elts);
}

Why the for-loop?

Whatever statistical function you need, look in a statistics book or use Google to find the functions and their purpose explained.

manish_regmi

hi,
What does median do. you want to find the median of inventory or median of num_elts.

regards manish

shadiestone

ASKER

inventory

ASKER CERTIFIED SOLUTION

stefan73

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manish_regmi

Inventory then,

/*median of inventry.*/
void median( double inventory[], int num_elts)
{
int i,;
double median = 0;
for( i = 0; i < num_elts; i++;){
median += inventry[i]
}
median = median/num_elts;
printf("the median is %f. ", median) ;
}

regards.

leflon

Hi shadlestone,

@manish_regmi
> median = median/num_elts;
is not the median, but the average.

the median of a range of values is the value right in the middle of this values. there are as many values smaller than the median, as there are values bigger than it.

1 2 3 4 5 -> median 3
1 2 3 4 -> median (2+3)/2 = 2.5

double median(double inv[], int num)
{
if(num%2==0)
return (inv[num/2]+inv[1+(num/2)])/2;
else
return inv[(num+1)/2];
}

i haven't checked if the array indices are correct, would be better to do it so.

important, the inv[] array has to be sorted to get the proper median value.

leflon

Sjef Bosman

Is this another test question? Smells like it anyway... ;) Better give hints to find the solution, or the bug, but not complete solutions.

SOLUTION

twobitadder

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mark_at_acc

Some are saying he needs to sort, he is already, right? Looks like levlon got it right. Just send the "num_elts" value for "num" variable in levlon's function, and you got it. shadie, in your code you don't need the for loop, since you know how many there are --> num_elts.

Within your bubble sort, quick way is is to call levlon's median function after the sort, and instead of returning median, print it. Not "elegant", since I think main() should recieve all values, but it works.

homework should be to look up std dev. and variance, and try those yourself.