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Calcolare la Media, la Mediana e la Moda usando array Media Mediana – numero a meta della lista ordinata. –1, 2, 3, 4, 5 3 e la mediana Moda - numero che occorre piu spesso –1, 1, 1, 2, 3, 3, 4, 5 1 e la moda

1/* Fig. 6.16: fig06_16.c 2 Analisi di un insieme di dati. Calcolo della media, 3 della mediana e della moda di un insieme di dati */ 4#include 5#define SIZE void mean( const int [] ); 8void median( int [] ); 9void mode( int [], const int [] ) ; 10void bubbleSort( int [] ); 11void printArray( const int [] ); 12 13int main() 14{ 15 int frequency[ 10 ] = { 0 }; 16 int response[ SIZE ] = 17 { 6, 7, 8, 9, 8, 7, 8, 9, 8, 9, 18 7, 8, 9, 5, 9, 8, 7, 8, 7, 8, 19 6, 7, 8, 9, 3, 9, 8, 7, 8, 7, 20 7, 8, 9, 8, 9, 8, 9, 7, 8, 9, 21 6, 7, 8, 7, 8, 7, 9, 8, 9, 2, 22 7, 8, 9, 8, 9, 8, 9, 7, 5, 3, 23 5, 6, 7, 2, 5, 3, 9, 4, 6, 4, 24 7, 8, 9, 6, 8, 7, 8, 9, 7, 8, 25 7, 4, 4, 2, 5, 3, 8, 7, 5, 6, 26 4, 5, 6, 1, 6, 5, 7, 8, 7 }; 27

28 mean( response ); 29 median( response ); 30 mode( frequency, response ); 31 return 0; 32} ******** Mean ******** The mean is the average value of the data items. The mean is equal to the total of all the data items divided by the number of data items (99). The mean value for this run is: 681 / 99 = Output Vediamo prima loutput, poi la funzione…

33 34void mean( const int answer[] ) 35{ 36 int j, total = 0; printf( "%s\n%s\n%s\n", "********", " Mean", "********" ); for ( j = 0; j <= SIZE - 1; j++ ) 41 total += answer[ j ]; printf( "The mean is the average value of the data\n" 44 "items. The mean is equal to the total of\n" 45 "all the data items divided by the number\n" 46 "of data items ( %d ). The mean value for\n" 47 "this run is: %d / %d = %.4f\n\n", 48 SIZE, total, SIZE, ( double ) total / SIZE ); 49} 50

Output ******** Median ******** The unsorted array of responses is The sorted array is The median is element 49 of the sorted 99 element array. For this run the median is 7 Vediamo prima loutput, poi la funzione…

51void median( int answer[] ) 52{ 53 printf( "\n%s\n%s\n%s\n%s", 54 "********", " Median", "********", 55 "The unsorted array of responses is" ); printArray( answer ); 58 bubbleSort( answer ); 59 printf( "\n\nThe sorted array is" ); 60 printArray( answer ); 61 printf( "\n\nThe median is element %d of\n" 62 "the sorted %d element array.\n" 63 "For this run the median is %d\n\n", 64 SIZE / 2, SIZE, answer[ SIZE / 2 ] ); 65}

******** Mode ******** Response Frequency Histogram * 2 3 *** 3 4 **** 4 5 ***** 5 8 ******** 6 9 ********* 7 23 *********************** 8 27 *************************** 9 19 ******************* The mode is the most frequent value. For this run the mode is 8 which occurred 27 times. Vediamo prima loutput, poi la funzione…

66 67void mode( int freq[], const int answer[] ) 68{ 69 int rating, j, h, largest = 0, modeValue = 0; printf( "\n%s\n%s\n%s\n", 72 "********", " Mode", "********" ); for ( rating = 1; rating <= 9; rating++ ) 75 freq[ rating ] = 0; for ( j = 0; j <= SIZE - 1; j++ ) 78 ++freq[ answer[ j ] ]; printf( "%s%11s%19s\n\n%54s\n%54s\n\n", 81 "Response", "Frequency", "Histogram", 82 " ", " " ); for ( rating = 1; rating <= 9; rating++ ) { 85 printf( "%8d%11d ", rating, freq[ rating ] ); if ( freq[ rating ] > largest ) { 88 largest = freq[ rating ]; 89 modeValue = rating; 90 } Nota che lindice in frequency[] e il valore di un elemento in response[] ( answer[] )

95 printf( "\n" ); 96 } printf( "The mode is the most frequent value.\n" 99 "For this run the mode is %d which occurred" 100 " %d times.\n", modeValue, largest ); 101} void bubbleSort( int a[] ) 104{ 105 int pass, j, hold; for ( pass = 1; pass <= SIZE - 1; pass++ ) for ( j = 0; j <= SIZE - 2; j++ ) if ( a[ j ] > a[ j + 1 ] ) { 112 hold = a[ j ]; 113 a[ j ] = a[ j + 1 ]; 114 a[ j + 1 ] = hold; 115 } 116} for ( h = 1; h <= freq[ rating ]; h++ ) 93 printf( "*" ); 94 Scrive * a seconda del valore di frequency[] Bubble sort: se due elementi non sono in ordine, li scambia

printf( "%2d", a[ j ] ); 128 } 129} void printArray( const int a[] ) 119{ 120 int j; for ( j = 0; j <= SIZE - 1; j++ ) { if ( j % 20 == 0 ) 125 printf( "\n" );

Output ******** Mean ******** The mean is the average value of the data items. The mean is equal to the total of all the data items divided by the number of data items (99). The mean value for this run is: 681 / 99 = ******** Median ******** The unsorted array of responses is The sorted array is The median is element 49 of the sorted 99 element array. For this run the median is 7

******** Mode ******** Response Frequency Histogram * 2 3 *** 3 4 **** 4 5 ***** 5 8 ******** 6 9 ********* 7 23 *********************** 8 27 *************************** 9 19 ******************* The mode is the most frequent value. For this run the mode is 8 which occurred 27 times.

Ricerca su array: ricerca lineare e ricerca binaria Ricerca di un valore chiave in un array Ricerca lineare –Semplice –Confronta ogni elemento dellarray con la chiave –Utile nel caso di array piccoli e completamente non ordinati.

1./* Fig6_18.c Ricerca lineare in un array */ 2.#include 3.#define SIZE int linearSearch(int [], int, int); 5.main() 6.{ 7. int a[SIZE], x, searchKey, element; 8. for (x = 0; x <= SIZE - 1; x++) /* create some data */ 9. a[x] = 2 * x; 10. printf("Enter integer search key:\n"); 11. scanf("%d", &searchKey); 12. element = linearSearch(a, searchKey, SIZE); 13. if (element != -1) 14. printf("Found value in element %d\n", element); 15. else 16. printf("Value not found\n"); 17. return 0; 18.}

19.int linearSearch(int array[], int key, int size) 20.{ 21. int n; 22. for (n = 0; n <= size - 1; ++n) 23. if (array[n] == key) 24. return n; 25. return -1; 26.} Enter integer search key: 87 Value not found Enter integer search key: 88 Found value in element 44 Output

Ricerca su array: ricerca lineare e ricerca binaria Ricerca binaria –Su array ordinati –Confronta lelemento di mezzo (med) con la chiave Se uguali, elemento trovato Se chiave < med, guarda la prima meta dell array Se chiave > med, guarda la seconda meta Ripete –Molto veloce; al piu log n passi, dove n> 2 e il numero di elementi dellarray. Per un array di 30 elementi bastano 5 confronti

1./* Fig6_19.c Ricerca binaria in un array */ 2.#include 3.#define SIZE 15 4.int binarySearch(int [], int, int, int); 5.void printHeader(void); 6.void printRow(int [], int, int, int); 7.main() 8.{ 9. int a[SIZE], i, key, result; 10. for (i = 0; i <= SIZE - 1; i++) 11. a[i] = 2 * i; 12. printf("Enter a number between 0 and 28: "); 13. scanf("%d", &key); 14. printHeader(); 15. result = binarySearch(a, key, 0, SIZE - 1); 16. if (result != -1) 17. printf("\n%d found in array element %d\n", key, result); 18. else 19. printf("\n%d not found\n", key); 20. return 0; 21.}

22.int binarySearch(int b[], int searchKey, int low, int high) 23.{ 24. int middle; 25. while (low <= high) { 26. middle = (low + high) / 2; 27. printRow(b, low, middle, high); 28. if (searchKey == b[middle]) 29. return middle; 30. else if (searchKey < b[middle]) 31. high = middle - 1; 32. else 33. low = middle + 1; 34. } 35. return -1; /* chiave non trovata */ 36.}

37./* Stampa lintestazione per loutput */ 38.void printHeader(void) 39.{ 40. int i; 41. printf("\nSubscripts:\n"); 42. for (i = 0; i <= SIZE - 1; i++) 43. printf("%3d ", i); 44. printf("\n"); 45. for (i = 1; i <= 4 * SIZE; i++) 46. printf("-"); 47. printf("\n"); 48.}

49./* Stampa una riga delloutput mostrando la parte corrente dellarray su ci si sta lavorando. */ 50.void printRow(int b[], int low, int mid, int high) 51.{ 52. int i; 53. for (i = 0; i <= SIZE - 1; i++) 54. if (i high) 55. printf(" "); 56. else if (i == mid) 57. printf("%3d*", b[i]); /* segna il valore di mezzo */ 58. else 59. printf("%3d ", b[i]); 60. printf("\n"); 61.}

Output Enter a number between 0 and 28: 2 Subscripts: * * * 4 2 found in array element 1 Enter a number between 0 and 28: 24 Subscripts: * * * 28 24* 24 found in array element 12