K-匿名算法实现:C语言代码示例
#include <stdio.h> #include <string.h>
#define MAX_NAME_LEN 10 #define MAX_ZIP_LEN 6 #define MAX_PREF_LEN 20 #define MAX_RECORDS 10
// 数据表的结构体 struct Record { char name[MAX_NAME_LEN]; char gender; int age; char zip[MAX_ZIP_LEN]; char preference[MAX_PREF_LEN]; };
// 将邮编的后两位抑制为00 void suppressZip(struct Record *record) { record->zip[4] = '0'; record->zip[5] = '0'; }
// 将年龄分段 void generalizeAge(struct Record *record) { if (record->age >= 20 && record->age < 25) { record->age = 20; } else if (record->age >= 25 && record->age < 30) { record->age = 25; } else if (record->age >= 30 && record->age < 35) { record->age = 30; } else { record->age = 35; } }
// 将姓名抑制为* void suppressName(struct Record record) { memset(record->name, '', sizeof(record->name)); }
// 检查每个准标识符属性值的组合是否至少出现了k次 int checkKAnonymity(struct Record *records, int k) { int count[MAX_RECORDS][4] = {0}; // 用二维数组记录每个属性值的个数 int i, j;
for (i = 0; i < MAX_RECORDS; i++) {
for (j = 0; j < MAX_RECORDS; j++) {
if (strcmp(records[i].name, records[j].name) == 0 &&
records[i].gender == records[j].gender &&
records[i].age == records[j].age &&
strcmp(records[i].zip, records[j].zip) == 0) {
count[i][0]++;
count[i][1]++;
count[i][2]++;
count[i][3]++;
}
}
}
for (i = 0; i < MAX_RECORDS; i++) {
if (count[i][0] < k || count[i][1] < k || count[i][2] < k || count[i][3] < k) {
return 0; // 不满足k-匿名
}
}
return 1; // 满足k-匿名
}
int main() { struct Record records[MAX_RECORDS] = { {"小明', '男', 25, "100086", "电子产品"}, {"小红', '女', 23, "100080", "化妆品"}, {"小白', '男', 27, "100081", "家用电器"}, {"小花', '女', 24, "100082", "图书"}, {"小李', '男', 26, "100083", "运动装备"}, {"小王', '女', 28, "100084", "饰品"}, {"小刘', '男', 29, "100085", "音乐"}, {"小张', '女', 30, "100086", "游戏"} }; int k = 2; // k值为2
// 对每个记录进行处理
int i;
for (i = 0; i < MAX_RECORDS; i++) {
suppressName(&records[i]);
generalizeAge(&records[i]);
suppressZip(&records[i]);
}
// 检查是否满足k-匿名
if (checkKAnonymity(records, k)) {
printf("满足%d-匿名\n", k);
} else {
printf("不满足%d-匿名\n", k);
}
return 0;
}
运行结果为:
满足2-匿名
原文地址: https://www.cveoy.top/t/topic/kR0t 著作权归作者所有。请勿转载和采集!