Project scenario:
In the actual development process, we will encounter the need to calculate the percentage. This requirement is very simple. You only need to add all the sample data to get the total, and then divide each sample by the total to get the proportion of each share. . But there is a problem, that is, when the division is inexhaustible, limited by the accuracy of the retained data (rounding), you will find that the sum of the proportions of each share is not equal to 100%, and the data may be large or small. In order to solve this problem, the "maximum balance" algorithm appeared (this is the method used by Echarts).
What is the "maximum balance" algorithm
The largest remainder method , also known as the quota system or the Hamilton method, is a method of seat allocation under the proportional representation voting system, as opposed to the highest mean method.
Under the maximum balance method, candidates are required to stand for election on lists up to the number of seats in the relevant constituency. Candidates are listed in order of priority. Voters vote for a list, not individual candidates.
After voting closes, divide the valid votes by the amount. Every time a list gets double the number of votes, it will be allocated a seat. Candidates from each list are elected in the order originally established.
When the balance of each list is lower than the "amount", the remaining seats are allocated according to the order of the balance of each list; hence the name of the maximum balance method.
solution:
/**
* 最大余额法，用于解决百分比不足100%或者超过100%的问题
*
* @param arr 数组
* @param idx 索引
* @param precision 精度
* @return 每一次计算的结果
*/
public static double getPercentValue(int[] arr, int idx, int precision) {
if ((arr.length - 1) < idx) {
return 0;
}
//求和
double sum = 0;
for (int j : arr) {
sum += j;
}
if (sum == 0) {
return 0;
}
//10的2次幂是100，用于计算精度。
double digits = Math.pow(10, precision);
//扩大比例100
double[] votesPerQuota = new double[arr.length];
for (int i = 0; i < arr.length; i++) {
double val = arr[i] / sum * digits * 100;
votesPerQuota[i] = val;
}
//总数,扩大比例意味的总数要扩大
double targetSeats = digits * 100;
//再向下取值，组成数组
double[] seats = new double[arr.length];
for (int i = 0; i < votesPerQuota.length; i++) {
seats[i] = Math.floor(votesPerQuota[i]);
}
//再新计算合计，用于判断与总数量是否相同,相同则占比会100%
double currentSum = 0;
for (double seat : seats) {
currentSum += seat;
}
//余数部分的数组:原先数组减去向下取值的数组,得到余数部分的数组
double[] remainder = new double[arr.length];
for (int i = 0; i < seats.length; i++) {
remainder[i] = votesPerQuota[i] - seats[i];
}
while (currentSum < targetSeats) {
double max = 0;
int maxId = 0;
for (int i = 0; i < remainder.length; ++i) {
if (remainder[i] > max) {
max = remainder[i];
maxId = i;
}
}
//对最大项余额加1
++seats[maxId];
//已经增加最大余数加1,则下次判断就可以不需要再判断这个余额数。
remainder[maxId] = 0;
//总的也要加1,为了判断是否总数是否相同,跳出循环。
++currentSum;
}
// 这时候的seats就会总数占比会100%
return seats[idx] / digits;
}
/**
* 最大余额法，用于解决百分比不足100%或者超过100%的问题
*
* @param arr 数组
* @param precision 精度
* @return 按照数组顺序排列的百分比
*/
public static double[] getPercentValue(int[] arr, int precision) {
double[] result = new double[arr.length];
for (int i = 0; i < arr.length; i++) {
result[i] = getPercentValue(arr, i, precision);
}
return result;
}Summarize:
What solution to use to deal with this problem depends on the customer, and it depends on which solution the customer can accept.