【发布时间】：2021-03-27 22:29:54
【问题描述】：
所以我要做的是取一个数字向量，找出每个元素出现的次数，然后计算其在向量中出现次数的百分比。
它在一个只有一个数字的几个元素的向量上工作得非常好，但在一个不同元素的向量（准确地说是 2 个）出现多次的向量上完全不行。
这是一个详细的输出：
count = 6
pushed back 21
total = 6
percent = 6 / 6* 100
pushed back 100
Number:
21
Percent Chance
100
count = 5
pushed back 42
count = 5
pushed back 21
total = 5
total = 10
percent = 5 / 10* 100
pushed back 0
percent = 5 / 10* 100
pushed back 0
Number:
42
Percent Chance
0 // SHOULD BE 50
Number:
21
Percent Chance
0 // SHOULD BE 50
这里是代码：（这个函数的输出是用不同的 x 向量运行两次）
std::vector <std::string> findpercentages(std::vector <std::string> x)
{
int count{}; //for amount of times element occurs
std::vector <std::string> tempvec{}; //to hold the element/s
std::vector <int> counts{}; //to hold count for each element in tempvec
std::vector <std::string> finalvec{}; //elements and counts of elements
for (size_t i = 0; i < x.size(); i++)
{
//look for x[element] in tempvec, if not in tempvec, count occurances in x and add element to
// tempvec, add count to counts
std::vector <std::string>::iterator point { std::find(tempvec.begin(), tempvec.end(), x[i]) };
if (point == tempvec.end())
{
count = std::count(x.begin(), x.end(), x[i]);
std::cout << "count = " << count << '\n';
counts.push_back(count);
std::cout << "pushed back " << x[i] << '\n';
tempvec.push_back(x[i]);
}
}
int total{};
for (size_t n = 0; n < counts.size(); n++)
{
total += counts[n]; //total for percentage calculation
std::cout << "total = " << total << '\n';
}
for (size_t y = 0; y < tempvec.size(); y++)
{
finalvec.push_back(tempvec[y]);
//percent calculation. used unsigned_int64 because got arithmatic overflow warning if i just use
// int
unsigned __int64 percentage = static_cast <unsigned __int64> (round((counts[y] / total) * 100));
std::cout << "percent = " << counts[y] << " / " << total << "* 100" << '\n';
finalvec.push_back(std::to_string(percentage));
std::cout << "pushed back " << percentage << '\n';
}
return finalvec;
}
【问题讨论】：
标签： c++ math vector percentage