直方图程序给出奇怪的输出C ++

Question

I have been writing code to produce a horizontal histogram. This program takes user input of any range of numbers into a vector. Then it asks the user for the lowest value they want the histogram to begin at, and how big they want each bin to be. For example:

if lowestValue = 1 and binSize = 20 and vector is filled with values {1, 2, 3, 20, 30, 40, 50} it would print something like:

(bin)   (bars)  (num)(percent)
[ 1-21) ####      4    57%
[21-41) ##        2    28%
[41-61) ##        2    28%

Here is most of the code that does so:

void printHistogram(int lowestValue, int binSize, vector<double> v)
{
    int binFloor = lowestValue, binCeiling = 0;
    int numBins = amountOfBins(binSize, (int)range(v));
    for (int i = 0; i<=numBins; i++)
    {
        binCeiling = binFloor+binSize;
        int amoInBin = amountInBin(v,binFloor, binSize);
        double perInBin = percentInBin(v, amoInBin);
        if (binFloor < 10)
        {
            cout << "[ " << binFloor << '-' << binCeiling << ") " << setw(20) << left << formatBars(perInBin) << ' ' << amoInBin << ' '<< setprecision(4) << perInBin << '%' << endl;
            binFloor += binSize;
        }

        else
        {
            cout << '[' << binFloor << '-' << binCeiling << ") " << setw(20) << left << formatBars(perInBin) << ' ' << amoInBin << ' '<< setprecision(4) << perInBin << '%' << endl;
            binFloor += binSize;
        }
    }
}

and the function that counts how many terms are in each bin:

int amountInBin(vector<double> v, int lowestBinValue, int binSize)
{
     int count = 0;
     for (size_t i; i<v.size(); i++)
     {
         if (v[i] >= lowestBinValue && v[i] < (lowestBinValue+binSize))
             count += 1;
     }
     return count;
}

Now my issue:

For some reason, it is not counting values between 20-40. At least as far as I can see from my testing. Here is an image of a run:

Any help is appreciated.

Answer 1

I would suggest a different approach. Making two passes, first calculating the number of bins, then another pass to add them up, looks fragile, and error-prone. Not really surprise to see you trying to figure out a bug of this kind. I think your original approach is too complicated.

As the saying goes "the more you overthink the plumbing, the easier it is to stop up the drain". Find the simplest way to do something, and it will have the least amount of surprises and gotchas, to deal with.

I think it's simpler to make a single pass over the values, calculating which bin each value belongs to, and counting the number of values seen per bin. Let's use a std::map , keyed by bin number, with the value being the number of values in each bin.

void printHistogram(int lowestValue, int binSize, const std::vector<double> &v)
{
    std::map<int, size_t> histogram;

    for (auto value:v)
    {
       int bin_number= value < lowestValue ? 0:(value-lowestValue)/binSize;

       ++histogram[bin_number];
    }

And ...that's it. histogram is now your histogram. histogram[0] is now the number of values in the first bin, [lowestValue, lowestValue+binSize) , which also includes all values less than lowestValue . histogram[1] will be the number of values found for the next bin, and so on.

Now, you just have to iterate over the histogram map, and generate your actual histogram.

Now, the tricky part here is that the histogram map will only include keys for which at least 1 value was found. If no value was dropped into the bin, the map will not include the bin number. So, if there were no values in the first bin, histogram[0] won't even exist, the first value in the map will be the bin for the lowest value in the vector.

This isn't such a difficult problem to solve, by iterating over the map with a little bit of extra intelligence:

int next_bin_number=0;

for (auto b=histogram.begin(); b != histogram.end(); b++)
{
    while (next_bin_number < b->first)
    {
         // next_bin_number had 0 values. Print the histogram row
         // for bin #next_bin_number, showing 0 values in it.

         ++next_bin_number;
    }

    int n_values=b->second;

    // Bin #n_next_number, with n_values, print its histogram row

    ++next_bin_number;
}

Answer 2

循环中的代码不会初始化i ，因此结果充其量是不可预测的。