gpt4 book ai didi

arrays - OpenCv - 从直方图数组中获取值

转载 作者:太空宇宙 更新时间:2023-11-03 22:15:01 24 4
gpt4 key购买 nike

我需要在整个数组中搜索数组(MatND 对象)以获取 MAX 值的 5%。 minMaxLoc 函数返回 MAX 值,但我不知道如何自行搜索。

有什么想法吗?

最佳答案

如果是uchar垫

/**
* @param : input image
* @hist : histogram
* @nmin : total minimum pixels number
* @nmax : total maximum pixels number
* @channel : channel number
*
* ex : images with 1000 pixels, 50 equal to 5% of it
*/
std::pair<size_t, size_t> get_quantile_uchar(cv::Mat &input, cv::MatND &hist, size_t nmin, size_t nmax, int channel)
{
int const hist_size = std::numeric_limits<uchar>::max() + 1;
float const hranges[2] = {0, 255};
float const *ranges[] = {hranges};

//compute and cumulate the histogram
cv::calcHist(&input, 1, &channel, cv::Mat(), hist, 1, &hist_size, ranges);
auto *hist_ptr = hist.ptr<float>(0);
for(size_t i = 1; i != hist_size; ++i){
hist_ptr[i] += hist_ptr[i - 1];
}

// get the new min/max
std::pair<size_t, size_t> min_max(0, hist_size - 1);
while(min_max.first != (hist_size - 1) && hist_ptr[min_max.first] <= nmin){
++min_max.first; // the corresponding histogram value is the current cell position
}

while(min_max.second > 0 && hist_ptr[min_max.second] > nmax){
--min_max.second; // the corresponding histogram value is the current cell position
}

if (min_max.second < hist_size - 2)
++min_max.second;

return min_max;
}

例如,如果有一个 Mat(100 * 100) 的值在 0~255 之间,你可以测量顶部像这样的 5% 百分位和最低 3% 百分位

auto const result = get_quantile(input, hist, input.total * 0.03, input.total * 0.95, 0); 

如果不是uchar Mat,那么你可以先对你要测量的 channel 进行排序

/**
* @brief generic algorithm for other channel types except of uchar
* @param input the input image
* @param output the output image
* @param smin total number of minimum pixels
* @param smax total number maximum pixels
* @param channel the channel used to compute the histogram
*
* This algorithm only support uchar channel and float channel by now
*/
template<typename T>
std::pair<T, T> get_quantile(cv::Mat &input, size_t smin, size_t smax, int channel)
{
std::vector<float> temp_input = copy_to_one_dim_array_ch<float>(input, channel);
std::sort(std::begin(temp_input), std::end(temp_input));

return std::pair<T, T>(temp_input[smin], temp_input[temp_input.size() - 1 - smax]);
}

接下来的问题是如何实现函数copy_to_one_dim_array_ch

/*
* experimental version for cv::Mat, try to alleviate the problem
* of code bloat.User should make sure the space of begin point to
* have enough of spaces.
*/
template<typename T, typename InputIter>
void copy_to_one_dim_array_ch(cv::Mat const &src, InputIter begin, int channel)
{
int const channel_number = src.channels();
if(channel_number <= channel || channel < 0){
throw std::out_of_range("channel value is invalid\n" + std::string(__FUNCTION__) +
"\n" + std::string(__FILE__));
}

for(int row = 0; row != src.rows; ++row){
auto ptr = src.ptr<T>(row) + channel;
for(int col = 0; col != src.cols; ++col){
*begin = *ptr;
++begin;
ptr += channel_number;
}
}
}

template<typename T>
std::vector<T> const copy_to_one_dim_array_ch(cv::Mat const &src, int channel)
{
std::vector<T> result(src.total());
copy_to_one_dim_array_ch<T>(src, std::begin(result), channel);

return result;
}

部分功能需要c++11支持,函数copy_to_one_dim_array_ch不支持非字节图像

如果你想让它变得更容易使用,你可以1 : 将这些函数包装在一个类中。
2:在 uchar Mat 上应用完全特化3 : 将类包装在一个函数中

关于arrays - OpenCv - 从直方图数组中获取值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16251866/

24 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com