我正在使用每样本16位的图像。
有没有一种(简单)的方法来执行这些图像的直方图均衡化(转换为8位/像素是不可选的)?
我正在使用每样本16位的图像。
有没有一种(简单)的方法来执行这些图像的直方图均衡化(转换为8位/像素是不可选的)?
equalizeHist
在OpenCV中只处理8位数据。
但是OpenCV中的图像归一化并不仅限于8位数据。请参阅此处的描述。在您的情况下,函数调用应如下所示:
normalize(src_image, dst_image, 0, 65535, NORM_MINMAX);
如果您想提高图像对比度,首先尝试规范化,只有在这种方法不起作用时再尝试均衡化。规范化更快且破坏性较小。
参考:http://answers.opencv.org/question/3176/improve-contrast-of-a-16u-image/
目前,OpenCV的equalizeHist仅支持8位图像。我基于OpenCV实现(此处),创建了一个16位直方图均衡化函数,代码在这里。
void equalizeHist16Bit(const cv::Mat &_src, cv::Mat &_dst)
{
_dst = _src.clone();
const int hist_sz = 65536;
int *hist = new int[hist_sz] {};
int *lut = new int[hist_sz] {};
for (int y = 0; y < _src.rows; y++)
for (int x = 0; x < _src.cols; x++)
hist[(int)_src.at<unsigned short int>(y, x)]++;
auto i = 0;
while (!hist[i]) ++i;
auto total = (int)_src.total();
if (hist[i] == total)
{
_dst.setTo(i);
return;
}
float scale = (hist_sz - 1.f) / (total - hist[i]);
auto sum = 0;
for (lut[i++] = 0; i < hist_sz; ++i)
{
sum += hist[i];
lut[i] = cv::saturate_cast<ushort>(sum * scale);
}
for (int y = 0; y < _src.rows; y++)
for (int x = 0; x < _src.cols; x++)
{
_dst.at<unsigned short int>(y, x) = lut[(int)_src.at<unsigned short int>(y, x)];
}
}
Python的简单实现 参考:https://github.com/torywalker/histogram-equalizer/blob/master/HistogramEqualization.ipynb
import cv2
import numpy as np
import matplotlib.pyplot as plt
img_tif=cv2.imread("scan.tif",cv2.IMREAD_ANYDEPTH)
img = np.asarray(img_tif)
flat = img.flatten()
hist = get_histogram(flat,65536)
#plt.plot(hist)
cs = cumsum(hist)
# re-normalize cumsum values to be between 0-255
# numerator & denomenator
nj = (cs - cs.min()) * 65535
N = cs.max() - cs.min()
# re-normalize the cdf
cs = nj / N
cs = cs.astype('uint16')
img_new = cs[flat]
#plt.hist(img_new, bins=65536)
#plt.show(block=True)
img_new = np.reshape(img_new, img.shape)
cv2.imwrite("contrast.tif",img_new)