Python 2.x 中两个图像的直方图匹配?

2024-02-07

我正在尝试匹配两个图像的直方图(在 MATLAB 中,这可以使用imhistmatch http://www.mathworks.com/help/images/ref/imhistmatch.html)。标准 Python 库中是否有等效的函数?我看过 OpenCV、scipy 和 numpy,但没有看到任何类似的功能。


我之前写过一个答案here https://stackoverflow.com/a/31493356/1461210解释如何对图像直方图进行分段线性插值,以强制执行高光/中间调/阴影的特定比率。

相同的基本原则直方图匹配 https://en.wikipedia.org/wiki/Histogram_matching两个图像之间。本质上,您计算源图像和模板图像的累积直方图,然后进行线性插值以查找模板图像中与源图像中唯一像素值的分位数最接近的唯一像素值:

import numpy as np

def hist_match(source, template):
    """
    Adjust the pixel values of a grayscale image such that its histogram
    matches that of a target image

    Arguments:
    -----------
        source: np.ndarray
            Image to transform; the histogram is computed over the flattened
            array
        template: np.ndarray
            Template image; can have different dimensions to source
    Returns:
    -----------
        matched: np.ndarray
            The transformed output image
    """

    oldshape = source.shape
    source = source.ravel()
    template = template.ravel()

    # get the set of unique pixel values and their corresponding indices and
    # counts
    s_values, bin_idx, s_counts = np.unique(source, return_inverse=True,
                                            return_counts=True)
    t_values, t_counts = np.unique(template, return_counts=True)

    # take the cumsum of the counts and normalize by the number of pixels to
    # get the empirical cumulative distribution functions for the source and
    # template images (maps pixel value --> quantile)
    s_quantiles = np.cumsum(s_counts).astype(np.float64)
    s_quantiles /= s_quantiles[-1]
    t_quantiles = np.cumsum(t_counts).astype(np.float64)
    t_quantiles /= t_quantiles[-1]

    # interpolate linearly to find the pixel values in the template image
    # that correspond most closely to the quantiles in the source image
    interp_t_values = np.interp(s_quantiles, t_quantiles, t_values)

    return interp_t_values[bin_idx].reshape(oldshape)

例如:

from matplotlib import pyplot as plt
from scipy.misc import lena, ascent

source = lena()
template = ascent()
matched = hist_match(source, template)

def ecdf(x):
    """convenience function for computing the empirical CDF"""
    vals, counts = np.unique(x, return_counts=True)
    ecdf = np.cumsum(counts).astype(np.float64)
    ecdf /= ecdf[-1]
    return vals, ecdf

x1, y1 = ecdf(source.ravel())
x2, y2 = ecdf(template.ravel())
x3, y3 = ecdf(matched.ravel())

fig = plt.figure()
gs = plt.GridSpec(2, 3)
ax1 = fig.add_subplot(gs[0, 0])
ax2 = fig.add_subplot(gs[0, 1], sharex=ax1, sharey=ax1)
ax3 = fig.add_subplot(gs[0, 2], sharex=ax1, sharey=ax1)
ax4 = fig.add_subplot(gs[1, :])
for aa in (ax1, ax2, ax3):
    aa.set_axis_off()

ax1.imshow(source, cmap=plt.cm.gray)
ax1.set_title('Source')
ax2.imshow(template, cmap=plt.cm.gray)
ax2.set_title('template')
ax3.imshow(matched, cmap=plt.cm.gray)
ax3.set_title('Matched')

ax4.plot(x1, y1 * 100, '-r', lw=3, label='Source')
ax4.plot(x2, y2 * 100, '-k', lw=3, label='Template')
ax4.plot(x3, y3 * 100, '--r', lw=3, label='Matched')
ax4.set_xlim(x1[0], x1[-1])
ax4.set_xlabel('Pixel value')
ax4.set_ylabel('Cumulative %')
ax4.legend(loc=5)

对于一对 RGB 图像,您可以将此函数单独应用于每个通道。根据您想要实现的效果,您可能需要首先将图像转换为不同的色彩空间。例如,您可以转换为HSV空间 https://i.stack.imgur.com/GWtt1.jpg然后,如果您想匹配亮度,而不是色调或饱和度,则仅在 V 通道上进行匹配。

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