python2.x中两幅图像的直方图匹配？

Question

我正在尝试匹配两个图像的直方图（在MATLAB中，这是可以做到的）
使用
imhistmatch).
标准Python库中是否有可用的等效函数？我已经
查看了OpenCV、scipy和numpy，但没有看到任何类似的功能。

Answer 1

我以前写过一个答案
这里解释如何做
图像直方图的分段线性插值
高光/中音/阴影的特定比率。
相同的基本原则是[直方图]的基础
匹配两人之间
图像。基本上，你要计算出你的源和目标的累积直方图
模板图像，然后线性插值，以找到唯一的像素值
与唯一像素的分位数最接近的模板图像
源图像中的值：

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)

For example:

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，您可以将此函数分别应用于每个图像频道。根据你想要达到的效果，你可能想要
首先将图像转换为不同的颜色空间。例如，你可以转换成HSV空间然后呢如果你想匹配亮度，就在V频道上进行匹配，但不是色调或饱和度。