Question

我正在尝试匹配两个图像的直方图（在MATLAB中，这可以使用imhistmatch完成）。是否有标准Python库提供的等效函数？我看过OpenCV，scipy和numpy，但没有看到任何类似的功能。

Answer 1

我之前写了一个答案here，解释了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/阴影的特定比例。

两张图像之间的histogram matching基本原则相同。基本上，您计算源图像和模板图像的累积直方图，然后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匹配的唯一像素值：

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，您可以将此功能分别应用于每个颜色通道。

Answer 2

这是另一种基于this和scikit-image exposure的{{1}}函数的实现，该函数使用cumulative_distribution类似于ali_m的实现。假定输入和模板图像是灰度图像，并且像素值是[0,255]中的整数。

np.interp

输出如下所示：

Answer 3

我想在上述两种解决方案中添加一个小的补充。如果有人计划将此作为全局函数（例如用于灰度图像），则将最终匹配的数组转换为其相应的格式（numpy.uint8）是一个好主意。这可能有助于将来进行图像转换而不会产生冲突。

def hist_norm(source, template):

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

    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)
    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]
    interp_t_values = np.interp(s_quantiles, t_quantiles, t_values)
    interp_t_values = interp_t_values.astype(olddtype)

    return interp_t_values[bin_idx].reshape(oldshape)

Python 2.x中两个图像的直方图匹配？

3 个答案: