我正在尝试基于像素强度值对2D MR图像的不同区域进行自动图像分割。第一步是在图像的直方图上实现高斯混合模型。

我需要将从该score_samples方法获得的结果高斯绘制到直方图上。我已尝试按照答案（了解高斯混合模型）中的代码进行操作。

但是，所得的高斯根本无法匹配直方图。如何使高斯与直方图匹配？

import numpy as np
import cv2
import matplotlib.pyplot as plt
from sklearn.mixture import GaussianMixture

# Read image
img = cv2.imread("test.jpg",0)

hist = cv2.calcHist([img],[0],None,[256],[0,256])
hist[0] = 0     # Removes background pixels

# Fit GMM
gmm = GaussianMixture(n_components = 3)
gmm = gmm.fit(hist)

# Evaluate GMM
gmm_x = np.linspace(0,255,256)
gmm_y = np.exp(gmm.score_samples(gmm_x.reshape(-1,1)))


# Plot histograms and gaussian curves
fig, ax = plt.subplots()
ax.hist(img.ravel(),255,[1,256])
ax.plot(gmm_x, gmm_y, color="crimson", lw=4, label="GMM")

ax.set_ylabel("Frequency")
ax.set_xlabel("Pixel Intensity")

plt.legend()

plt.show()

我还尝试用总和手动构造高斯。

import numpy as np
import cv2
import matplotlib.pyplot as plt
from sklearn.mixture import GaussianMixture

def gauss_function(x, amp, x0, sigma):
    return amp * np.exp(-(x - x0) ** 2. / (2. * sigma ** 2.))

# Read image
img = cv2.imread("test.jpg",0)

hist = cv2.calcHist([img],[0],None,[256],[0,256])
hist[0] = 0     # Removes background pixels

# Fit GMM
gmm = GaussianMixture(n_components = 3)
gmm = gmm.fit(hist)

# Evaluate GMM
gmm_x = np.linspace(0,255,256)
gmm_y = np.exp(gmm.score_samples(gmm_x.reshape(-1,1)))

# Construct function manually as sum of gaussians
gmm_y_sum = np.full_like(gmm_x, fill_value=0, dtype=np.float32)
for m, c, w in zip(gmm.means_.ravel(), gmm.covariances_.ravel(), gmm.weights_.ravel()):
    gauss = gauss_function(x=gmm_x, amp=1, x0=m, sigma=np.sqrt(c))
    gmm_y_sum += gauss / np.trapz(gauss, gmm_x) * w

# Plot histograms and gaussian curves
fig, ax = plt.subplots()
ax.hist(img.ravel(),255,[1,256])
ax.plot(gmm_x, gmm_y, color="crimson", lw=4, label="GMM")
ax.plot(gmm_x, gmm_y_sum, color="black", lw=4, label="Gauss_sum", linestyle="dashed")

ax.set_ylabel("Frequency")
ax.set_xlabel("Pixel Intensity")

plt.legend()

plt.show()

用 ax.hist(img.ravel(),255,[1,256], normed=True)