您尝试执行的特征提取是一项计算机密集型任务。我通过只为整个图像计算一次共现图来加快您的方法，而不是在滑动窗口的重叠位置上一遍又一遍地计算共现图。
共现图是与原始图像大小相同的图像堆栈，其中 - 对于每个像素 - 强度级别由整数替换，该整数编码两个强度的共现，即Ii在该像素处和Ij在偏移像素。共现图的层数与我们考虑的偏移量一样多（即所有可能的距离-角度对）。通过保留共现图，您不需要从头开始计算滑动窗口的每个位置的 GLCM，因为您可以重用先前计算的共现图来获得每个距离的邻接矩阵（GLCM） -角度对。这种方法为您提供了显着的速度增益。
我想出的解决方案依赖于以下功能：
import numpy as np
from skimage import io
from scipy import stats
from skimage.feature import greycoprops
def offset(length, angle):
"""Return the offset in pixels for a given length and angle"""
dv = length * np.sign(-np.sin(angle)).astype(np.int32)
dh = length * np.sign(np.cos(angle)).astype(np.int32)
return dv, dh
def crop(img, center, win):
"""Return a square crop of img centered at center (side = 2*win + 1)"""
row, col = center
side = 2*win + 1
first_row = row - win
first_col = col - win
last_row = first_row + side
last_col = first_col + side
return img[first_row: last_row, first_col: last_col]
def cooc_maps(img, center, win, d=[1], theta=[0], levels=256):
"""
Return a set of co-occurrence maps for different d and theta in a square
crop centered at center (side = 2*w + 1)
"""
shape = (2*win + 1, 2*win + 1, len(d), len(theta))
cooc = np.zeros(shape=shape, dtype=np.int32)
row, col = center
Ii = crop(img, (row, col), win)
for d_index, length in enumerate(d):
for a_index, angle in enumerate(theta):
dv, dh = offset(length, angle)
Ij = crop(img, center=(row + dv, col + dh), win=win)
cooc[:, :, d_index, a_index] = encode_cooccurrence(Ii, Ij, levels)
return cooc
def encode_cooccurrence(x, y, levels=256):
"""Return the code corresponding to co-occurrence of intensities x and y"""
return x*levels + y
def decode_cooccurrence(code, levels=256):
"""Return the intensities x, y corresponding to code"""
return code//levels, np.mod(code, levels)
def compute_glcms(cooccurrence_maps, levels=256):
"""Compute the cooccurrence frequencies of the cooccurrence maps"""
Nr, Na = cooccurrence_maps.shape[2:]
glcms = np.zeros(shape=(levels, levels, Nr, Na), dtype=np.float64)
for r in range(Nr):
for a in range(Na):
table = stats.itemfreq(cooccurrence_maps[:, :, r, a])
codes = table[:, 0]
freqs = table[:, 1]/float(table[:, 1].sum())
i, j = decode_cooccurrence(codes, levels=levels)
glcms[i, j, r, a] = freqs
return glcms
def compute_props(glcms, props=('contrast',)):
"""Return a feature vector corresponding to a set of GLCM"""
Nr, Na = glcms.shape[2:]
features = np.zeros(shape=(Nr, Na, len(props)))
for index, prop_name in enumerate(props):
features[:, :, index] = greycoprops(glcms, prop_name)
return features.ravel()
def haralick_features(img, win, d, theta, levels, props):
"""Return a map of Haralick features (one feature vector per pixel)"""
rows, cols = img.shape
margin = win + max(d)
arr = np.pad(img, margin, mode='reflect')
n_features = len(d) * len(theta) * len(props)
feature_map = np.zeros(shape=(rows, cols, n_features), dtype=np.float64)
for m in xrange(rows):
for n in xrange(cols):
coocs = cooc_maps(arr, (m + margin, n + margin), win, d, theta, levels)
glcms = compute_glcms(coocs, levels)
feature_map[m, n, :] = compute_props(glcms, props)
return feature_map
演示
以下结果对应于(250, 200)来自 Landsat 图像的像素裁剪。我已经考虑了两个距离、四个角度和两个 GLCM 属性。这会为每个像素生成一个 16 维的特征向量。请注意，滑动窗口是正方形的，它的边是2*win + 1像素（在这个测试中win = 19，使用了 的值）。此示例运行大约需要 6 分钟，这比“永远”要短得多；-)
In [331]: img.shape
Out[331]: (250L, 200L)
In [332]: img.dtype
Out[332]: dtype('uint8')
In [333]: d = (1, 2)
In [334]: theta = (0, np.pi/4, np.pi/2, 3*np.pi/4)
In [335]: props = ('contrast', 'homogeneity')
In [336]: levels = 256
In [337]: win = 19
In [338]: %time feature_map = haralick_features(img, win, d, theta, levels, props)
Wall time: 5min 53s
In [339]: feature_map.shape
Out[339]: (250L, 200L, 16L)
In [340]: feature_map[0, 0, :]
Out[340]:
array([ 10.3314, 0.3477, 25.1499, 0.2738, 25.1499, 0.2738,
25.1499, 0.2738, 23.5043, 0.2755, 43.5523, 0.1882,
43.5523, 0.1882, 43.5523, 0.1882])
In [341]: io.imshow(img)
Out[341]: <matplotlib.image.AxesImage at 0xce4d160>
