统一圆形LBP人脸识别实施
我正在尝试使用统一圆形LBP（1个单位半径邻域中的8个点）来实现基本的人脸识别系统。 我正在拍摄一张图像，将其重新调整到200 x 200像素，然后将图像分成8x8的小图像 。 然后，我计算每个小图像的直方图并获取直方图列表 。 为了比较2张图像 ，我计算了相应直方图之间的卡方距离并生成一个分数。
这是我的统一LBP实施：
import numpy as np
import math
uniform = {0: 0, 1: 1, 2: 2, 3: 3, 4: 4, 5: 58, 6: 5, 7: 6, 8: 7, 9: 58, 10: 58, 11: 58, 12: 8, 13: 58, 14: 9, 15: 10, 16: 11, 17: 58, 18: 58, 19: 58, 20: 58, 21: 58, 22: 58, 23: 58, 24: 12, 25: 58, 26: 58, 27: 58, 28: 13, 29: 58, 30: 14, 31: 15, 32: 16, 33: 58, 34: 58, 35: 58, 36: 58, 37: 58, 38: 58, 39: 58, 40: 58, 41: 58, 42: 58, 43: 58, 44: 58, 45: 58, 46: 58, 47: 58, 48: 17, 49: 58, 50: 58, 51: 58, 52: 58, 53: 58, 54: 58, 55: 58, 56: 18, 57: 58, 58: 58, 59: 58, 60: 19, 61: 58, 62: 20, 63: 21, 64: 22, 65: 58, 66: 58, 67: 58, 68: 58, 69: 58, 70: 58, 71: 58, 72: 58, 73: 58, 74: 58, 75: 58, 76: 58, 77: 58, 78: 58, 79: 58, 80: 58, 81: 58, 82: 58, 83: 58, 84: 58, 85: 58, 86: 58, 87: 58, 88: 58, 89: 58, 90: 58, 91: 58, 92: 58, 93: 58, 94: 58, 95: 58, 96: 23, 97: 58, 98: 58, 99: 58, 100: 58, 101: 58, 102: 58, 103: 58, 104: 58, 105: 58, 106: 58, 107: 58, 108: 58, 109: 58, 110: 58, 111: 58, 112: 24, 113: 58, 114: 58, 115: 58, 116: 58, 117: 58, 118: 58, 119: 58, 120: 25, 121: 58, 122: 58, 123: 58, 124: 26, 125: 58, 126: 27, 127: 28, 128: 29, 129: 30, 130: 58, 131: 31, 132: 58, 133: 58, 134: 58, 135: 32, 136: 58, 137: 58, 138: 58, 139: 58, 140: 58, 141: 58, 142: 58, 143: 33, 144: 58, 145: 58, 146: 58, 147: 58, 148: 58, 149: 58, 150: 58, 151: 58, 152: 58, 153: 58, 154: 58, 155: 58, 156: 58, 157: 58, 158: 58, 159: 34, 160: 58, 161: 58, 162: 58, 163: 58, 164: 58, 165: 58, 166: 58, 167: 58, 168: 58, 169: 58, 170: 58, 171: 58, 172: 58, 173: 58, 174: 58, 175: 58, 176: 58, 177: 58, 178: 58, 179: 58, 180: 58, 181: 58, 182: 58, 183: 58, 184: 58, 185: 58, 186: 58, 187: 58, 188: 58, 189: 58, 190: 58, 191: 35, 192: 36, 193: 37, 194: 58, 195: 38, 196: 58, 197: 58, 198: 58, 199: 39, 200: 58, 201: 58, 202: 58, 203: 58, 204: 58, 205: 58, 206: 58, 207: 40, 208: 58, 209: 58, 210: 58, 211: 58, 212: 58, 213: 58, 214: 58, 215: 58, 216: 58, 217: 58, 218: 58, 219: 58, 220: 58, 221: 58, 222: 58, 223: 41, 224: 42, 225: 43, 226: 58, 227: 44, 228: 58, 229: 58, 230: 58, 231: 45, 232: 58, 233: 58, 234: 58, 235: 58, 236: 58, 237: 58, 238: 58, 239: 46, 240: 47, 241: 48, 242: 58, 243: 49, 244: 58, 245: 58, 246: 58, 247: 50, 248: 51, 249: 52, 250: 58, 251: 53, 252: 54, 253: 55, 254: 56, 255: 57}
def bilinear_interpolation(i, j, y, x, img):
fy, fx = int(y), int(x)
cy, cx = math.ceil(y), math.ceil(x)
# calculate the fractional parts
ty = y - fy
tx = x - fx
w1 = (1 - tx) * (1 - ty)
w2 = tx * (1 - ty)
w3 = (1 - tx) * ty
w4 = tx * ty
return w1 * img[i + fy, j + fx] + w2 * img[i + fy, j + cx] +
w3 * img[i + cy, j + fx] + w4 * img[i + cy, j + cx]
def thresholded(center, pixels):
out = []
for a in pixels:
if a > center:
out.append(1)
else:
out.append(0)
return out
def uniform_circular(img, P, R):
ysize, xsize = img.shape
transformed_img = np.zeros((ysize - 2 * R,xsize - 2 * R), dtype=np.uint8)
for y in range(R, len(img) - R):
for x in range(R, len(img[0]) - R):
center = img[y,x]
pixels = []
for point in range(0, P):
r = R * math.cos(2 * math.pi * point / P)
c = R * math.sin(2 * math.pi * point / P)
pixels.append(bilinear_interpolation(y, x, r, c, img))
values = thresholded(center, pixels)
res = 0
for a in range(0, P):
res += values[a] << a
transformed_img.itemset((y - R,x - R), uniform[res])
transformed_img = transformed_img[R:-R,R:-R]
return transformed_img
我对AT＆T数据库进行了一项实验，每个主题采用2个图库图像和8个探针图像。 该实验的ROC出现了：
在上述ROC中， x轴表示错误接受率 ， y轴表示真实接受率 。 根据统一的LBP标准，精度似乎很差。 我相信我的实施有问题。 如果有人能帮助我，那将会很棒。 谢谢阅读。
编辑：
我想我在上面的代码中犯了一个错误。 我顺时针旋转，而LBP上的纸张表明我应该在分配重量的同时逆时针旋转。 该行： c = R * math.sin(2 * math.pi * point / P)应该是c = -R * math.sin(2 * math.pi * point / P) 。 编辑后的结果更差。 这表明我的代码有些问题。 我想我选择插值坐标的方式是搞砸了。
编辑：接下来我尝试在这里复制@byfish的代码，并使用统一的hashmap来实现统一圆形LBP。
def uniform_circular(img, P, R):
ysize, xsize = img.shape
transformed_img = np.zeros((ysize - 2 * R,xsize - 2 * R), dtype=np.uint8)
for point in range(0, P):
x = R * math.cos(2 * math.pi * point / P)
y = -R * math.sin(2 * math.pi * point / P)
fy, fx = int(y), int(x)
cy, cx = math.ceil(y), math.ceil(x)
# calculate the fractional parts
ty = y - fy
tx = x - fx
w1 = (1 - tx) * (1 - ty)
w2 = tx * (1 - ty)
w3 = (1 - tx) * ty
w4 = tx * ty
for i in range(R, ysize - R):
for j in range(R, xsize - R):
t = w1 * img[i + fy, j + fx] + w2 * img[i + fy, j + cx] +
w3 * img[i + cy, j + fx] + w4 * img[i + cy, j + cx]
center = img[i,j]
pixels = []
res = 0
transformed_img[i - R,j - R] += (t > center) << point
for i in range(R, ysize - R):
for j in range(R, xsize - R):
transformed_img[i - R,j - R] = uniform[transformed_img[i - R,j - R]]
以下是中华民国同样的情况：
我试图在C ++中实现相同的代码。 代码如下：
#include <stdio.h>
#include <stdlib.h>
#include <opencv2/opencv.hpp>
using namespace cv;
int* uniform_circular_LBP_histogram(Mat& src) {
int i, j;
int radius = 1;
int neighbours = 8;
Size size = src.size();
int *hist_array = (int *)calloc(59,sizeof(int));
int uniform[] = {0,1,2,3,4,58,5,6,7,58,58,58,8,58,9,10,11,58,58,58,58,58,58,58,12,58,58,58,13,58,14,15,16,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,17,58,58,58,58,58,58,58,18,58,58,58,19,58,20,21,22,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,23,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,24,58,58,58,58,58,58,58,25,58,58,58,26,58,27,28,29,30,58,31,58,58,58,32,58,58,58,58,58,58,58,33,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,34,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,35,36,37,58,38,58,58,58,39,58,58,58,58,58,58,58,40,58,58,58,58,58,58,58,58,58,58,58,58,58,58,58,41,42,43,58,44,58,58,58,45,58,58,58,58,58,58,58,46,47,48,58,49,58,58,58,50,51,52,58,53,54,55,56,57};
Mat dst = Mat::zeros(size.height - 2 * radius, size.width - 2 * radius, CV_8UC1);
for (int n = 0; n < neighbours; n++) {
float x = static_cast<float>(radius) * cos(2.0 * M_PI * n / static_cast<float>(neighbours));
float y = static_cast<float>(radius) * -sin(2.0 * M_PI * n / static_cast<float>(neighbours));
int fx = static_cast<int>(floor(x));
int fy = static_cast<int>(floor(y));
int cx = static_cast<int>(ceil(x));
int cy = static_cast<int>(ceil(x));
float ty = y - fy;
float tx = y - fx;
float w1 = (1 - tx) * (1 - ty);
float w2 = tx * (1 - ty);
float w3 = (1 - tx) * ty;
float w4 = 1 - w1 - w2 - w3;
for (i = 0; i < 59; i++) {
hist_array[i] = 0;
}
for (i = radius; i < size.height - radius; i++) {
for (j = radius; j < size.width - radius; j++) {
float t = w1 * src.at<uchar>(i + fy, j + fx) +
w2 * src.at<uchar>(i + fy, j + cx) +
w3 * src.at<uchar>(i + cy, j + fx) +
w4 * src.at<uchar>(i + cy, j + cx);
dst.at<uchar>(i - radius, j - radius) += ((t > src.at<uchar>(i,j)) &&
(abs(t - src.at<uchar>(i,j)) > std::numeric_limits<float>::epsilon())) << n;
}
}
}
for (i = radius; i < size.height - radius; i++) {
for (j = radius; j < size.width - radius; j++) {
int val = uniform[dst.at<uchar>(i - radius, j - radius)];
dst.at<uchar>(i - radius, j - radius) = val;
hist_array[val] += 1;
}
}
return hist_array;
}
int main( int argc, char** argv )
{
Mat src;
int i,j;
src = imread( argv[1], 0 );
if( argc != 2 || !src.data )
{
printf( "No image data n" );
return -1;
}
const int width = 200;
const int height = 200;
Size size = src.size();
Size new_size = Size();
new_size.height = 200;
new_size.width = 200;
Mat resized_src;
resize(src, resized_src, new_size, 0, 0, INTER_CUBIC);
int count = 1;
for (i = 0; i <= width - 8; i += 25) {
for (j = 0; j <= height - 8; j += 25) {
Mat new_mat = resized_src.rowRange(i, i + 25).colRange(j, j + 25);
int *hist = uniform_circular_LBP_histogram(new_mat);
int z;
for (z = 0; z < 58; z++) {
std::cout << hist[z] << ",";
}
std::cout << hist[z] << "n";
count += 1;
}
}
return 0;
}
ROC的相同：
我也做了一个基于排名的实验。 得到了这个CMC曲线。
关于CMC曲线的一些细节：X轴表示等级。 （1-10），Y轴表示精度（0-1）。 所以，我获得了80％+ Rank1的准确性。
我不知道python，但很可能你的代码坏了。
我的建议是，按照这2个链接，并尝试将一个C ++代码移植到python。 第一个链接还包含一些关于LBP的信息。
http://www.bytefish.de/blog/local_binary_patterns/
https://github.com/berak/uniform-lbp
还有一件事我可以说，你说你正在将图像大小调整为200x200。 你为什么这样做？ 据我所知，AT＆T的图像比这还小，只是让图像变大，但我认为它不会帮助你，而且它可能会对性能产生负面影响。
我会说一些事情尝试他们
（1）ROC曲线介于错误接受率和错误拒绝率之间。
错误接受情况良好，但在上述说明中应为真正的拒绝率，而不是正确的接受率。
检查曲线正在绘制的参数。 我无法帮助你了解代码。 我不知道Python。
（2）如果结果不好，请检查LBP是否能有效解决您尝试解决的问题。 LBP主要用于纹理分析。
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