均匀圆LBP面部识别实现 [英] Uniform Circular LBP face recognition implementation
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问题描述
我正在尝试使用统一圆LBP（1个单位半径邻域中的8个点）实现基本的人脸识别系统。
我正在拍摄图片，将其调整为200 x 200 像素，然后在8x8小图片中分割图片。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个图片，我计算相应直方图之间的卡方距离，并生成分数。
这是我的统一LBP实施：
import numpy as np
import math
uniform = {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： 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： 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： 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： 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： 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： 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）
＃计算小数部分
ty = y - fy
tx = x - fx
w1 =（1 - tx）*（1 - ty）
w2 = tx * b $ b 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像素）：
out = []
for a（以像素为单位）：
如果> center：
out.append（1）
else：
out.append（0）
return out
def uniform_circular ，P，R）：
ysize，xsize = img.shape
transformed_img = np.zeros（（ysize - 2 * R，xsize - 2 * R），dtype = np.uint8）$ b $范围（R，len（img [0]） - R）中的x的
：
center = img [y，x]
pixels = []
对于范围内的点（0，P）：
r = R * math.cos（2 * math.pi * point / P）
c = R * math.sin（2 * math.pi * point / P）
pixels.append（bilinear_interpolation（y，x，r，c，img））
values = ）
res = 0
for a in range（0，P）：
res + = values [a]< a
transformed_img.itemset（（y - R，x - R），uniform [res]）
transformed_img = transformed_img [R：-R，R：-R]
return transformed_img
我在，并使用统一的hashmap来实现统一循环LBP。
def uniform_circular（img，P，R）：
ysize，xsize = img.shape
transformed_img = np.zeros（（ysize - 2 * R，xsize - 2 * R） ，dtype = np.uint8）
对于范围（0，P）中的点：$ b $ bx = 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）
＃计算小数部分
ty = y - fy
tx = x - fx
w1 =（1 - tx）*范围（R，R）中的i的1-ty）
w2 = tx *（1-ty）
w3 =（1-tx）* ty
w4 = tx * ty
ysize-R）：
for range（R，xsize-R）中的j：
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]
像素= []
res = 0
transformed_img [i-R，j-R] + =（t>中心）<< （R，xsize-R）中的j的
：
converted_img [i-R，j-R ] = uniform [transformed_img [i-R，j-R]]
：
/ p>
我试图在C ++中实现相同的代码。这是代码：
#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 neighbors = 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
float y = static_cast< float>（radius）* -sin（2.0 * M_PI * n / static_cast< float>（neighbors）
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
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））&& \
（t-src.at< uchar>（i，j））> std :: numeric_limits< float> :: epsilon（））） n;
}
}
}
for（i = radius; i
int val = uniform [dst.at （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 \\\
）;
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] \\\
;
count + = 1;
}
}
return 0;
}
ROC同样：
>
也做了一个基于排名的实验。并获得此CMC曲线。
有关CMC曲线的一些详细信息：X轴表示秩。 （1-10），Y轴表示精度（0-1）。所以，我得到了80％+ Rank1的准确性。
解决方案
我不知道python， 。
我的建议是，按照这两个链接，尝试将一个C ++代码移植到python。第一个链接还包含有关LBP的一些信息。
http://www.bytefish.de/blog/local_binary_patterns/
https://github.com/berak/uniform-lbp
还有一件事我可以说，你说您正在将图片调整为200x200。你为什么这样做？据我所知，AT& T图像小于那些，你只是使图像更大，但我不认为它会帮助你，而且它可能对性能有负面影响。
I am trying to implement a basic face recognition system using Uniform Circular LBP (8 Points in 1 unit radius neighborhood). I am taking an image, re-sizing it to 200 x 200 pixels and then splitting the image in 8x8 little images. I then compute the histogram for each little image and get a list of histograms. To compare 2 images, I compute chi-squared distance between the corresponding histograms and generate a score.
Here's my Uniform LBP implementation:
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
I did an experiment on AT&T database taking 2 gallery images and 8 probe images per subject. The ROC for the experiment came out to be:
In the above ROC, x axis denotes the false accept rate and the y axis denotes the genuine accept rate. The accuracy seems to be poor according to Uniform LBP standards. I am sure there is something wrong with my implementation. It would great if someone could help me with it. Thanks for reading.
EDIT:
I think I made a mistake in the above code. I am going clockwise while the paper on LBP suggest that I should go anticlockwise while assigning weights. The line: c = R * math.sin(2 * math.pi * point / P) should be c = -R * math.sin(2 * math.pi * point / P). Results after the edit are even worse. This suggests something is way wrong with my code. I guess the way I am choosing the coordinates for interpolation is messed up.
Edit: next I tried to replicate @bytefish's code here and used the uniform hashmap to make the implementation Uniform Circular 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]]
Here's the ROC for the same:
I tried to implement the same code in C++. Here is the code:
#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 for the same:
I also did a rank based experiment. And got this CMC curve.
Some details about the CMC curve: X axis represents ranks. (1-10) and Y axis represents the accuracy (0-1). So, I got a 80%+ Rank1 accuracy.
解决方案
I don't know about python but most probably your code is broken.
My advice is, follow these 2 links, and try to port one of the C++ codes to python. First link also contains some information about LBP.
http://www.bytefish.de/blog/local_binary_patterns/
https://github.com/berak/uniform-lbp
And one other thing I can say, you said you are resizing images into 200x200. Why are you doing that? As far as I know AT&T images are smaller than that, your are just making images bigger but I don't think it is going to help you, moreover it may have a negative effect in performance.
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