I am trying to implement a basic face recognition system using Uniform Circular LBP (8 points in the neighborhood of a unit radius). I take an image by redrawing it to 200 x 200 pixels, and then breaking the image into 8x8 small images . Then I calculate a histogram for each small image and get a list of histograms . To compare 2 images , I calculate the chi-square distance between the corresponding histograms and generate an account.
Here is my unified 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 the AT & T database, taking 2 gallery images and 8 test images per object. The ROC for the experiment came out as follows:
In the above ROC, the x axis indicates the rate of false reception , and the y axis indicates the actual rate of reception . Accuracy is apparently poor according to LBP standards. I am sure something is wrong with my implementation. It would be great if someone helped me with this. Thanks for reading.
EDIT:
I think I made a mistake in the code above. I am going clockwise, and the paper on the LBP suggests that I should go counterclockwise when assigning the weights. Line: c = R * math.sin(2 * math.pi * point / P) should be c = -R * math.sin(2 * math.pi * point / P) . The results after editing are even worse. This suggests that something is wrong with my code. I assume that the way to select the coordinates for interpolation is corrupted.
Edit: next I tried to replicate @bytefish code here and used a single hash file to implement 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 ranking based experiment. And got that CMC curve.
Some details about the CMC curve: the X axis represents the series. (1-10), and the Y axis is accuracy (0-1). So, I got an accuracy of 80% + Rank1.