YOLO K-means获取anchors大小代码详解
预备知识：应该了解yolo的基本操作，详见YOLO v1，YOLO v2，YOLO v3。
首先应该了解yolo标签文件的格式，其格式为：图片的位置 框的4个坐标和1个类别ID (xmin,ymin,xmax,ymax,id) …。示例如下：
/home/aift/CV/detect/yolo3-keras-master/VOCdevkit/VOC2007/JPEGImages/000012.jpg 156,97,351,270,6
/home/aift/CV/detect/yolo3-keras-master/VOCdevkit/VOC2007/JPEGImages/000017.jpg 185,62,279,199,14 90,78,403,336,12
/home/aift/CV/detect/yolo3-keras-master/VOCdevkit/VOC2007/JPEGImages/000023.jpg 9,230,245,500,1 230,220,334,500,1 2,1,117,369,14 3,2,243,462,14 225,1,334,486,14
/home/aift/CV/detect/yolo3-keras-master/VOCdevkit/VOC2007/JPEGImages/000026.jpg 90,125,337,212,6
/home/aift/CV/detect/yolo3-keras-master/VOCdevkit/VOC2007/JPEGImages/000032.jpg 104,78,375,183,0 133,88,197,123,0 195,180,213,229,14 26,189,44,238,14
/home/aift/CV/detect/yolo3-keras-master/VOCdevkit/VOC2007/JPEGImages/000033.jpg 9,107,499,263,0 421,200,482,226,0 325,188,411,223,0
/home/aift/CV/detect/yolo3-keras-master/VOCdevkit/VOC2007/JPEGImages/000034.jpg 116,167,360,400,18 141,153,333,229,18
/home/aift/CV/detect/yolo3-keras-master/VOCdevkit/VOC2007/JPEGImages/000035.jpg 1,96,191,361,14 218,98,465,318,14
……
实现代码如下：（注释已经很详细了）
import numpy as np
'''
k-means拿到数据里所有的目标框，得到所有的宽和高，在这里面随机取得9个随机中心，之后以9个点为中心得到9个族，不断计算其他点到中点的距离调整
每个点所归属的族和中心，直到9个中心不再变即可。这9个中心的x，y就是整个数据的9个合适的anchors==框的宽和高。
部分注释来源：https://www.jianshu.com/p/3fddf7c08a58
建议单步调试看运行结果
'''
class YOLO_Kmeans:
def __init__(self, cluster_number, in_filename, out_filename):
# 读取kmeans的中心数
self.cluster_number = cluster_number
# 标签文件的文件名
self.in_filename = in_filename # "2007_train.txt"
self.out_filename = out_filename # "2007_train_clusters.txt"
# 这里注意对每个box算其与9个clusters的iou
def iou(self, boxes, clusters): # 1 box -> k clusters
# boxes ： 所有的[[width, height], [width, height], …… ]
# clusters : 9个随机的中心点[width, height]
n = boxes.shape[0] # 6301
k = self.cluster_number # 9
# 所有的boxes的面积
box_area = boxes[:, 0] * boxes[:, 1] # w * h (6301,)
# 将box_area的每个元素重复k次
box_area = box_area.repeat(k) # (56709,)
box_area = np.reshape(box_area, (n, k)) # (6301, 9)
# 计算9个中点的面积
cluster_area = clusters[:, 0] * clusters[:, 1] # w * h (9,)
# 对cluster_area进行复制n份
cluster_area = np.tile(cluster_area, [1, n]) # (1, 56709)
cluster_area = np.reshape(cluster_area, (n, k)) # (6301, 9)
# 获取box和中心的交叉w的宽 w
box_w_matrix = np.reshape(boxes[:, 0].repeat(k), (n, k)) # (6301, 9)
cluster_w_matrix = np.reshape(np.tile(clusters[:, 0], (1, n)), (n, k)) # (6301, 9)
min_w_matrix = np.minimum(cluster_w_matrix, box_w_matrix) # (6301, 9)
# 获取box和中心的交叉h的高 h
box_h_matrix = np.reshape(boxes[:, 1].repeat(k), (n, k))
cluster_h_matrix = np.reshape(np.tile(clusters[:, 1], (1, n)), (n, k))
min_h_matrix = np.minimum(cluster_h_matrix, box_h_matrix)
# 交叉点的面积
inter_area = np.multiply(min_w_matrix, min_h_matrix) # (6301, 9) 6301个boxes与9个clusters的交集面积
# 9个交叉点和所有的boxes的iou值
result = inter_area / (box_area + cluster_area - inter_area)
return result # (6301, 9) 6301个boxes与9个clusters的iou
def avg_iou(self, boxes, clusters):
# 计算9个中点与所有的boxes总的iou，n个点的平均iou
accuracy = np.mean([np.max(self.iou(boxes, clusters), axis=1)])
return accuracy
def kmeans(self, boxes, k, dist=np.median):
# boxes = [[宽， 高], [宽， 高], …… ]
# k 中心点数
# np.median 求众数
box_number = boxes.shape[0] # VOC2007: 6301
distances = np.empty((box_number, k)) # (6301, 9)
# 每个boxes属于哪个cluster
last_nearest = np.zeros((box_number,)) # (6301,)
np.random.seed()
# 从所有的boxe中选取9个随机中心点(w, h) https://www.cnblogs.com/cloud-ken/p/9931273.html
clusters = boxes[np.random.choice(box_number, k, replace=False)] # init k clusters (9, 2)
while True:
# 计算所有的boxes和clusters的值（n，k）
distances = 1 - self.iou(boxes, clusters) # (6301, 9) 6301个boxes与9个clusters的iou距离(1 - iou) (6301, 9)
# 选取iou值最小的点（n，）
current_nearest = np.argmin(distances, axis=1) # (6301,)
# 中心点未改变，跳出
if (last_nearest == current_nearest).all():
break # clusters won't change
# 计算每个群组的中心或者众数
for cluster in range(k):
clusters[cluster] = dist(boxes[current_nearest == cluster], axis=0) # update clusters
# 改变中心点（每个boxes属于哪个cluster）
last_nearest = current_nearest
return clusters
def result2txt(self, result):
# 把9个中心点，写入txt文件
f = open(self.out_filename, 'w')
row = np.shape(result)[0] # 9
for i in range(row):
if i == 0:
x_y = "%d,%d" % (result[i][0], result[i][1])
else:
x_y = ", %d,%d" % (result[i][0], result[i][1])
f.write(x_y)
f.close()
def txt2boxes(self):
# 打开文件
f = open(self.in_filename, 'r')
dataSet = [] # list
# 读取文件
for line in f:
infos = line.split(" ")
length = len(infos)
# infons[0] 为图片的名称
for i in range(1, length):
# 获取文件的宽和高 这里说明Bbox的标注信息是左上角和右下角标注
width = int(infos[i].split(",")[2]) - \
int(infos[i].split(",")[0])
height = int(infos[i].split(",")[3]) - \
int(infos[i].split(",")[1])
dataSet.append([width, height]) # [(w, h), (w, h), ……]
result = np.array(dataSet)
f.close()
return result
def txt2clusters(self):
# 获取所有的文件目标的宽和高,width, height
all_boxes = self.txt2boxes() # ndarray: [(w, h), (w, h), ……]
# result 9个聚类中心点
result = self.kmeans(all_boxes, k=self.cluster_number) # (9, 2)
# 按第一列顺序排序(w) https://www.cnblogs.com/hellcat/p/6874175.html#_label0_0
result = result[np.lexsort(result.T[0, None])] # (9, 2) result.T[0, None]: (1, 9)
# 把结果写入txt文件
self.result2txt(result)
print("K anchors:\n {}".format(result))
# 计算9个中点与所有的boxes总的iou，n个点的平均iou
print("Accuracy: {:.2f}%".format(
self.avg_iou(all_boxes, result) * 100))
if __name__ == "__main__":
cluster_number = 9 # clusters数量
in_filename = '2007_train.txt' # 输入标签文件
out_filename = '2007_train_clusters.txt' # 输出clusters文件
kmeans = YOLO_Kmeans(cluster_number, in_filename, out_filename)
kmeans.txt2clusters()
done~