YOLOv1和YOLOv2简单看了一下，详细看了看YOLOv3,刚看的时候是蒙圈的，经过一番研究，分步记录一下几个关键的点：
v2和v3中加入了anchors和Faster rcnn有一定区别，这个anchors如何理解呢？
个人理解白话篇：
（1）就是有一批标注数据，里面有中心坐标和宽高，对应格式就是voc数据集标xml注格式即可。
代码提取标注数据里的宽高并用图像的宽高进行归一化：
def load_dataset(path):
dataset = []
for xml_file in glob.glob("{}/*xml".format(path)):
tree = ET.parse(xml_file)
height = int(tree.findtext("./size/height"))
width = int(tree.findtext("./size/width"))
for obj in tree.iter("object"):
xmin = int(obj.findtext("bndbox/xmin")) / width
ymin = int(obj.findtext("bndbox/ymin")) / height
xmax = int(obj.findtext("bndbox/xmax")) / width
ymax = int(obj.findtext("bndbox/ymax")) / height
dataset.append([xmax - xmin, ymax - ymin])
return np.array(dataset)
（2）具体怎么分的呢？就是用K-means对所有标注的bbox数据根据宽高进行分堆，voc数据被分为9个堆，距离是用的distance = 1-iou
import numpy as np ''' (1)k-means拿到数据里所有的目标框N个，得到所有的宽和高，在这里面随机取得9个作为随机中心
(2)然后其他所有的bbox根据这9个宽高依据iou(作为距离)进行计算，计算出N行9列个distance吧
(3)找到每一行中最小的那个即所有的bbox都被分到了9个当中的一个,然后计算9个族中所有bbox的中位数更新中心点。
(4）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。
''' def iou(box, clusters): """ Calculates the Intersection over Union (IoU) between a box and k clusters. :param box: tuple or array, shifted to the origin (i. e. width and height) :param clusters: numpy array of shape (k, 2) where k is the number of clusters :return: numpy array of shape (k, 0) where k is the number of clusters """ #计算每个box与9个clusters的iou # boxes ： 所有的[[width, height], [width, height], …… ] # clusters : 9个随机的中心点[width, height] x = np.minimum(clusters[:, 0], box[0]) y = np.minimum(clusters[:, 1], box[1]) if np.count_nonzero(x == 0) > 0 or np.count_nonzero(y == 0) > 0: raise ValueError("Box has no area") intersection = x * y # 所有的boxes的面积 box_area = box[0] * box[1] cluster_area = clusters[:, 0] * clusters[:, 1] iou_ = intersection / (box_area + cluster_area - intersection) return iou_ def avg_iou(boxes, clusters): """ Calculates the average Intersection over Union (IoU) between a numpy array of boxes and k clusters. :param boxes: numpy array of shape (r, 2), where r is the number of rows :param clusters: numpy array of shape (k, 2) where k is the number of clusters :return: average IoU as a single float """ return np.mean([np.max(iou(boxes[i], clusters)) for i in range(boxes.shape[0])]) def translate_boxes(boxes): """ Translates all the boxes to the origin. :param boxes: numpy array of shape (r, 4) :return: numpy array of shape (r, 2) """ new_boxes = boxes.copy() for row in range(new_boxes.shape[0]): new_boxes[row][2] = np.abs(new_boxes[row][2] - new_boxes[row][0]) new_boxes[row][3] = np.abs(new_boxes[row][3] - new_boxes[row][1]) return np.delete(new_boxes, [0, 1], axis=1) def kmeans(boxes, k, dist=np.median): """ Calculates k-means clustering with the Intersection over Union (IoU) metric. :param boxes: numpy array of shape (r, 2), where r is the number of rows :param k: number of clusters :param dist: distance function :return: numpy array of shape (k, 2) """ rows = boxes.shape[0] distances = np.empty((rows, k)) last_clusters = np.zeros((rows,)) np.random.seed() # the Forgy method will fail if the whole array contains the same rows #初始化k个聚类中心（从原始数据集中随机选择k个） clusters = boxes[np.random.choice(rows, k, replace=False)] while True: for row in range(rows): # 定义的距离度量公式：d(box,centroid)=1-IOU(box,centroid)。到聚类中心的距离越小越好， # 但IOU值是越大越好，所以使用 1 - IOU，这样就保证距离越小，IOU值越大。 # 计算所有的boxes和clusters的值（row，k） distances[row] = 1 - iou(boxes[row], clusters) #print(distances) # 将标注框分配给“距离”最近的聚类中心（也就是这里代码就是选出（对于每一个box）距离最小的那个聚类中心）。 nearest_clusters = np.argmin(distances, axis=1) # 直到聚类中心改变量为0（也就是聚类中心不变了）。 if (last_clusters == nearest_clusters).all(): break # 计算每个群的中心（这里把每一个类的中位数作为新的聚类中心） for cluster in range(k): #这一句是把所有的boxes分到k堆数据中,比较别扭，就是分好了k堆数据，每堆求它的中位数作为新的点 clusters[cluster] = dist(boxes[nearest_clusters == cluster], axis=0) last_clusters = nearest_clusters return clusters
运行代码：
import glob
import xml.etree.ElementTree as ET
import numpy as np
from kmeans import kmeans, avg_iou
#ANNOTATIONS_PATH = "Annotations"
CLUSTERS = 9
def load_dataset(path):
dataset = []
for xml_file in glob.glob("{}/*xml".format(path)):
tree = ET.parse(xml_file)
height = int(tree.findtext("./size/height"))
width = int(tree.findtext("./size/width"))
for obj in tree.iter("object"):
xmin = int(obj.findtext("bndbox/xmin")) / width
ymin = int(obj.findtext("bndbox/ymin")) / height
xmax = int(obj.findtext("bndbox/xmax")) / width
ymax = int(obj.findtext("bndbox/ymax")) / height
dataset.append([xmax - xmin, ymax - ymin])
return np.array(dataset)
ANNOTATIONS_PATH ="自己数据路径"
data = load_dataset(ANNOTATIONS_PATH)
out = kmeans(data, k=CLUSTERS)
print("Accuracy: {:.2f}%".format(avg_iou(data, out) * 100))
#print("Boxes:\n {}".format(out))
print("Boxes:\n {}-{}".format(out[:, 0]*416, out[:, 1]*416))
ratios = np.around(out[:, 0] / out[:, 1], decimals=2).tolist()
print("Ratios:\n {}".format(sorted(ratios)))
自己计算的VOC2007数据集总共9963个标签数据，跟论文中给到的有些许出入，可能是coco和voc2007的区别吧,
计算如下：
Accuracy:
67.22%
Boxes（自己修改的格式 都4舍5入了，ratios有些许对不上）:
[347,327 40,40 76,77 184,277 89,207 162,134 14,27 44,128 23,72]
Ratios:
[0.32, 0.35, 0.43, 0.55, 0.67, 0.99, 1.02, 1.06, 1.21]