介绍
YOLOv2作者说 anchors 是使用 K-MEANS 计算得出的，计算出哪种长宽的框比较多，使用这样 长宽 的 anchors 可以加快收敛速度
YOLOv2 vs YOLOv3
YOLOv2 的anchor大小
作者用最后一层feature map的相对大小来定义anchor大小。
也就是说，在YOLOv2中的神经网络，最后一层feature map大小为 13X13（不同的神经网络，最后的feature map大小不一样的），那么anchor大小范围就在（0x0，13x13]。
如果一个anchor大小是9x9，那么其在原图上的实际大小是288x288。
YOLOv3 的anchor大小
在 YOLOv3 中，作者改用 输入图片的大小 来定义anchor，anchor的大小为（0x0，input_w x input_h]。
所以，在YOLOv2 和 YOLOv3 的 cfg文件中，anchor的大小有明显的区别。
代码
根据 https://github.com/lars76/kmeans-anchor-boxes 工程修改
example.py
修改下面常量：
ANNOTATIONS_PATH：xml文件路径CLUSTERS：计算得出的 anchors 数量，YOLOv2用5个anchor，所以是5
修改13：
print("Boxes:\n {}".format(out*13))
如果是YOLOv2，archors大小是根据最后一层的feature map大小计算的（YOLOv2 默认最后一层的feature map大小是13，所以这里 乘以
13）
在具体操作时，要根据神经网络的最后一层的feature map大小来设置如果是YOLOv3，archors大小是根据输入图片的大小计算的，这里要改成输入图片的大小
import glob
import xml.etree.ElementTree as ET
import numpy as np
from kmeans import kmeans, avg_iou
ANNOTATIONS_PATH = "/Users/mym/Desktop/ai/yolov2-tf2/data/train/annotation"
CLUSTERS = 5
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)
data = load_dataset(ANNOTATIONS_PATH)
out = kmeans(data, k=CLUSTERS)
print("Accuracy: {:.2f}%".format(avg_iou(data, out) * 100))
print("Boxes:\n {}".format(out*13))
ratios = np.around(out[:, 0] / out[:, 1], decimals=2).tolist()
print("Ratios:\n {}".format(sorted(ratios)))
执行结果如下：
Accuracy: 62.08%
Boxes:
[[0.46875 0.40625 ]
[4.90625 5.109375]
[2.59375 3.046875]
[1.25 1.28125 ]
[0.8125 0.59375 ]]
Ratios:
[0.85, 0.96, 0.98, 1.15, 1.37]
kmeans.py
kmeans.py算法实现，不需要改动
import numpy as np
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
"""
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
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
clusters = boxes[np.random.choice(rows, k, replace=False)]
while True:
for row in range(rows):
distances[row] = 1 - iou(boxes[row], clusters)
nearest_clusters = np.argmin(distances, axis=1)
if (last_clusters == nearest_clusters).all():
break
for cluster in range(k):
clusters[cluster] = dist(boxes[nearest_clusters == cluster], axis=0)
last_clusters = nearest_clusters
return clusters