YOLO v3 中关于 anchor 的 k-means 聚类代码
文章目录
- 1. k-means 聚类代码
- 2. YOLOv3 中默认的 anchor
- 3. github 上的代码复制如下：
- 4. 有用请点赞，谢谢
1. k-means 聚类代码
我使用的代码是：https://github.com/lars76/kmeans-anchor-boxes
其他的k-means 代码（没用过）是：
- https://github.com/qqwweee/keras-yolo3/blob/master/kmeans.py
- https://github.com/AlexeyAB/darknet/blob/master/scripts/gen_anchors.py
输入是存放 xml标签文件的文件夹：
只需要更改 example.py 中的一行代码：
ANNOTATIONS_PATH = "xmlLabel/train" # 更改自己的路径（存放训练标签 xml 的文件路径）
- 运行
example.py计算当前数据集的需要设置的 anchor 的大小（相对于416输入而言）
在我的数据集上的输出结果如下：
rows = 8607 # 我的 label 目标的数量
[[0.01416016 0.015625 ] # 每一个 anchor的宽/图像的宽 ，高/高
[0.00830078 0.00927734]
[0.06542969 0.06982422]
[0.03417969 0.03662109]
[0.01123047 0.01220703]
[0.02685547 0.02832031]
[0.01757812 0.01953125]
[0.04443359 0.04833984]
[0.02148438 0.0234375 ]]
Accuracy: 83.41%
Boxes:
[ 5.890625 3.453125 27.21875 14.21875 4.671875 11.171875 7.3125 18.484375 8.9375 ]- # 每个 anchor 的宽
[ 6.5 3.859375 29.046875 15.234375 5.078125 11.78125 8.125 20.109375 9.75 ] # # 每个 anchor 的高
Ratios:
[0.89, 0.9, 0.91, 0.92, 0.92, 0.92, 0.93, 0.94, 0.95] # 每个 anchor 的 宽/高
- 对输入anchor 进行排序后的结果是（取整数是为了好看）：
[3, 4, 5, 7, 8, 11, 14, 18, 27]
[3, 5, 6, 8, 9, 11, 15, 20, 29]
anchor_416 = 3, 3, 4, 5, 5, 6, 7, 8, 8, 9, 11, 11, 14, 15, 18, 20, 27, 29
anchor_416_2 = 6, 7, 9, 10, 11, 13, 14, 16, 17, 19, 22, 23, 28, 30, 36, 40, 54, 58
anchor_416_3 = 10, 11, 14, 15, 17, 19, 21, 24, 26, 29, 33, 35, 42, 45, 55, 60, 81, 87
anchor_416_4 = 13, 15, 18, 20, 23, 26, 29,32, 35,39, 44,47, 56,60, 73,80, 108,116
anchor_416_5 = 17, 19, 23, 25, 29, 32, 36, 40, 44, 48, 55, 58, 71, 76, 92, 100, 136, 145
将 anchor 排序的代码如下（自己写的）：
import numpy as np
# anchors = [10,13, 16,30, 33,23, 30,61, 62,45, 59,119, 116,90, 156,198, 373,326]
# for i in range(0, len(anchors), 2):
# print(anchors[i] * anchors[i + 1])
x = [5.890625, 3.453125, 27.21875, 14.21875, 4.671875, 11.171875, 7.3125, 18.484375, 8.9375]
y = [6.5, 3.859375, 29.046875, 15.234375, 5.078125, 11.78125, 8.125, 20.109375, 9.75 ]
area = []
for i in range(len(x)):
area.append(x[i] * y[i])
print(area)
print(np.argsort(area))
new_x = [0 for _ in range(len(x))]
new_y = [0 for _ in range(len(y))]
for i in range(len(np.argsort(area))):
new_x[i] = int(x[np.argsort(area)[i]])
new_y[i] = int(y[np.argsort(area)[i]])
anchors = []
for i in range(len(new_x)):
anchors.append(new_x[i])
anchors.append(new_y[i])
print(anchors)
for i in range(len(new_x)):
print(new_x[i] * new_y[i])
2. YOLOv3 中默认的 anchor
anchors = 10,13, 16,30, 33,23, 30,61, 62,45, 59,119, 116,90, 156,198, 373,326
一共有 18个数字，9个anchor，每一个anchor的大小，面积依次是：
130, 480,759, 1830, 2790, 7021, 10440, 30888, 121598
3. github 上的代码复制如下：
example.py
import glob
import xml.etree.ElementTree as ET
import numpy as np
from kmeans import kmeans, avg_iou
ANNOTATIONS_PATH = "Annotations"
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))
ratios = np.around(out[:, 0] / out[:, 1], decimals=2).tolist()
print("Ratios:\n {}".format(sorted(ratios)))
kmeans.py
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
4. 有用请点赞，谢谢
完