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YOLO2 Get Anchors
# -*- coding: utf-8 -*-
from __future__ import absolute_import
from __future__ import division
from __future__ import print_function
import argparse
import numpy as np
import os
import random
from tqdm import tqdm
import sklearn.cluster as cluster
def iou(x, centroids):
dists = []
for centroid in centroids:
c_w, c_h = centroid
w, h = x
if c_w >= w and c_h >= h:
dist = w * h / (c_w * c_h)
elif c_w >= w and c_h <= h:
dist = w * c_h / (w * h + (c_w - w) * c_h)
elif c_w <= w and c_h >= h:
dist = c_w * h / (w * h + c_w * (c_h - h))
else: # means both w,h are bigger than c_w and c_h respectively
dist = (c_w * c_h) / (w * h)
dists.append(dist)
return np.array(dists)
def avg_iou(x, centroids):
n, d = x.shape
sums = 0.
for i in range(x.shape[0]):
# note IOU() will return array which contains IoU for each centroid and X[i]
# slightly ineffective, but I am too lazy
sums += max(iou(x[i], centroids))
return sums / n
def write_anchors_to_file(centroids, distance, anchor_file):
anchors = centroids * 416 / 32 # I do not know whi it is 416/32
anchors = [str(i) for i in anchors.ravel()]
print(
"\n",
"Cluster Result:\n",
"Clusters:", len(centroids), "\n",
"Average IoU:", distance, "\n",
"Anchors:\n",
", ".join(anchors)
)
with open(anchor_file, 'w') as f:
f.write(", ".join(anchors))
f.write('\n%f\n' % distance)
def k_means(x, n_clusters, eps):
init_index = [random.randrange(x.shape[0]) for _ in range(n_clusters)]
centroids = x[init_index]
d = old_d = []
iterations = 0
diff = 1e10
c, dim = centroids.shape
while True:
iterations += 1
d = np.array([1 - iou(i, centroids) for i in x])
if len(old_d) > 0:
diff = np.sum(np.abs(d - old_d))
print('diff = %f' % diff)
if diff < eps or iterations > 1000:
print("Number of iterations took = %d" % iterations)
print("Centroids = ", centroids)
return centroids
# assign samples to centroids
belonging_centroids = np.argmin(d, axis=1)
# calculate the new centroids
centroid_sums = np.zeros((c, dim), np.float)
for i in range(belonging_centroids.shape[0]):
centroid_sums[belonging_centroids[i]] += x[i]
for j in range(c):
centroids[j] = centroid_sums[j] / np.sum(belonging_centroids == j)
old_d = d.copy()
def get_file_content(fnm):
with open(fnm) as f:
return [line.strip() for line in f]
def main(args):
print("Reading Data ...")
file_list = []
for f in args.file_list:
file_list.extend(get_file_content(f))
data = []
for one_file in tqdm(file_list):
one_file = one_file.replace('images', 'labels') \
.replace('JPEGImages', 'labels') \
.replace('.png', '.txt') \
.replace('.jpg', '.txt')
for line in get_file_content(one_file):
clazz, xx, yy, w, h = line.split()
data.append([float(w),float(h)])
data = np.array(data)
if args.engine.startswith("sklearn"):
if args.engine == "sklearn":
km = cluster.KMeans(n_clusters=args.num_clusters, tol=args.tol, verbose=True)
elif args.engine == "sklearn-mini":
km = cluster.MiniBatchKMeans(n_clusters=args.num_clusters, tol=args.tol, verbose=True)
km.fit(data)
result = km.cluster_centers_
# distance = km.inertia_ / data.shape[0]
distance = avg_iou(data, result)
else:
result = k_means(data, args.num_clusters, args.tol)
distance = avg_iou(data, result)
write_anchors_to_file(result, distance, args.output)
if "__main__" == __name__:
parser = argparse.ArgumentParser()
parser.add_argument('file_list', nargs='+', help='TrainList')
parser.add_argument('--num_clusters', '-n', default=5, type=int, help='Number of Clusters')
parser.add_argument('--output', '-o', default='../results/anchor.txt', type=str, help='Result Output File')
parser.add_argument('--tol', '-t', default=0.005, type=float, help='Tolerate')
parser.add_argument('--engine', '-m', default='sklearn', type=str,
choices=['original', 'sklearn', 'sklearn-mini'], help='Method to use')
args = parser.parse_args()
main(args)
@jinyu121

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@jinyu121 jinyu121 commented Oct 17, 2017

实际上这里“使用SkLearm来计算”是有问题的。应该是使用长宽比/IoU来进行计算的。

但是实验发现,两种方法生成的anchors差别不大,精度也大致相同。(原谅我对数字不太敏感),所以个人认为是可以通用的。

@DC-Shi

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@DC-Shi DC-Shi commented Dec 4, 2017

anchors差别可能还和数据有关,我在自己的标注图片上跑的,用同一种算法得到的结果差别不大(和随机初始值有关,但是结果也就差0.1不到),但是不同算法得到的就不一样了

engine result
sklearn 11.22762995, 10.73226759, 10.68771405, 9.1692398, 7.452993003, 6.555998014, 6.299477413, 4.831884219, 3.57714225, 3.068625678
original 10.83653525, 10.64363488, 10.54793825, 7.023316928, 6.781054383, 6.024040211, 6.005671164, 4.968083672, 3.417490904, 1.448678908

另外 anchors = centroids * 416 / 32 这一行,是因为输入维度是416,里边有5次pooling,也就是缩小了32

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@PythonImageDeveloper PythonImageDeveloper commented Mar 12, 2018

Hi ,
this file generate 10 values of anchors , i have question about these values , as we have 5 anchors and this generator generate 10 values, more likely a first two of 10 values related to first anchor box , right ? if so , what are means of these two values ? W , H for first anchors for aspect ratio and scale for that anchor?

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@jinyu121 jinyu121 commented Mar 28, 2018

@zeynali

The 10 values can be grouped as 5 pairs. For example, 11.22762995, 10.73226759, 10.68771405, 9.1692398, 7.452993003, 6.555998014, 6.299477413, 4.831884219, 3.57714225, 3.068625678 means (11.22762995, 10.73226759), (10.68771405, 9.1692398), (7.452993003, 6.555998014), (6.299477413, 4.831884219), (3.57714225, 3.068625678)

In my view, the values is H and W in some scale. (So we can just multiply or add them to the output of the net)

@muzi1012

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@muzi1012 muzi1012 commented Apr 17, 2018

大家谁能告诉我那个file_list的参数应该怎么写啊

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@jinyu121 jinyu121 commented Apr 18, 2018

@muzi1012

The txt file can generated by this file. Each file contains multi lines, each line is a full path of one image.

For example, if you have the training list(s) like this:

001
002
003

and

101
102
103

After the processing by voc_label.py, you may get files like

train_part_1.txt

path_to_voc/VOC2007/JPEGImages/001.png
path_to_voc/VOC2007/JPEGImages/002.png
path_to_voc/VOC2007/JPEGImages/003.png
....

train_part_2.txt

path_to_voc/VOC2007/JPEGImages/101.png
path_to_voc/VOC2007/JPEGImages/102.png
path_to_voc/VOC2007/JPEGImages/103.png
....

Then, you can use python ./get_anchor.py train_part_1.txt train_part_2.txt to get anchors.

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@xieyufei1993 xieyufei1993 commented Nov 2, 2018

得到的这10个值,两两相除,得到的就是需要设置的ratios吗?

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