# --------------------------------------------------------
# Fast/er R-CNN
# Licensed under The MIT License [see LICENSE for details]
# Written by Bharath Hariharan
# --------------------------------------------------------
from __future__ import absolute_import
from __future__ import division
from __future__ import print_function
import xml.etree.ElementTree as ET
import os
import pickle
import numpy as np
'''
评估函数
'''
# 读取xml文件
def parse_rec(filename):
""" Parse a PASCAL VOC xml file """
tree = ET.parse(filename)
objects = []
for obj in tree.findall('object'):
obj_struct = {}
obj_struct['name'] = obj.find('name').text
obj_struct['pose'] = obj.find('pose').text
obj_struct['truncated'] = int(obj.find('truncated').text)
obj_struct['difficult'] = int(obj.find('difficult').text)
bbox = obj.find('bndbox')
obj_struct['bbox'] = [int(bbox.find('xmin').text),
int(bbox.find('ymin').text),
int(bbox.find('xmax').text),
int(bbox.find('ymax').text)]
objects.append(obj_struct)
return objects
# 计算AP的函数
def voc_ap(rec, prec, use_07_metric=False):
""" ap = voc_ap(rec, prec, [use_07_metric])
Compute VOC AP given precision and recall.
If use_07_metric is true, uses the
VOC 07 11 point method (default:False).
计算AP值，若use_07_metric=true,则用11个点采样的方法，将rec从0-1分成11个点，这些点prec值求平均近似表示AP
若use_07_metric=false,则采用更为精确的逐点积分方法
"""
if use_07_metric:
# 11 point metric
ap = 0.
for t in np.arange(0., 1.1, 0.1):
if np.sum(rec >= t) == 0:
p = 0
else:
p = np.max(prec[rec >= t])
ap = ap + p / 11.
else:
# correct AP calculation
# first append sentinel values at the end
mrec = np.concatenate(([0.], rec, [1.]))
mpre = np.concatenate(([0.], prec, [0.]))
# compute the precision envelope
for i in range(mpre.size - 1, 0, -1):
mpre[i - 1] = np.maximum(mpre[i - 1], mpre[i])
# to calculate area under PR curve, look for points
# where X axis (recall) changes value
i = np.where(mrec[1:] != mrec[:-1])[0]
# and sum (\Delta recall) * prec
ap = np.sum((mrec[i + 1] - mrec[i]) * mpre[i + 1])
return ap
def voc_eval(detpath,
annopath,
imagesetfile,
classname,
cachedir,
ovthresh=0.5,
use_07_metric=False,
use_diff=False):
"""rec, prec, ap = voc_eval(detpath,
annopath,
imagesetfile,
classname,
[ovthresh],
[use_07_metric])
Top level function that does the PASCAL VOC evaluation.
detpath: Path to detections
detpath.format(classname) should produce the detection results file.
annopath: Path to annotations
annopath.format(imagename) should be the xml annotations file.
#imagesetfile，路径VOCdevkit/VOC20xx/ImageSets/Main/test.txt这里假设测试图像1000张，那么该txt文件1000行。
imagesetfile: Text file containing the list of images, one image per line.
classname: Category name (duh)
cachedir: Directory for caching the annotations
[ovthresh]: Overlap threshold (default = 0.5)
[use_07_metric]: Whether to use VOC07's 11 point AP computation
(default False)
"""
# assumes detections are in detpath.format(classname)
# assumes annotations are in annopath.format(imagename)
# assumes imagesetfile is a text file with each line an image name
# cachedir caches the annotations in a pickle file
# first load gt
# 读取真实的标签
# if cachedir不存在就创建一个
if not os.path.isdir(cachedir):
os.mkdir(cachedir)
cachefile = os.path.join(cachedir, '%s_annots.pkl' % imagesetfile)
# read list of images
with open(imagesetfile, 'r') as f:
lines = f.readlines() # 读取所有图片名
imagenames = [x.strip() for x in lines] # x.strip()代表去除开头和结尾的'\n'或者'\t'
# 如果缓存路径对应的文件没有，则读取annotations
if not os.path.isfile(cachefile):
# load annotations
# 这是一个字典
recs = {}
for i, imagename in enumerate(imagenames):
# parse_rec用于读取xml文件
recs[imagename] = parse_rec(annopath.format(imagename))
if i % 100 == 0:
print('Reading annotation for {:d}/{:d}'.format(
i + 1, len(imagenames)))
# save
print('Saving cached annotations to {:s}'.format(cachefile))
with open(cachefile, 'wb') as f:
pickle.dump(recs, f) # dump是序列化保存，load是序列化解析
else: # 如果已经有了cachefile缓存文件，直接读取
# load
with open(cachefile, 'rb') as f:
try:
recs = pickle.load(f)
except:
recs = pickle.load(f, encoding='bytes')
# extract gt objects for this class
#
class_recs = {} # 当前类别的标注
npos = 0
for imagename in imagenames:
# recs[imagename]是保存了图片的object里面的所有属性，是个字典
# 值保留指定类别的项
R = [obj for obj in recs[imagename] if obj['name'] == classname]
# 获得所有的bbox,里面保存了xmin,ymin,xmax,ymax
bbox = np.array([x['bbox'] for x in R])
if use_diff: # 如果使用difficult(难检测的),所有的值都是false
difficult = np.array([False for x in R]).astype(np.bool)
else: # 否则里面的内容有1有0
difficult = np.array([x['difficult'] for x in R]).astype(np.bool)
# len(R)就是当前类别的个数
# 开辟一个全为False长度是len(R)的数组
det = [False] * len(R)
# 我测试~difficult的意思是取相反数之后再减1，这是什么意思。。。
npos = npos + sum(~difficult)
class_recs[imagename] = {'bbox': bbox,
'difficult': difficult,
'det': det}
# read dets
# dets是检测结果的路径，读取出来就是图片名字、得分、bbox四个值。
detfile = detpath.format(classname)
# 读取txt里面的所有内容
with open(detfile, 'r') as f:
lines = f.readlines()
# 这里的操作x.strip().split(' ')首先去掉了每行开头和结尾的'\n'或'\t'然后去除每行的' '
# 之前再运行代码时，报错时一直以为空格会影响文件的读入，还专门写了函数去去除所有额外的空格。。多余了
splitlines = [x.strip().split(' ') for x in lines]
# 以图片的名称作为下标
image_ids = [x[0] for x in splitlines] # x[0]为名称
confidence = np.array([float(x[1]) for x in splitlines]) # x[1]为得分
BB = np.array([[float(z) for z in x[2:]] for x in splitlines]) # x[2]为bbox的4个值
# 统计检测的目标数量
nd = len(image_ids)
# tp:正类预测为正类->A预测成A
# fp:正类预测为负类->A预测成B
tp = np.zeros(nd)
fp = np.zeros(nd)
if BB.shape[0] > 0: # 行数>0
# sort by confidence
# 按照分数从大到小排序，返回下标
sorted_ind = np.argsort(-confidence)
# 按照分数从大到小排序，返回分数
# 下面也没有用到该变量，其实sorted_ind得到之后，直接根据confidence[sorted_ind[i]]就可以得到sorted_scores
sorted_scores = np.sort(-confidence)
# 对BB也重排一下
BB = BB[sorted_ind, :]
# image_ids也重排
image_ids = [image_ids[x] for x in sorted_ind]
# 上面这些操作就是为了下标对应起来，后面好操作
# go down dets and mark TPs and FPs
for d in range(nd):
'''
class_recs[imagename] = {'bbox': bbox,
'difficult': difficult,
'det': det}
'''
# 由image_ids[d]获取名称。然后得到R
R = class_recs[image_ids[d]]
# dets是检测结果的路径,BB通过dets获取每一行的数据，然后得到对应的BB值(也就是4个属性)
bb = BB[d, :].astype(float)
# 设置一个负无穷
ovmax = -np.inf
# BBGT是真实的坐标
BBGT = R['bbox'].astype(float)
if BBGT.size > 0: # 如果存在GT计算交并比，如果不存在，首先就是检测错误
# compute overlaps
# intersection
# 得到重叠区域，也就是左上角坐标取最大，右下角坐标取最小
ixmin = np.maximum(BBGT[:, 0], bb[0])
iymin = np.maximum(BBGT[:, 1], bb[1])
ixmax = np.minimum(BBGT[:, 2], bb[2])
iymax = np.minimum(BBGT[:, 3], bb[3])
iw = np.maximum(ixmax - ixmin + 1., 0.)
ih = np.maximum(iymax - iymin + 1., 0.)
# 计算重叠区域的面积
inters = iw * ih
# union
# 并集面积就是两个区域的面积减去重叠区域的面积
uni = ((bb[2] - bb[0] + 1.) * (bb[3] - bb[1] + 1.) +
(BBGT[:, 2] - BBGT[:, 0] + 1.) *
(BBGT[:, 3] - BBGT[:, 1] + 1.) - inters)
# 计算IOU，注意这个overlaps不一定是一个数值，可能是一个列表的,所有后面才有np.max和np.argmax
overlaps = inters / uni
# 保留最大的IOU
ovmax = np.max(overlaps)
# 保留最大IOU的下标
jmax = np.argmax(overlaps)
if ovmax > ovthresh: # 这个阙值默认0.5
if not R['difficult'][jmax]: # 这个后面是不是少个else，如果是难测样本呢？？
# R = class_recs[image_ids[d]]
if not R['det'][jmax]: # R['det']初始值全为False,意思应该是如果该位置第一次使用，才可以。那也会出现tp[d]=fp[d]=1的情况啊
# 下面都是标记
tp[d] = 1.
R['det'][jmax] = 1
else:
fp[d] = 1.
else:
fp[d] = 1.
# compute precision recall
# 我测试cumsum是前缀和的意思？？？？ 不懂
fp = np.cumsum(fp)
tp = np.cumsum(tp)
rec = tp / float(npos)
# avoid divide by zero in case the first detection matches a difficult
# ground truth
prec = tp / np.maximum(tp + fp, np.finfo(np.float64).eps)
ap = voc_ap(rec, prec, use_07_metric)
return rec, prec, ap
Faster Rcnn 代码解读之 voc_eval.py
猜你喜欢
转载自blog.csdn.net/qq_33193309/article/details/98616019
今日推荐
周排行