# --------------------------------------------------------
# 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
'''
評估函數
'''
# 讀取xml檔案
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
# 計算AP的函數
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).
計算AP值，若use_07_metric=true,則用11個點采樣的方法，将rec從0-1分成11個點，這些點prec值求平均近似表示AP
若use_07_metric=false,則采用更為精确的逐點積分方法
"""
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，路徑VOCdevkit/VOC20xx/ImageSets/Main/test.txt這裡假設測試圖像1000張，那麼該txt檔案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
# 讀取真實的标簽
# if cachedir不存在就建立一個
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() # 讀取所有圖檔名
imagenames = [x.strip() for x in lines] # x.strip()代表去除開頭和結尾的'\n'或者'\t'
# 如果緩存路徑對應的檔案沒有，則讀取annotations
if not os.path.isfile(cachefile):
# load annotations
# 這是一個字典
recs = {}
for i, imagename in enumerate(imagenames):
# parse_rec用于讀取xml檔案
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是序列化儲存，load是序列化解析
else: # 如果已經有了cachefile緩存檔案，直接讀取
# 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 = {} # 目前類别的标注
npos = 0
for imagename in imagenames:
# recs[imagename]是儲存了圖檔的object裡面的所有屬性，是個字典
# 值保留指定類别的項
R = [obj for obj in recs[imagename] if obj['name'] == classname]
# 獲得所有的bbox,裡面儲存了xmin,ymin,xmax,ymax
bbox = np.array([x['bbox'] for x in R])
if use_diff: # 如果使用difficult(難檢測的),所有的值都是false
difficult = np.array([False for x in R]).astype(np.bool)
else: # 否則裡面的内容有1有0
difficult = np.array([x['difficult'] for x in R]).astype(np.bool)
# len(R)就是目前類别的個數
# 開辟一個全為False長度是len(R)的數組
det = [False] * len(R)
# 我測試~difficult的意思是取相反數之後再減1，這是什麼意思。。。
npos = npos + sum(~difficult)
class_recs[imagename] = {'bbox': bbox,
'difficult': difficult,
'det': det}
# read dets
# dets是檢測結果的路徑，讀取出來就是圖檔名字、得分、bbox四個值。
detfile = detpath.format(classname)
# 讀取txt裡面的所有内容
with open(detfile, 'r') as f:
lines = f.readlines()
# 這裡的操作x.strip().split(' ')首先去掉了每行開頭和結尾的'\n'或'\t'然後去除每行的' '
# 之前再運作代碼時，報錯時一直以為空格會影響檔案的讀入，還專門寫了函數去去除所有額外的空格。。多餘了
splitlines = [x.strip().split(' ') for x in lines]
# 以圖檔的名稱作為下标
image_ids = [x[0] for x in splitlines] # x[0]為名稱
confidence = np.array([float(x[1]) for x in splitlines]) # x[1]為得分
BB = np.array([[float(z) for z in x[2:]] for x in splitlines]) # x[2]為bbox的4個值
# 統計檢測的目标數量
nd = len(image_ids)
# tp:正類預測為正類->A預測成A
# fp:正類預測為負類->A預測成B
tp = np.zeros(nd)
fp = np.zeros(nd)
if BB.shape[0] > 0: # 行數>0
# sort by confidence
# 按照分數從大到小排序，傳回下标
sorted_ind = np.argsort(-confidence)
# 按照分數從大到小排序，傳回分數
# 下面也沒有用到該變量，其實sorted_ind得到之後，直接根據confidence[sorted_ind[i]]就可以得到sorted_scores
sorted_scores = np.sort(-confidence)
# 對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]
# 上面這些操作就是為了下标對應起來，後面好操作
# 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]擷取名稱。然後得到R
R = class_recs[image_ids[d]]
# dets是檢測結果的路徑,BB通過dets擷取每一行的資料，然後得到對應的BB值(也就是4個屬性)
bb = BB[d, :].astype(float)
# 設定一個負無窮
ovmax = -np.inf
# BBGT是真實的坐标
BBGT = R['bbox'].astype(float)
if BBGT.size > 0: # 如果存在GT計算交并比，如果不存在，首先就是檢測錯誤
# compute overlaps
# intersection
# 得到重疊區域，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.)
# 計算重疊區域的面積
inters = iw * ih
# union
# 并集面積就是兩個區域的面積減去重疊區域的面積
uni = ((bb[2] - bb[0] + 1.) * (bb[3] - bb[1] + 1.) +
(BBGT[:, 2] - BBGT[:, 0] + 1.) *
(BBGT[:, 3] - BBGT[:, 1] + 1.) - inters)
# 計算IOU，注意這個overlaps不一定是一個數值，可能是一個清單的,所有後面才有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: # 這個阙值預設0.5
if not R['difficult'][jmax]: # 這個後面是不是少個else，如果是難測樣本呢？？
# R = class_recs[image_ids[d]]
if not R['det'][jmax]: # R['det']初始值全為False,意思應該是如果該位置第一次使用，才可以。那也會出現tp[d]=fp[d]=1的情況啊
# 下面都是标記
tp[d] = 1.
R['det'][jmax] = 1
else:
fp[d] = 1.
else:
fp[d] = 1.
# compute precision recall
# 我測試cumsum是字首和的意思？？？？ 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生成全圖anchors[圖]](data:image/gif;base64,R0lGODlhAQABAIAAAP///wAAACwAAAAAAQABAAACAkQBADs=)