Open kemaloksuz opened 4 years ago
@kemaloksuz I came across the same problem.
How to solve it ?
Hello,
As I mentioned, my solution is not an elegant one, just a workaround. However this is how I do it. These are the relevant lines from summarize() function (I want to see all the thresholds):
def summarize(self):
'''
Compute and display summary metrics for evaluation results.
Note this functin can *only* be applied on the default parameter setting
'''
def _summarize( ap=1, iouThr=None, areaRng='all', maxDets=100 ):
p = self.params
iStr = ' {:<18} {} @[ IoU={:<9} | area={:>6s} | maxDets={:>3d} ] = {:0.3f}'
titleStr = 'Average Precision' if ap == 1 else 'Average Recall'
typeStr = '(AP)' if ap==1 else '(AR)'
iouStr = '{:0.2f}:{:0.2f}'.format(p.iouThrs[0], p.iouThrs[-1]) \
if iouThr is None else '{:0.2f}'.format(iouThr)
aind = [i for i, aRng in enumerate(p.areaRngLbl) if aRng == areaRng]
mind = [i for i, mDet in enumerate(p.maxDets) if mDet == maxDets]
if ap == 1:
# dimension of precision: [TxRxKxAxM]
s = self.eval['precision']
# IoU
if iouThr is not None:
if iouThr == 0.9:
t = [8]
else:
t = np.where(iouThr == p.iouThrs)[0]
s = s[t]
s = s[:,:,:,aind,mind]
else:
# dimension of recall: [TxKxAxM]
s = self.eval['recall']
if iouThr is not None:
if iouThr == 0.9:
t = [8]
else:
t = np.where(iouThr == p.iouThrs)[0]
s = s[t]
s = s[:,:,aind,mind]
if len(s[s>-1])==0:
mean_s = -1
else:
mean_s = np.mean(s[s>-1])
print(iStr.format(titleStr, typeStr, iouStr, areaRng, maxDets, mean_s))
return mean_s
def _summarizeDets():
stats = np.zeros((20,))
stats[0] = _summarize(1)
stats[1] = _summarize(1, iouThr=.5, maxDets=self.params.maxDets[2])
stats[2] = _summarize(1, iouThr=.55, maxDets=self.params.maxDets[2])
stats[3] = _summarize(1, iouThr=.6, maxDets=self.params.maxDets[2])
stats[4] = _summarize(1, iouThr=.65, maxDets=self.params.maxDets[2])
stats[5] = _summarize(1, iouThr=.7, maxDets=self.params.maxDets[2])
stats[6] = _summarize(1, iouThr=.75, maxDets=self.params.maxDets[2])
stats[7] = _summarize(1, iouThr=.8, maxDets=self.params.maxDets[2])
stats[8] = _summarize(1, iouThr=.85, maxDets=self.params.maxDets[2])
stats[9] = _summarize(1, iouThr=.9, maxDets=self.params.maxDets[2])
stats[10] = _summarize(1, iouThr=.95, maxDets=self.params.maxDets[2])
stats[11] = _summarize(1, areaRng='small', maxDets=self.params.maxDets[2])
stats[12] = _summarize(1, areaRng='medium', maxDets=self.params.maxDets[2])
stats[13] = _summarize(1, areaRng='large', maxDets=self.params.maxDets[2])
stats[14] = _summarize(0, maxDets=self.params.maxDets[0])
stats[15] = _summarize(0, maxDets=self.params.maxDets[1])
stats[16] = _summarize(0, maxDets=self.params.maxDets[2])
stats[17] = _summarize(0, areaRng='small', maxDets=self.params.maxDets[2])
stats[18] = _summarize(0, areaRng='medium', maxDets=self.params.maxDets[2])
stats[19] = _summarize(0, areaRng='large', maxDets=self.params.maxDets[2])
return stats
The rest of the function is the same.
@kemaloksuz thanks for the reply :)
I met the same issue, and I found that the reason is the generated ndarray p.iouThrs
by np.linspace
is not exact.
It is supposed to be [0.5, 0.55, 0.6, 0.65, 0.7, 0.75, 0.8, 0.85, 0.9, 0.95]
, but in fact it is [0.5, 0.55, 0.6, 0.65, 0.7, 0.75, 0.8, 0.85, 0.8999999999999999, 0.95]
in my machine with numpy version=1.20.2, so IoU with 0.9 can't work. This is an inherent problem of IEEE float point, and we should always avoid doing float comparisons with ==
.
One simple way is to change the line:
t = np.where(iouThr == p.iouThrs)[0]
to:
t = np.where(np.isclose(iouThr, p.iouThrs))[0]
.
Note that t = np.where(iouThr == p.iouThrs)[0]
appeares twice in the code block of https://github.com/cocodataset/cocoapi/issues/409#issuecomment-632591018, so change both places.
I met the same issue, and I found that the reason is the generated ndarray
p.iouThrs
bynp.linspace
is not exact.It is supposed to be
[0.5, 0.55, 0.6, 0.65, 0.7, 0.75, 0.8, 0.85, 0.9, 0.95]
, but in fact it is[0.5, 0.55, 0.6, 0.65, 0.7, 0.75, 0.8, 0.85, 0.8999999999999999, 0.95]
in my machine with numpy version=1.20.2, so IoU with 0.9 can't work. This is an inherent problem of IEEE float point, and we should always avoid doing float comparisons with==
.One simple way is to change the line:
t = np.where(iouThr == p.iouThrs)[0]
to:t = np.where(np.isclose(iouThr, p.iouThrs))[0]
.Note that
t = np.where(iouThr == p.iouThrs)[0]
appeares twice in the code block of #409 (comment), so change both places.
Thanks
Hi,
I wanted to check other thresholds than 0.5 and 0.75 for mAP. To do that, I simply added several lines for each IoU such as:
stats[9] = _summarize(1, iouThr=.9, maxDets=self.params.maxDets[2])
However, there is a bug only for 0.9, which causes to return -1 as AP@90. Here is an example output of 3 thresholds:
Average Precision (AP) @[ IoU=0.85 | area= all | maxDets=100 ] = 0.230 Average Precision (AP) @[ IoU=0.90 | area= all | maxDets=100 ] = -1.000 Average Precision (AP) @[ IoU=0.95 | area= all | maxDets=100 ] = 0.018
This surprised me since AP@0.95 > AP@0.90. When I debug, I noticed that p.iouThrs[8] is not exactly 0.9 but 0.89999. Here is the output from debugger in the _summarize function:
(Pdb) p p.iouThrs[8] 0.8999999999999999
For me, I solved this bug by a small workaround, but I think it should be systematically handled in this repo.
Regards,
Kemal