How to compute the gaussian_radius?Who can tell me the formula about it?Thank you!

[qiuhaining]

No description provided.

[fenglv12345]

i have the same question. i have tryed to derive the mathematical formulas.but mine is different with the author's code

[fenglv12345]

here is one of my mathematical formulas for (r3)

[qiuhaining]

here is one of my mathematical formulas for (r3)

I guess that it is the two peaks of redius value。The minimum value should be obtained from the two peaks. But，i cannot prove my guess. Your answer is the one of the peaks.

[fenglv12345]

for r1

for r2

for r3

conclusion:
with great possibility,the authour's mathmatical formulas are not correct and are not corresponding to the mind in papers.

[fenglv12345]

the correct formulas as below

[qiuhaining]

the correct formulas as below

what is the redius of the object as below??

[fenglv12345]

compute as below

by the way,i am improving the method in this way.

[qiuhaining]

compute as below

by the way,i am improving the method in this way.

Yes，this is the same as my idea，i look forward to the result, but,we need a dataset including various aspect ratios to confirm the idea!

[fenglv12345]

just the previous datasets like COCO is ok

[qiuhaining]

just the previous datasets like COCO is ok

Its my code in CenterNet and it can improve the performance a little! The formula can get by MATLAB Simulink. I need time to do this.

a = min((1 - min_overlap) / (1 + min_overlap) * width, 0.5 * width)
b = min((1 - min_overlap) / (1 + min_overlap) * height, 0.5 * height)
return a, b

[fenglv12345]

@qiuhaining ok,can you underline the code you have changed or just a picture,because i can not easily find the diversity

[gtwell]

[ggsggs]

Was this a small bug? Has someone noticed any improvement after re-typing the formulas?

[fenglv12345]

@ggsggs i have tried.the result is that AP is almost the same,but AR is higher. i haven't found the reason why

[ggsggs]

@ggsggs i have tried.the result is that AP is almost the same,but AR is higher. i haven't found the reason why

Oh, interesting. Was it a significant improvement in AR?

[fenglv12345]

@ggsggs the result as below:

[fenglv12345]

@ggsggs by the way,my work is based on the paper centernet(objects as points),using the model resdcn-18

[ggsggs]

@ggsggs the result as below:

@fenglv12345 Thank you for the experiments. ~1-3 points higher! It's not bad at all :)

@ggsggs by the way,my work is based on the paper centernet(objects as points),using the model resdcn-18

I am also working with CenterNet(objects as points one), actually. So it is good to know.

[fenglv12345]

@ggsggs thanks for your reminding which make me deep into the problem,perhaps i found the reason why. And i have a idea to make the AP higher while the AP is higher. By the way,there is some other method which has the problem (COCO detection challenge . focus on the MSRA)

[CF2220160244]

@fenglv12345 @ggsggs , In CenterNet I use the modified gaussian_radius you mentioned above, and keep other training parameters invariable,I use DLA34dcn, but both the AP and AR reduce. Can you tell me how you get the AR increase? Do you change any other hyper-parameter?

[fenglv12345]

@CF2220160244 my work in based on resdcn18 ,i didn't change any hyper-parameter,maybe there is some wrong with your implement. the formula as below.or you can show me your code

[CF2220160244]

@fenglv12345 ,my code is:
r = 0.5*(1-np.sqrt(0.7))*np.sqrt(width ** 2 + height ** 2)
return r

[fenglv12345]

what about a and b？ please understand the formula,you should also change the gaussian function,rather than just the r

[fenglv12345]

(1)get r
(2)get a and b
(3)change the gasssian function as above (when you change this,you will find some details should be changed)

[fenglv12345]

by the way,you don't have to calculate this :np.sqrt(width ** 2 + height ** 2),just use w and h

[CF2220160244]

Thank you very much, I will try! @fenglv12345

[CF2220160244]

@fenglv12345
Is it right? I change the code to the following,

#def gaussian2D(shape, sigma=1):
def gaussian2D(shape, a=1,b=1):
m, n = [(ss - 1.) / 2. for ss in shape]
y, x = np.ogrid[-m:m+1,-n:n+1]
#h = np.exp(-(x * x + y * y) / (2 * sigma * sigma))
h = np.exp(-(x * x ) / (2 * a/3 * a/3)-(y * y) / (2 * b/3 * b/3))
h[h < np.finfo(h.dtype).eps * h.max()] = 0
return h

def draw_umich_gaussian(heatmap, center, radius, k=1):
diameter = 2 * radius + 1
height, width = heatmap.shape[0:2]
gaussian = gaussian2D((diameter, diameter), a=0.1155 * width, b=0.1155 * height)
#gaussian = gaussian2D((diameter, diameter), sigma=diameter / 6)
x, y = int(center[0]), int(center[1])
#height, width = heatmap.shape[0:2]
....

def gaussian_radius(det_size, min_overlap=0.7):
height, width = det_size
r = 0.5*(1-np.sqrt(min_overlap))*np.sqrt(width ** 2 + height ** 2)
return r

[CF2220160244]

@fenglv12345
After i change to the oval gauss，the AR is almost the same，but the AP drop obviously, there are more false positive.

[seekFire]

@fenglv12345
How to deduce the formula "a = sqrt(2)r1a / sqrt(a^2 + b^2)" when using oval gaussian?

[fenglv12345]

@seekFire as below

[fenglv12345]

@CF2220160244 which model you have trained?dla34 or res18?res18 worked.and dla34 didn't work,i am trying to fix the problem,not the formula but other things, i will upload my code two or three days later on my repositories

[seekFire]

@fenglv12345 Good! Thank you very much!

[CF2220160244]

@fenglv12345 I train DLA. Looking forward to your result!

[seekFire]

@fenglv12345
By the way, does the function "draw_dense_reg" in image.py also need to be rectified?

[fenglv12345]

@seekFire we don't use draw_dense_reg

[CF2220160244]

@fenglv12345 Do you have any progress about the training?

[lcchust]

@fenglv12345

In your code "gaussian_radius1", I think 'ra' represent 'rh' and 'rb' represent 'w', but in code "draw_umich_gaussian1" you send 'ra' to 'rw' and 'rb' to 'rh'

I'm confused with "ra", "rb" and "rh", "rw", could you explain it?

[Dantju]

@qiuhaining i follow your code,but get error while training,have u change code expect the image u show

[AlphaPlusTT]

which size should i use to calculate the radius of the gaussion? i believe the shape should be the size of the gt bbox divided by the stride, but the code seem to use the original size of the gt bbox ? did i miss something ?

[ranjiewwen]

def gaussian2D(shape, sigma=1):
    m, n = [(ss - 1.) / 2. for ss in shape]
    y, x = np.ogrid[-m:m+1,-n:n+1]

    h = np.exp(-(x * x + y * y) / (2 * sigma * sigma))
    h[h < np.finfo(h.dtype).eps * h.max()] = 0
    return h

def draw_umich_gaussian(heatmap, center, radius, k=1):
  diameter = 2 * radius + 1
  gaussian = gaussian2D((diameter, diameter), sigma=diameter / 6)

when the radius = 0, maybe small object cause?

[AlphaPlusTT]

@ranjiewwen yes, the stride in centernet is 4, If the short edge of the object only occupies 3 pixels, floor(3/4)=0

[Apostatee]

@fenglv12345

In your code "gaussian_radius1", I think 'ra' represent 'rh' and 'rb' represent 'w', but in code "draw_umich_gaussian1" you send 'ra' to 'rw' and 'rb' to 'rh'

I'm confused with "ra", "rb" and "rh", "rw", could you explain it?

Must be a mistake

[JyunYuLai]

@CF2220160244 my work in based on resdcn18 ,i didn't change any hyper-parameter,maybe there is some wrong with your implement. the formula as below.or you can show me your code

Hi @fenglv12345 ,
I'm curious about why you use r2 for CenterNet.
In CornerNet, because r3 > r1 > r2 so you choose r2.
But in CenterNet, the only case that happen is r1.
I am not sure if my idea is correct.
Looking forward to your reply.
Thank you.

[18804601171]

@fenglv12345 你好，请问你在前面放的你的代码图里，ra和rb与公式相反了，是否是写错了呢

[urbaneman]

@CF2220160244 my work in based on resdcn18 ,i didn't change any hyper-parameter,maybe there is some wrong with your implement. the formula as below.or you can show me your code

Hi @fenglv12345 ,
I'm curious about why you use r2 for CenterNet.
In CornerNet, because r3 > r1 > r2 so you choose r2.
But in CenterNet, the only case that happen is r1.
I am not sure if my idea is correct.
Looking forward to your reply.
Thank you.

Hi @JyunYuLai ,
I agree with you. CenterNet only need to find one point(center), not two corners. So, the only case that happens is r1.
What do you think? @fenglv12345.
Looking forward to your reply. Thanks.

[igo312]

In my experiment useing right method to generate gaussian heatmap will make result bad surprisingly

Intuitively it should improve the perfomance, so really upset about that. Let's named origin generating method as G1 and rectified method as G2

my dataset is aerial image.here's example of G1 and G2, first one is G1, second one is G2

though the model trained by G2 get smaller loss but the map is only 67 in training dataset and 58 in validation dataset

the model trained by G1 get 89 in training dataset and 75 in validation dataset

[wangx1996]

In my experiment useing right method to generate gaussian heatmap will make result bad surprisingly

Intuitively it should improve the perfomance, so really upset about that. Let's named origin generating method as G1 and rectified method as G2

my dataset is aerial image.here's example of G1 and G2, first one is G1, second one is G2

though the model trained by G2 get smaller loss but the map is only 67 in training dataset and 58 in validation dataset

the model trained by G1 get 89 in training dataset and 75 in validation dataset

以这种方式生成的高斯分布是带有方向性的，看你的数据应该是旋转box，但是只依靠上述的高斯分布生成方法只能是对正矩形有效吧

[liuzw-cyy]

In my experiment useing right method to generate gaussian heatmap will make result bad surprisingly

Intuitively it should improve the perfomance, so really upset about that. Let's named origin generating method as G1 and rectified method as G2

my dataset is aerial image.here's example of G1 and G2, first one is G1, second one is G2

though the model trained by G2 get smaller loss but the map is only 67 in training dataset and 58 in validation dataset

the model trained by G1 get 89 in training dataset and 75 in validation dataset

Hello! Excuse me, can you share the code for drawing these two diagrams?

[Jayden9912]

Hi.

Given this

$$ IOU = \frac{(w-2r_2\cos\theta)(h-2r_2\sin\theta)}{wh}$$

$$ (4\sin\theta \cos\theta)r_2^2 - 2(w\sin\theta+h\cos\theta)r_2+wh(1-IOU) = 0$$

Using quadratic formula,
$$r_2 = \frac{2(w\sin\theta +h\cos\theta)\pm \sqrt{4(w\sin\theta+h\cos\theta)^2-16\sin\theta \cos\theta(wh)(1-IOU)}}{8\sin\theta \cos\theta} $$

How should I proceed to simplify the equation to get

$$ r_2 = \frac{(1-\sqrt{IOU})\sqrt{w^2+h^2}}{2} $$

which is calculated by using the concept of overlapping in the diagonal line of the 2 bounding boxes

$$ \sqrt{IOU} = \frac{\sqrt{(w^2+h^2)}-2r_2}{\sqrt{w^2+h^2}} $$

[qiuhaining]

收到了！谢谢！