center_scores in pred_square seems to scalar values.

[navervision]

center_scores in pred_square seems to scalar values.

[geonm]

Sorry for the late reply. I was on vacation.

Yes. This was intended.

center_scores is used for finding a rectangle lying in the center of the image.

I simply designed center_scores as a vector array for how far the center points of the squares are from the center point of the image.

square_centers are the center points of all detected rectangles and centers (should be center 😄 ) is the center of the image.

As shown in

mlsd/utils.py

Lines 461 to 466 in 53a0161

	###################################### CENTER SCORES
	centers = np.array([[256 // 2, 256 // 2]], dtype='float32') # [1, 2]
	# squares: [n, 4, 2]
	square_centers = np.mean(squares, axis=1) # [n, 2]
	center2center = np.sqrt(np.sum((centers - square_centers) ** 2))
	center_scores = center2center / (map_size / np.sqrt(2.0))

, this was intended.

NOTE: center_scores is not a scala value.

[geonm]

It has been fixed #22.

mlsd/utils.py

Line 465 in 1c25206

center2center = np.sqrt(np.sum((centers - square_centers) ** 2, axis=1))

Thanks!

[whwnsdlr1]

thank to reply
But if center_array is a vector and not a scalar then axis=1 isn't what you need?
center2center = np.sqrt(np.sum((centers - square_centers) ** 2))
-> a = (centers - square_centers) ** 2 # (n, 2)
-> b = np.sum(a) # scalar, np.sum reduce 0d
-> center2center = np.sqrt(b) # scalar

If np.sum with axis=1
-> a = (centers - square_centers) ** 2 # (n, 2)
-> b = np.sum(a, axis = 1) # (n,)
-> center2center = np.sqrt(b) # (n,)

[geonm]

@whwnsdlr1

Thanks for your comment.

Yes, you are right.

The line 466

mlsd/utils.py

Line 465 in 53a0161

center2center = np.sqrt(np.sum((centers - square_centers) ** 2))

should be modified to follow above my comment.

I will fix it. Thank you again.