我要评论本编码方式来自论文
Enhanced Center Coding for Cell Detection with Convolutional Neural Networks
https://arxiv.org/abs/1904.08864?context=cs.CV
公式：
效果：
实现：
'''
The algorithm comes from the paper
"Enhanced Center Coding for Cell Detection with Convolutional Neural Networks"
https://arxiv.org/abs/1904.08864?context=cs.CV
'''
import numpy as np
from skimage.draw import disk as sk_disk
def draw_repel_code_ori(im: np.ndarray, pts: np.ndarray, r, A=1):
'''
:param im: [H, W, 1] The image to be drawn.
:param pts: [N, yx] Input points. You need to ensure that the each point is unique.
:param r: The radius of the drawing area.
:param A: Control the gradual change.
:return:
'''
if not isinstance(pts, np.ndarray):
pts = np.asarray(pts, np.int)
assert im.ndim == 3 and im.shape[2] == 1
wait_check_pixs = []
for pt in pts:
rr, cc = sk_disk(pt, r, shape=im.shape[:2])
pts2 = np.stack([rr, cc], 1)
pts2 = np.asarray(pts2, np.int32)
wait_check_pixs.extend(pts2)
check_pixs = np.unique(wait_check_pixs, axis=0)
# # 源代码，很慢
# for pix in check_pixs:
# ds = np.linalg.norm(pix[None] - pts, 2, 1)
# seq_id = np.argsort(ds)[:2]
# # close_2pts = list(pts[seq_id])
# close_2pts_ds = list(ds[seq_id])
# C = 0
# if len(close_2pts_ds) == 2:
# D = close_2pts_ds[0] * (1 + close_2pts_ds[0] / (np.clip(close_2pts_ds[1], 1e-8, None)))
# if D < r:
# C = 1 / (1 + D * A)
# elif len(close_2pts_ds) == 1:
# D = close_2pts_ds[0]
# if D < r:
# C = 1 / (1 + D * A)
# im[pix[0], pix[1]] = C
# # 这里为上面的并行加速代码。但还有待优化，内存占用太大了。
# big_ds = np.linalg.norm(np.asarray(check_pixs[:, None] - pts[None], np.float32), 2, 2)
# big_seq_id = np.argsort(big_ds, axis=-1)[:, :2]
# big_close_2pts_ds = np.take_along_axis(big_ds, big_seq_id, 1)
#
# if big_close_2pts_ds.shape[1] == 2:
# D = big_close_2pts_ds[:, 0] * (1 + big_close_2pts_ds[:, 0] / (np.clip(big_close_2pts_ds[:, 1], 1e-8, None)))
# C = np.where(D < r, 1 / (1 + D * A), 0)
# elif big_close_2pts_ds.shape[1] == 1:
# D = big_close_2pts_ds[:, 0]
# C = np.where(D < r, 1 / (1 + D * A), 0)
# else:
# return im
# im[check_pixs[:, 0], check_pixs[:, 1]] = C[:, None]
# 这里为上面的并行加速代码的分段代码，一次性查全部点，占用内存过于吓人。
n_each_draw = 5000
draw_count = int(np.ceil(len(check_pixs) / n_each_draw))
for i in range(draw_count):
cur_check_pixs = check_pixs[i*n_each_draw:(i+1)*n_each_draw]
big_ds = np.linalg.norm(np.asarray(cur_check_pixs[:, None] - pts[None], np.float32), 2, 2)
big_seq_id = np.argsort(big_ds, axis=-1)[:, :2]
big_close_2pts_ds = np.take_along_axis(big_ds, big_seq_id, 1)
if big_close_2pts_ds.shape[1] == 2:
D = big_close_2pts_ds[:, 0] * (1 + big_close_2pts_ds[:, 0] / (np.clip(big_close_2pts_ds[:, 1], 1e-8, None)))
C = np.where(D < r, 1 / (1 + D * A), 0)
elif big_close_2pts_ds.shape[1] == 1:
D = big_close_2pts_ds[:, 0]
C = np.where(D < r, 1 / (1 + D * A), 0)
else:
# 这里是一次性检查全部点，如果没有，可以认为后面也没有了，可以直接退出
break
im[cur_check_pixs[:, 0], cur_check_pixs[:, 1]] = C[:, None]
return im
def draw_repel_code_fast(im: np.ndarray, pts: np.ndarray, r, A=1):
'''
Another faster acceleration method code of the above function, but there are accuracy problems.
:param im: [H, W, 1] The image to be drawn.
:param pts: [N, yx] Input points. You need to ensure that the each point is unique.
:param r: The radius of the drawing area.
:param A: Control the gradual change.
:return:
'''
assert im.ndim == 3 and im.shape[2] == 1
for pt in pts:
ds = np.linalg.norm(pt[None,] - pts, 2, 1)
# 注意，这个点集已经包含了待查询点自身
# 该处r*2是为了增加更多的检查点，减少不准确问题
close_pts = pts[ds <= r * 2]
rr, cc = sk_disk(pt, r, shape=im.shape[:2])
pts2 = np.stack([rr, cc], 1)
pts2 = np.asarray(pts2, np.int32)
big_ds = np.linalg.norm(np.asarray(pts2[:, None] - close_pts[None], np.float32), 2, 2)
# big_ds = np.linalg.norm(np.asarray(close_pts[None,] - pts2[:, None], np.float32), 2, 2)
big_seq_id = np.argsort(big_ds, axis=-1)[:, :2]
big_close_2pts_ds = np.take_along_axis(big_ds, big_seq_id, 1)
if big_close_2pts_ds.shape[1] == 2:
D = big_close_2pts_ds[:, 0] * (1 + big_close_2pts_ds[:, 0] / (np.clip(big_close_2pts_ds[:, 1], 1e-8, None)))
C = np.where(D < r, 1 / (1 + D * A), 0)
elif big_close_2pts_ds.shape[1] == 1:
D = big_close_2pts_ds[:, 0]
C = np.where(D < r, 1 / (1 + D * A), 0)
else:
# return im
continue
im[pts2[:, 0], pts2[:, 1]] = C[:, None]
return im
if __name__ == '__main__':
import cv2
im = np.zeros([512, 512, 1], np.float32)
# 注意输入的pt坐标顺序为yx，而不是xy
pts = np.random.randint(0, 512, [100, 2], np.int)
# 不允许有重复的坐标
pts = np.unique(pts, axis=0)
oim1 = draw_repel_code_ori(im, pts, 9)
oim2 = draw_repel_code_fast(im, pts, 9)
cv2.imshow('oim1', oim1)
cv2.imshow('oim2', oim2)
cv2.waitKey(0)
本文地址：https://blog.csdn.net/ONE_SIX_MIX/article/details/107666163
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