O
@kidswong999 显示要填的确定位置啊，而且绘制的文字不是要求字符串的吗？
我想在这个机器人寻线的例程里在识别到的最近的黑色色块矩形中绘制出矩形的长度要怎么改啊？
Black Grayscale Line Following Example
跟随机器人做一条线需要很多努力。 本示例脚本显示了如何执行跟随机器人的
机器视觉部分。 您可以使用此脚本的输出来驱动差分驱动机器人遵循一条线。
这个脚本只是产生一个转动的值，告诉你的机器人向左或向右。
为了使这个脚本正常工作，你应该把摄像机指向45度左右的一条线。请确保只有线在相机的视野内。
import sensor, image, time, math
Tracks a black line. Use [(128, 255)] for a tracking a white line.
GRAYSCALE_THRESHOLD = [(0, 64)]
#设置阈值，如果是黑线，GRAYSCALE_THRESHOLD = [(0, 64)]；
#如果是白线，GRAYSCALE_THRESHOLD = [(128，255)]
Each roi is (x, y, w, h). The line detection algorithm will try to find the
centroid of the largest blob in each roi. The x position of the centroids
will then be averaged with different weights where the most weight is assigned
to the roi near the bottom of the image and less to the next roi and so on.
ROIS = [ # [ROI, weight]
(0, 100, 160, 20, 0.7), # You'll need to tweak the weights for you app
(0, 050, 160, 20, 0.3), # depending on how your robot is setup.
(0, 000, 160, 20, 0.1)
]
#roi代表三个取样区域，（x,y,w,h,weight）,代表左上顶点（x,y）宽高分别为w和h的矩形，
#weight为当前矩形的权值。注意本例程采用的QQVGA图像大小为160x120，roi即把图像横分成三个矩形。
#三个矩形的阈值要根据实际情况进行调整，离机器人视野最近的矩形权值要最大，
#如上图的最下方的矩形，即(0, 100, 160, 20, 0.7)
Compute the weight divisor (we're computing this so you don't have to make weights add to 1).
weight_sum = 0 #权值和初始化
for r in ROIS: weight_sum += r[4] # r[4] is the roi weight.
#计算权值和。遍历上面的三个矩形，r[4]即每个矩形的权值。
Camera setup...
sensor.reset() # Initialize the camera sensor.
sensor.set_pixformat(sensor.GRAYSCALE) # use grayscale.
sensor.set_framesize(sensor.QQVGA) # use QQVGA for speed.
sensor.skip_frames(30) # Let new settings take affect.
sensor.set_auto_gain(False) # must be turned off for color tracking
sensor.set_auto_whitebal(False) # must be turned off for color tracking
#关闭白平衡
clock = time.clock() # Tracks FPS.
while(True):
clock.tick() # Track elapsed milliseconds between snapshots().
img = sensor.snapshot() # Take a picture and return the image.
centroid_sum = 0
#利用颜色识别分别寻找三个矩形区域内的线段
for r in ROIS:
blobs = img.find_blobs(GRAYSCALE_THRESHOLD, roi=r[0:4], merge=True)
# r[0:4] is roi tuple.
#找到视野中的线,merge=true,将找到的图像区域合并成一个
#目标区域找到直线
if blobs:
# Find the index of the blob with the most pixels.
most_pixels = 0
largest_blob = 0
for i in range(len(blobs)):
#目标区域找到的颜色块（线段块）可能不止一个，找到最大的一个，作为本区域内的目标直线
if blobs[i].pixels() > most_pixels:
most_pixels = blobs[i].pixels()
#merged_blobs[i][4]是这个颜色块的像素总数，如果此颜色块像素总数大于 #most_pixels，则把本区域作为像素总数最大的颜色块。更新most_pixels和largest_blob
largest_blob = i
# Draw a rect around the blob.
img.draw_rectangle(blobs[largest_blob].rect())
#将此区域的像素数最大的颜色块画矩形和十字形标记出来
img.draw_cross(blobs[largest_blob].cx(),
blobs[largest_blob].cy())
centroid_sum += blobs[largest_blob].cx() * r[4] # r[4] is the roi weight.
#计算centroid_sum，centroid_sum等于每个区域的最大颜色块的中心点的x坐标值乘本区域的权值
center_pos = (centroid_sum / weight_sum) # Determine center of line.
#中间公式
# Convert the center_pos to a deflection angle. We're using a non-linear
# operation so that the response gets stronger the farther off the line we
# are. Non-linear operations are good to use on the output of algorithms
# like this to cause a response "trigger".
deflection_angle = 0
#机器人应该转的角度
# The 80 is from half the X res, the 60 is from half the Y res. The
# equation below is just computing the angle of a triangle where the
# opposite side of the triangle is the deviation of the center position
# from the center and the adjacent side is half the Y res. This limits
# the angle output to around -45 to 45. (It's not quite -45 and 45).
deflection_angle = -math.atan((center_pos-80)/60)
#角度计算.80 60 分别为图像宽和高的一半，图像大小为QQVGA 160x120.
#注意计算得到的是弧度值
# Convert angle in radians to degrees.
deflection_angle = math.degrees(deflection_angle)
#将计算结果的弧度值转化为角度值
# Now you have an angle telling you how much to turn the robot by which
# incorporates the part of the line nearest to the robot and parts of
# the line farther away from the robot for a better prediction.
print("Turn Angle: %f" % deflection_angle)
#将结果打印在terminal中
print(clock.fps()) # Note: Your OpenMV Cam runs about half as fast while
# connected to your computer. The FPS should increase once disconnected.