Question

我是新来的，我需要你的帮助!!!! 我正在制作一个项目（我的学校的手势识别），所以未读的图像就是我的手。我想在点之间找到角度（链码）

提前谢谢你：）

import cv2
import cv2.cv as cv
import numpy as np

# Create display windows
cv2.namedWindow("input", cv.CV_WINDOW_AUTOSIZE)
cv2.namedWindow("output", cv.CV_WINDOW_AUTOSIZE)

# Parameters
blur_ksize    = 5
thresh_Tlower = 100
kernel = np.ones((5,5),np.uint8) 

# Imread the image
img = cv2.imread ("0_dani_mask.png",0)

# Funding the contours of the hand
contours, hierarchy = cv2.findContours(img.copy(), cv2.RETR_TREE,   cv.CV_CHAIN_APPROX_NONE)

# contour = the biggest (area)
big_contour = contours[0]
num_points_cnt = len( big_contour )
print "num_points_cnt = ", num_points_cnt

theta = np.zeros( num_points_cnt, np.uint8 )

P0 = big_contour[0]
x0 = P0[0,0]
y0 = P0[0,1]

for n in range( 1, num_points_cnt ):
    P = big_contour[n]
    x = P[0,0]
    y = P[0,1]

    dX = x - x0
    dY = y - y0
    angulo = np.arctan( dY/dX )
    theta[n] = angulo

    x0 = x
    y0 = y

P = big_contour[0]
x = P[0,0]
y = P[0,1]
dX = x - x0
dY = y - y0

angulo = np.arctan(dY/dX)
theta[0] = angulo

print theta

该计划告诉我：

num_points_cnt = 2031 [0 0 0 ...，0 0 0] /Users/dani/Desktop/myproject/src/kasksasa.py:47：运行时警告：在int_scalars中遇到零除 angulo = np.tanh（dY / dX）

Answer 1

要查找给定RewriteEngine on RewriteRule ^about/$ about.php [NC,L] RewriteCond %{THE_REQUEST} ^[A-Z]{3,9}\\ /about\\.php [NC] RewriteRule ^about.php$ /about-us/ [L,R=301]和dY的角度，请使用numpy.arctan2(dY, dX)。它正确处理dX。

python，openCv，链码 - ＆gt;我想找到点之间的角度

1 个答案: