Question

我尝试了Estimations: 2 -> 1: [ 527.15670903 222.57196904] 3 -> 1: [ 527.21339222 221.86819147] 4 -> 1: [ 527.63122722 222.30614892] Avg loc: [ 527.33377616 222.24876981] Est loc: [527 222] True loc: [527 222]但有时会返回5个，有时会返回6个数字。我希望它一直是6的长度。我更喜欢它的格式如import numpy as np import cv2 # read images, taken from http://kahlan.eps.surrey.ac.uk/featurespace/web/ img1 = cv2.imread("img1.png", 1) img2 = cv2.imread("img2.png", 1) img3 = cv2.imread("img3.png", 1) img4 = cv2.imread("img4.png", 1) # true locations of the chicken's crossed eye (labeled myself) loc1 = np.array([527, 222, 1]) loc2 = np.array([449, 241, 1]) loc3 = np.array([476, 275, 1]) loc4 = np.array([385, 236, 1]) # define ground truth homographies, also from http://kahlan.eps.surrey.ac.uk/featurespace/web/ H12 = np.array([ [8.7976964e-01, 3.1245438e-01, -3.9430589e+01], [-1.8389418e-01, 9.3847198e-01, 1.5315784e+02], [1.9641425e-04, -1.6015275e-05, 1.0000000e+00]]) H13 = np.array([ [7.6285898e-01, -2.9922929e-01, 2.2567123e+02], [3.3443473e-01, 1.0143901e+00, -7.6999973e+01], [3.4663091e-04, -1.4364524e-05, 1.0000000e+00]]) H14 = np.array([ [6.6378505e-01, 6.8003334e-01, -3.1230335e+01], [-1.4495500e-01, 9.7128304e-01, 1.4877420e+02], [4.2518504e-04, -1.3930359e-05, 1.0000000e+00]]) # need the homographies going the other direction H21 = np.linalg.inv(H12) H31 = np.linalg.inv(H13) H41 = np.linalg.inv(H14) # ensure they are homogeneous by dividing by the last entry H21 = H21/H21[-1,-1] H31 = H31/H31[-1,-1] H41 = H41/H41[-1,-1] # warp the locations loc2, loc3, loc4 to the coordinates of img1 est21 = np.matmul(H21, loc2) est31 = np.matmul(H31, loc3) est41 = np.matmul(H41, loc4) # make homogeneous, toss the final 1 est21 = est21[:-1]/est21[-1] est31 = est31[:-1]/est31[-1] est41 = est41[:-1]/est41[-1] # remove the last coordinate, take an average avgest = (est21 + est31 + est41)/3 estloc = np.around(avgest).astype(int) # output print("Estimations:" "\n2 -> 1: ", est21, "\n3 -> 1: ", est31, "\n4 -> 1: ", est41, "\nAvg loc: ", avgest, "\nEst loc: ", estloc, "\nTrue loc: ", loc1[:-1]) # show images cv2.circle(img1, (estloc[0], estloc[1]), 2, (0,0,255), -1) # filled cv2.circle(img1, (estloc[0], estloc[1]), 20, (255,255,255)) # outline cv2.imshow('img1-est', img1) cv2.waitKey(0) cv2.circle(img2, (loc2[0], loc2[1]), 2, (0,0,255), -1) # filled cv2.circle(img2, (loc2[0], loc2[1]), 20, (255,255,255)) # outline cv2.imshow('img2-loc', img2) cv2.waitKey(0) cv2.circle(img3, (loc3[0], loc3[1]), 2, (0,0,255), -1) # filled cv2.circle(img3, (loc3[0], loc3[1]), 20, (255,255,255)) # outline cv2.imshow('img3-log', img3) cv2.waitKey(0) cv2.circle(img4, (loc4[0], loc4[1]), 2, (0,0,255), -1) # filled cv2.circle(img4, (loc4[0], loc4[1]), 20, (255,255,255)) # outline cv2.imshow('img4-log', img4) cv2.waitKey(0) cv2.destroyAllWindows()而不使用像SecureRandom.random_number(9**6)这样的语法，因此在我的控制器测试中更容易被删除。

Answer 1

你可以用数学做到这一点：

(SecureRandom.random_number(9e5) + 1e5).to_i

然后验证：

100000.times.map do
  (SecureRandom.random_number(9e5) + 1e5).to_i
end.map { |v| v.to_s.length }.uniq
# => [6]

这将产生100000..999999范围内的值：

10000000.times.map do
  (SecureRandom.random_number(9e5) + 1e5).to_i
end.minmax
# => [100000, 999999]

如果您需要更简洁的格式，只需将其转换为方法：

def six_digit_rand
  (SecureRandom.random_number(9e5) + 1e5).to_i
end

Answer 2

要生成一个随机的6位数字字符串：

copy_to

这是正在发生的事情的更多细节，通过将单行分成多行并解释变量来显示：

# This generates a 6-digit string, where the
# minimum possible value is "000000", and the
# maximum possible value is "999999"
SecureRandom.random_number(10**6).to_s.rjust(6, '0')

Answer 3

SecureRandom.random_number（n）给出0到n之间的随机值。你可以使用rand函数来实现它。

2.3.1 :025 > rand(10**5..10**6-1)
=> 742840

rand（a..b）给出a和b之间的随机数。在这里，你总是得到一个介于10 ^ 5和10 ^ 6-1之间的6位数随机数。

使用Ruby中的SecureRandom生成长度为6的随机数

3 个答案: