How to use _find_homography method in Airtest

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sift.py

Source:sift.py Github

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...134 """特征点匹配点对数目>=4个，可使用单矩阵映射,求出识别的目标区域."""135 sch_pts, img_pts = np.float32([kp_sch[m.queryIdx].pt for m in good]).reshape(136 -1, 1, 2), np.float32([kp_src[m.trainIdx].pt for m in good]).reshape(-1, 1, 2)137 # M是转化矩阵138 M, mask = _find_homography(sch_pts, img_pts)139 matches_mask = mask.ravel().tolist()140 # 从good中间筛选出更精确的点(假设good中大部分点为正确的，由ratio=0.7保障)141 selected = [v for k, v in enumerate(good) if matches_mask[k]]142 # 针对所有的selected点再次计算出更精确的转化矩阵M来143 sch_pts, img_pts = np.float32([kp_sch[m.queryIdx].pt for m in selected]).reshape(144 -1, 1, 2), np.float32([kp_src[m.trainIdx].pt for m in selected]).reshape(-1, 1, 2)145 M, mask = _find_homography(sch_pts, img_pts)146 # 计算四个角矩阵变换后的坐标，也就是在大图中的目标区域的顶点坐标:147 h, w = im_search.shape[:2]148 h_s, w_s = im_source.shape[:2]149 pts = np.float32([[0, 0], [0, h - 1], [w - 1, h - 1], [w - 1, 0]]).reshape(-1, 1, 2)150 dst = cv2.perspectiveTransform(pts, M)151 # trans numpy arrary to python list: [(a, b), (a1, b1), ...]152 def cal_rect_pts(dst):153 return [tuple(npt[0]) for npt in dst.astype(int).tolist()]154 pypts = cal_rect_pts(dst)155 # 注意：虽然4个角点有可能越出source图边界，但是(根据精确化映射单映射矩阵M线性机制)中点不会越出边界156 lt, br = pypts[0], pypts[2]157 middle_point = int((lt[0] + br[0]) / 2), int((lt[1] + br[1]) / 2)158 # 考虑到算出的目标矩阵有可能是翻转的情况，必须进行一次处理，确保映射后的“左上角”在图片中也是左上角点：159 x_min, x_max = min(lt[0], br[0]), max(lt[0], br[0])160 y_min, y_max = min(lt[1], br[1]), max(lt[1], br[1])161 # 挑选出目标矩形区域可能会有越界情况，越界时直接将其置为边界：162 # 超出左边界取0，超出右边界取w_s-1，超出下边界取0，超出上边界取h_s-1163 # 当x_min小于0时，取0。 x_max小于0时，取0。164 x_min, x_max = int(max(x_min, 0)), int(max(x_max, 0))165 # 当x_min大于w_s时，取值w_s-1。 x_max大于w_s-1时，取w_s-1。166 x_min, x_max = int(min(x_min, w_s - 1)), int(min(x_max, w_s - 1))167 # 当y_min小于0时，取0。 y_max小于0时，取0。168 y_min, y_max = int(max(y_min, 0)), int(max(y_max, 0))169 # 当y_min大于h_s时，取值h_s-1。 y_max大于h_s-1时，取h_s-1。170 y_min, y_max = int(min(y_min, h_s - 1)), int(min(y_max, h_s - 1))171 # 目标区域的角点，按左上、左下、右下、右上点序：(x_min,y_min)(x_min,y_max)(x_max,y_max)(x_max,y_min)172 pts = np.float32([[x_min, y_min], [x_min, y_max], [173 x_max, y_max], [x_max, y_min]]).reshape(-1, 1, 2)174 pypts = cal_rect_pts(pts)175 return middle_point, pypts, [x_min, x_max, y_min, y_max, w, h]176def _two_good_points(pts_sch1, pts_sch2, pts_src1, pts_src2, im_search, im_source):177 """返回两对匹配特征点情形下的识别结果."""178 # 先算出中心点(在im_source中的坐标)：179 middle_point = [int((pts_src1[0] + pts_src2[0]) / 2), int((pts_src1[1] + pts_src2[1]) / 2)]180 pypts = []181 # 如果特征点同x轴或同y轴(无论src还是sch中)，均不能计算出目标矩形区域来，此时返回值同good=1情形182 if pts_sch1[0] == pts_sch2[0] or pts_sch1[1] == pts_sch2[1] or pts_src1[0] == pts_src2[0] or pts_src1[1] == pts_src2[1]:183 confidence = ONE_POINT_CONFI184 one_match = generate_result(middle_point, pypts, confidence)185 return one_match186 # 计算x,y轴的缩放比例：x_scale、y_scale，从middle点扩张出目标区域:(注意整数计算要转成浮点数结果!)187 h, w = im_search.shape[:2]188 h_s, w_s = im_source.shape[:2]189 x_scale = abs(1.0 * (pts_src2[0] - pts_src1[0]) / (pts_sch2[0] - pts_sch1[0]))190 y_scale = abs(1.0 * (pts_src2[1] - pts_src1[1]) / (pts_sch2[1] - pts_sch1[1]))191 # 得到scale后需要对middle_point进行校正，并非特征点中点，而是映射矩阵的中点。192 sch_middle_point = int((pts_sch1[0] + pts_sch2[0]) / 2), int((pts_sch1[1] + pts_sch2[1]) / 2)193 middle_point[0] = middle_point[0] - int((sch_middle_point[0] - w / 2) * x_scale)194 middle_point[1] = middle_point[1] - int((sch_middle_point[1] - h / 2) * y_scale)195 middle_point[0] = max(middle_point[0], 0) # 超出左边界取0 (图像左上角坐标为0,0)196 middle_point[0] = min(middle_point[0], w_s - 1) # 超出右边界取w_s-1197 middle_point[1] = max(middle_point[1], 0) # 超出上边界取0198 middle_point[1] = min(middle_point[1], h_s - 1) # 超出下边界取h_s-1199 # 计算出来rectangle角点的顺序：左上角->左下角->右下角->右上角， 注意：暂不考虑图片转动200 # 超出左边界取0, 超出右边界取w_s-1, 超出下边界取0, 超出上边界取h_s-1201 x_min, x_max = int(max(middle_point[0] - (w * x_scale) / 2, 0)), int(202 min(middle_point[0] + (w * x_scale) / 2, w_s - 1))203 y_min, y_max = int(max(middle_point[1] - (h * y_scale) / 2, 0)), int(204 min(middle_point[1] + (h * y_scale) / 2, h_s - 1))205 # 目标矩形的角点按左上、左下、右下、右上的点序：(x_min,y_min)(x_min,y_max)(x_max,y_max)(x_max,y_min)206 pts = np.float32([[x_min, y_min], [x_min, y_max], [x_max, y_max], [x_max, y_min]]).reshape(-1, 1, 2)207 for npt in pts.astype(int).tolist():208 pypts.append(tuple(npt[0]))209 return middle_point, pypts, [x_min, x_max, y_min, y_max, w, h]210def _find_homography(sch_pts, src_pts):211 """多组特征点对时，求取单向性矩阵."""212 try:213 M, mask = cv2.findHomography(sch_pts, src_pts, cv2.RANSAC, 5.0)214 except Exception:215 import traceback216 traceback.print_exc()217 raise HomographyError("OpenCV error in _find_homography()...")218 else:219 if mask is None:220 raise HomographyError("In _find_homography(), find no mask...")221 else:222 return M, mask223def _target_error_check(w_h_range):224 """校验识别结果区域是否符合常理."""225 x_min, x_max, y_min, y_max, w, h = w_h_range226 tar_width, tar_height = x_max - x_min, y_max - y_min227 # 如果src_img中的矩形识别区域的宽和高的像素数＜5，则判定识别失效。认为提取区域待不可能小于5个像素。(截图一般不可能小于5像素)228 if tar_width < 5 or tar_height < 5:229 raise SiftResultCheckError("In src_image, Taget area: width or height < 5 pixel.")230 # 如果矩形识别区域的宽和高，与sch_img的宽高差距超过5倍(屏幕像素差不可能有5倍)，认定为识别错误。231 if tar_width < 0.2 * w or tar_width > 5 * w or tar_height < 0.2 * h or tar_height > 5 * h:232 raise SiftResultCheckError("Target area is 5 times bigger or 0.2 times smaller than sch_img.")233def _cal_sift_confidence(im_search, resize_img, rgb=False):234 if rgb:...

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sift_test.py

Source:sift_test.py Github

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...44good = good_diff45# print(good)46# for i in good:47# print(i.distance)48def _find_homography(sch_pts, src_pts):49 """多组特征点对时，求取单向性矩阵."""50 try:51 M, mask = cv2.findHomography(sch_pts, src_pts, cv2.RANSAC, 5.0)52 except Exception:53 import traceback54 traceback.print_exc()55 raise Exception("OpenCV error in _find_homography()...")56 else:57 if mask is None:58 raise Exception("In _find_homography(), find no mask...")59 else:60 return M, mask61def _many_good_pts(im_source, im_search, kp_sch, kp_src, good):62 """特征点匹配点对数目>=4个，可使用单矩阵映射,求出识别的目标区域."""63 sch_pts, img_pts = np.float32([kp_sch[m.queryIdx].pt for m in good]).reshape(64 -1, 1, 2), np.float32([kp_src[m.trainIdx].pt for m in good]).reshape(-1, 1, 2)65 # M是转化矩阵66 M, mask = _find_homography(sch_pts, img_pts)67 matches_mask = mask.ravel().tolist()68 # 从good中间筛选出更精确的点(假设good中大部分点为正确的，由ratio=0.7保障)69 selected = [v for k, v in enumerate(good) if matches_mask[k]]70 # 针对所有的selected点再次计算出更精确的转化矩阵M来71 sch_pts, img_pts = np.float32([kp_sch[m.queryIdx].pt for m in selected]).reshape(72 -1, 1, 2), np.float32([kp_src[m.trainIdx].pt for m in selected]).reshape(-1, 1, 2)73 M, mask = _find_homography(sch_pts, img_pts)74 # print(M, mask)75 # 计算四个角矩阵变换后的坐标，也就是在大图中的目标区域的顶点坐标:76 h, w = im_search.shape[:2]77 h_s, w_s = im_source.shape[:2]78 pts = np.float32([[0, 0], [0, h - 1], [w - 1, h - 1], [w - 1, 0]]).reshape(-1, 1, 2)79 dst = cv2.perspectiveTransform(pts, M)80 # trans numpy arrary to python list: [(a, b), (a1, b1), ...]81 def cal_rect_pts(dst):82 return [tuple(npt[0]) for npt in dst.astype(int).tolist()]83 pypts = cal_rect_pts(dst)84 # 注意：虽然4个角点有可能越出source图边界，但是(根据精确化映射单映射矩阵M线性机制)中点不会越出边界85 lt, br = pypts[0], pypts[2]86 middle_point = int((lt[0] + br[0]) / 2), int((lt[1] + br[1]) / 2)87 # 考虑到算出的目标矩阵有可能是翻转的情况，必须进行一次处理，确保映射后的“左上角”在图片中也是左上角点：...

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homography.py

Source:homography.py Github

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...6 def __init__(self, matcher: cv2.DescriptorMatcher, neighbours: int = 3):7 self.matcher = matcher8 self.matches: List[cv2.DMatch] = list()9 self.neighbours = neighbours10 def _find_homography(self,11 kp_frame: List[cv2.KeyPoint],12 kp_marker: List[cv2.KeyPoint],13 matches: List[cv2.DMatch],14 threshold: float = 5.001,15 maxiters: int = 10000000) -> Array[float]:16 src_pts = np.float32([kp_marker[m.queryIdx].pt for m in matches]).reshape(-1, 1, 2)17 dst_pts = np.float32([kp_frame[m.trainIdx].pt for m in matches]).reshape(-1, 1, 2)18 M, _ = cv2.findHomography(src_pts, dst_pts, cv2.RANSAC,19 ransacReprojThreshold=threshold, maxIters=maxiters)20 return M21 def __call__(self, 22 kp_frame: List[cv2.KeyPoint],23 kp_marker: List[cv2.KeyPoint],24 des_frame: List[cv2.DMatch],25 des_marker: List[cv2.DMatch],26 n_best: int = 100,27 threshold: float = 5.0) -> Array[float]:28 29 if isinstance(self.matcher, cv2.FlannBasedMatcher):30 self.matches = self.matcher.knnMatch(des_marker, des_frame, k=self.neighbours)31 else:32 self.matches = self.matcher.match(des_marker, des_frame)33 self.matches = sorted(self.matches, key=lambda x: x.distance)[:n_best]...

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