特征匹配+单应性

Question

我已经在python中关注了opencv的教程。 我正在使用OpenCV 3.2和Python 3.6.1。 代码是这样写的：

import cv2
import numpy as np
import matplotlib.pyplot as plt

MIN_MATCH_COUNT = 10

img1 = cv2.imread('test.jpg',0)
img2 = cv2.imread('hanapin_mo.jpg',0)

sift = cv2.xfeatures2d.SIFT_create()

kp1, des1 = sift.detectAndCompute(img1,None)
kp2, des2 = sift.detectAndCompute(img2,None)

FLANN_INDEX_KDTREE = 1
index_params = dict(algorithm = FLANN_INDEX_KDTREE, trees = 5)
search_params = dict(checks = 50)

flann = cv2.FlannBasedMatcher(index_params, search_params)

matches = flann.knnMatch(des1, des2, k = 2)

good = []
for m,n in matches:
    if m.distance <0.7*n.distance:
        good.append(m)

if len(good)>MIN_MATCH_COUNT:
    src_pts = np.float32([ kp1[m.queryIdx].pt for m in good ]).reshape(-1,1,2)
    dst_pts = np.float32([ kp2[m.trainIdx].pt for m in good ]).reshape(-1,1,2)

    M, mask = cv2.findHomography(src_pts, dst_pts, cv2.RANSAC, 5.0)
    matchesMask = mask.ravel().tolist()

    h,w,d = img1.shape
    pts = np.float32([ [0,0],[0,h-1],[w-1,h-1],[w-1,0] ]).reshape(-1,1,2)
    dst = cv2.perspectiveTransform(pts,M)

    img2 = cv2.polylines(img2,[np.int32(dst)], True, 255, 3, cv2.LINE_AA)

else:
    print("Not enough matches are found - {}/{}".format(len(good), MIN_MATCH_COUNT) )
    matchesMask = None

draw_params = dict(matchColor = (0,250,0),
                   singlePointColor = None,
                   matchesMask = matchesMask,
                   flags = 2)

img3 = cv2.drawMatchesKnn(img1, kp1, img2, kp2, good, None, **draw_params)

plt.imshow(img3, 'gray'),plt.show()

当我运行模块时，发生错误：

追溯（最近一次通话最近）：文件“ C：/ Users / Albert Eli Reyes / Desktop / SIFT / SIFT.py”，第35行，在h，w，d = img1.shape ValueError：没有足够的值可用于解包（预期3，得到2）

    h,w,d = img1.shape
    pts = np.float32([ [0,0],[0,h-1],[w-1,h-1],[w-1,0] ]).reshape(-1,1,2)
    dst = cv2.perspectiveTransform(pts,M)

    img2 = cv2.polylines(img2,[np.int32(dst)], True, 255, 3, cv2.LINE_AA)

说“没有足够的值要解包”是什么意思？ 我应该怎么做才能使其运行？

Answer 1

形状函数仅输出两个值的元组。 您试图获取三个值，而该函数仅返回两个值

h,w = img1.shape

上面的工作

Answer 2

好吧，这意味着img1.shape仅返回两个值。 因此，如果删除d变量初始化，则代码应该可以正常工作，因为我仍然看不到在代码中使用它。

h, w = img1.shape