When receiving "ValueError: not enough values to unpack (expected 2, got 1)", how can I force the program to ignore and continue?

When receiving "ValueError: not enough values to unpack (expected 2, got 1)", how can I force the program to ignore and continue?

我正在使用 Python (3) 和 OpenCV (3.3) 在网络摄像头上 运行 实时对象检测,使用样本图像,然后与视频流进行特征匹配。我已经使用 SIFT/SURF 让它工作,但我正在尝试使用 ORB 算法。

我在某些情况下收到以下错误导致程序崩溃:

for i, (m, n) in enumerate(matches):
ValueError: not enough values to unpack (expected 2, got 1)

我明白它崩溃的原因,有时图像之间有很好的匹配,有时没有,导致不匹配。

我的问题是,如何强制程序忽略并跳过没有足够值的情况并继续 运行ning。

问题代码的主要区域:

    for i, (m, n) in enumerate(matches):
        if m.distance < 0.7*n.distance:
            good.append(m)

示例 'matches' 输出:

[[<DMatch 0x11bdcc030>, <DMatch 0x11bbf20b0>], [<DMatch 0x11bbf2490>, <DMatch 0x11bbf24f0>], [<DMatch 0x11bbf2750>, <DMatch 0x11bbf25d0>], [<DMatch 0x11bbf2570>, <DMatch 0x11bbf2150>], etc etc

完整代码:

import numpy as np
import cv2
from matplotlib import pyplot as plt
import matplotlib.patches as mpatches
import os, os.path
import math
import time
from datetime import datetime
startTime = datetime.now()

MIN_MATCH_COUNT = 10   # default=10

img1 = cv2.imread('Pattern3_small.jpg',0)          # queryImage

# Create ORB object. You can specify params here or later.
orb = cv2.ORB_create()

cap = cv2.VideoCapture(0)
# cap = cv2.VideoCapture("output_H264_30.mov")

# find the keypoints and descriptors with SIFT
kp1, des1 = orb.detectAndCompute(img1,None)

pts_global = []
dst_global = []

position = []
heading = []
# plt.axis([0, 1280, 0, 720])

tbl_upper_horiz = 1539
tbl_lower_horiz = 343
tbl_upper_vert = 1008
tbl_lower_vert = 110

# cv2.namedWindow("Frame", cv2.WINDOW_NORMAL)
# cv2.resizeWindow("Frame", 600,350)

while True:
    _, img2 = cap.read()

    # Start timer
    timer = cv2.getTickCount()

    # find the keypoints and descriptors with SIFT
    # kp1, des1 = sift.detectAndCompute(img1,None)
    kp2, des2 = orb.detectAndCompute(img2,None)

    FLANN_INDEX_KDTREE = 0
    FLANN_INDEX_LSH = 6
    # index_params = dict(algorithm = FLANN_INDEX_KDTREE, trees = 5)
    index_params= dict(algorithm = FLANN_INDEX_LSH,
                   table_number = 6, # 12, 6
                   key_size = 12,     # 20, 12
                   multi_probe_level = 1) #2, 1
    search_params = dict(checks = 50)

    flann = cv2.FlannBasedMatcher(index_params, search_params)

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

    # print (matches)

    # Calculate Frames per second (FPS)
    fps = cv2.getTickFrequency() / (cv2.getTickCount() - timer);

    # store all the good matches as per Lowe's ratio test.
    good = []

    # ratio test as per Lowe's paper
    for i, (m, n) in enumerate(matches):
        if m.distance < 0.7*n.distance:
            good.append(m)

# Do something afterwards

感谢您的帮助。

matches的每个元素视为一个集合并使用异常处理:

for i, pair in enumerate(matches):
    try:
        m, n = pair
        if m.distance < 0.7*n.distance:
            good.append(m)

    except ValueError:
        pass

如果您这样做:

for i, (m, n) in enumerate(matches):

而且你不能保证所有的元组总是包含两个元素,那么你应该这样做:

for i, values in enumerate(matches): if len(values) < 2: continue # you don't have the second element to compare against ... # Do your usual processing here

上面的代码是明确的和可读的,即很明显您需要两个元素才能在处理链中进一步进行。它也不太容易出错,因为您要确保拥有正确的数据才能继续。它还允许您计算有多少元组被丢弃。