当收到“ValueError: not enough values to unpack

Question

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。

在某些情况下，我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)

我理解崩溃背后的原因，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。

有问题的主要代码区域：

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

        if m.distance < 0.7*n.distance:

            good.append(m)

“匹配”输出示例：

[[<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

Answer 1

将 的每个元素matches视为一个集合并使用异常处理：for i, pair in enumerate(matches):&nbsp; &nbsp; try:&nbsp; &nbsp; &nbsp; &nbsp; m, n = pair&nbsp; &nbsp; &nbsp; &nbsp; if m.distance < 0.7*n.distance:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; good.append(m)&nbsp; &nbsp; except ValueError:&nbsp; &nbsp; &nbsp; &nbsp; pass

Answer 2

如果你这样做：for i, (m, n) in enumerate(matches):并且您不能保证所有元组始终包含两个元素，那么您应该执行以下操作：for i, values in enumerate(matches): &nbsp; if len(values) < 2: &nbsp; &nbsp; continue # you don't have the second element to compare against &nbsp; ... &nbsp; # 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，因为您要确保有正确的数据可以继续。它还允许您计算丢弃了多少元组。