具有透明度的python opencv cv2 matchTemplate

Question

OpenCV 3.0.0 添加了在执行模板匹配时指定掩码的功能。当我指定掩码时，出现此错误:error: (-215) (depth == CV_8U || depth == CV_32F) && type == _templ.type() && _img.dims() <= 2 in function matchTemplateMask

模板图像(透明的 PNG):

源图片:

代码

# read the template emoji with the alpha channel
template = cv2.imread(imagePath, cv2.IMREAD_UNCHANGED)
channels = cv2.split(template)
zero_channel = np.zeros_like(channels[0])
mask = np.array(channels[3])

# all elements in alpha_channel that have value 0 are set to 1 in the mask matrix
mask[channels[3] == 0] = 1

# all elements in alpha_channel that have value 100 are set to 0 in the mask matrix
mask[channels[3] == 100] = 0

transparent_mask = cv2.merge([zero_channel, zero_channel, zero_channel, mask])

print image.shape, image.dtype  # (72, 232, 3) uint8
print template.shape, template.dtype  # (40, 40, 4) uint8
print transparent_mask.shape, transparent_mask.dtype    # (40, 40, 4) uint8

# find the matches
res = cv2.matchTemplate(image, template, cv2.TM_CCOEFF_NORMED, mask=transparent_mask)

图片类型有问题吗？我无法使用 matchTemplate 方法的新掩码参数找到任何示例(在 Python 中)。有谁知道如何制作面具？

Answer 1

我能够使用 Python 2.7.13 和 opencv-python==3.1.0.4 让它工作

这是它的代码。

import cv2
import numpy as np
import sys

if len(sys.argv) < 3:
    print 'Usage: python match.py <template.png> <image.png>'
    sys.exit()

template_path = sys.argv[1]
template = cv2.imread(template_path, cv2.IMREAD_UNCHANGED)
channels = cv2.split(template)
zero_channel = np.zeros_like(channels[0])
mask = np.array(channels[3])

image_path = sys.argv[2]
image = cv2.imread(image_path, cv2.IMREAD_UNCHANGED)

mask[channels[3] == 0] = 1
mask[channels[3] == 100] = 0

# transparent_mask = None
# According to http://www.devsplanet.com/question/35658323, we can only use
# cv2.TM_SQDIFF or cv2.TM_CCORR_NORMED
# All methods can be seen here:
# http://docs.opencv.org/2.4/doc/tutorials/imgproc/histograms/template_matching/template_matching.html#which-are-the-matching-methods-available-in-opencv
method = cv2.TM_SQDIFF  # R(x,y) = \sum _{x',y'} (T(x',y')-I(x+x',y+y'))^2 (essentially, sum of squared differences)

transparent_mask = cv2.merge([zero_channel, zero_channel, zero_channel, mask])
result = cv2.matchTemplate(image, template, method, mask=transparent_mask)
min_val, max_val, min_loc, max_loc = cv2.minMaxLoc(result)
print 'Lowest squared difference WITH mask', min_val

# Now we'll try it without the mask (should give a much larger error)
transparent_mask = None
result = cv2.matchTemplate(image, template, method, mask=transparent_mask)
min_val, max_val, min_loc, max_loc = cv2.minMaxLoc(result)
print 'Lowest squared difference WITHOUT mask', min_val

Here这是一个要点。

本质上，您需要确保使用正确的匹配方法。