Asked  6 Months ago    Answers:  5   Viewed   115 times

How can I crop images, like I've done before in PIL, using OpenCV.

Working example on PIL

im = Image.open('0.png').convert('L')
im = im.crop((1, 1, 98, 33))
im.save('_0.png')

But how I can do it on OpenCV?

This is what I tried:

im = cv.imread('0.png', cv.CV_LOAD_IMAGE_GRAYSCALE)
(thresh, im_bw) = cv.threshold(im, 128, 255, cv.THRESH_OTSU)
im = cv.getRectSubPix(im_bw, (98, 33), (1, 1))
cv.imshow('Img', im)
cv.waitKey(0)

But it doesn't work.

I think I incorrectly used getRectSubPix. If this is the case, please explain how I can correctly use this function.

 Answers

84

It's very simple. Use numpy slicing.

import cv2
img = cv2.imread("lenna.png")
crop_img = img[y:y+h, x:x+w]
cv2.imshow("cropped", crop_img)
cv2.waitKey(0)
Tuesday, June 1, 2021
 
Xatoo
answered 6 Months ago
16

I was able to get this to work using Python 2.7.13 and opencv-python==3.1.0.4

Here is the code for it.

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 it is as a gist.

Essentially, you need to make sure you're using the right matching method.

Wednesday, August 4, 2021
 
edsk
answered 4 Months ago
97

For future readers, the code does work it was just that my image was so blurry, the above code had no visible effect!

Saturday, August 14, 2021
 
e_i_pi
answered 4 Months ago
32

Use Graphics.DrawImage Method (Image, RectangleF, RectangleF, GraphicsUnit) method.

    Dim fileName = "C:file.jpg"
    Dim CropRect As New Rectangle(100, 0, 100, 100)
    Dim OriginalImage = Image.FromFile(fileName)
    Dim CropImage = New Bitmap(CropRect.Width, CropRect.Height)
    Using grp = Graphics.FromImage(CropImage)
        grp.DrawImage(OriginalImage, New Rectangle(0, 0, CropRect.Width, CropRect.Height), CropRect, GraphicsUnit.Pixel)
        OriginalImage.Dispose()
        CropImage.Save(fileName)
    End Using
Monday, August 23, 2021
 
Jasarien
answered 4 Months ago
92

MP4V or MPEG-4 part 2 is not supported by most browsers, you may want to try H.264 (MPEG-4 part 10) instead.

To do that, change:

fourcc = cv2.VideoWriter_fourcc(*'MP4V')

to

fourcc = cv2.VideoWriter_fourcc(*'H264')

If you are using Python 3, use the following hexadecimal code instead (there seems to be a bug when using the four bytes notation):

fourcc = 0x00000021

Run the script and you will likely get the following error message:

Failed to load OpenH264 library: openh264-1.6.0-win32msvc.dll Please check environment and/or download library: https://github.com/cisco/openh264/releases

You need to do as the message says and download the required library from github and place it somewhere accessible by your PATH.

Using H.264 compression you will also get a smaller file which is better for Web.

Monday, September 27, 2021
 
ANIS BOULILA
answered 2 Months ago
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