I want to ask how to Extract the a n×n image segment from an input image centered around the position (x,y), the image has format like this [[num,num,num],[num,numm,num],[num,num,num]........], size of image is about 10 * 10. Thank you !
I do really recommend using numpy, when working with multidimensional arrays/when you want to manipulate these arrays, but here you go, a numpy solution and a solution without numpy.
import numpy as np
img = np.array([[1,2,3,4,5],[2,3,4,5,6],[3,4,5,6,7],[4,5,6,7,8],[5,6,7,8,9]])
print('img:\n',img,'\n')
partOfImage = img[0:3,0:3]
print('partOfImage:\n',partOfImage)
img = [[1,2,3,4,5],[2,3,4,5,6],[3,4,5,6,7],[4,5,6,7,8],[5,6,7,8,9]]
noNumpy =[[img[i][j] for j in range(0,3)] for i in range(0,3)]
print(noNumpy)
Question. So you wish to find the center of an image, expand 1 pixel left, left down, down, right down, right, right up, up, and left up to get an image from the middle of your sample. Correct? I do not have a code example, but a conceptual comment.
Fixed spatial images normally start their addressing at upper left for 0,0. Because of their even numbers of pixels approach, it's near-impossible to find a "middle". So finding a "middle pixel" means you need to apply a percentage and choose the nearest "whole pixel" to the choice. Then expand to your 3x3 dimensions.
Related
I need to add a border to each image in a set of tightly cropped logos. The border color should either match the most commonly used color along the edge, OR, if there are too many different edge colors, it should choose some sort of average of the edge colors.
Here's an example. In this case, I would want the added border to be the same color as the image "background" (using that term in the lay sense). A significant majority of the pixels along the edges are that color, and there are only two other colors, so the decision algorithm would be able to select that rather drecky greenish tan for the added border (not saying anything bad about the organization behind the logo, mind you).
Does Pillow have any functions to simplify this task?
I found answers that show how to use Pillow to add borders and how to determine the average color of an entire image. But I couldn't find any code that looks only at the edges of an image and finds the predominant color, which color could then be used in the border-adding routine. Just in case someone has already done that work, please point me to it. ('Edges' meaning bands of pixels along the top/bottom/left/right margins of the image, whose height or width would be specified as a percentage of the image's total size.)
Short of pointing me to a gist that solves my whole problem, are there Pillow routines that look at edges and/or that count the colors in a pixel range and put them into an array or what not?
I see here that OpenCV can add a border the duplicates the color of the each outermost pixel along all four edges, but that looks funky—I want a solid-color border. And I'd prefer to stick with Pillow—unless another library can do the whole edge-color-analysis-and-add-border procedure in one step, more or less, in which case, please point it out.
Overwrite the center part of the image with some fixed color, that – most likely – won't be present within the edge. For that, maybe use a color with a certain alpha value. Then, there's a function getcolors, which exactly does, what you're looking for. Sort the resulting list, and get the color with the highest count. (That, often, will be the color we used to overwrite the center part. So check for that, and take the second entry, if needed.) Finally, use ImageOps.expand to add the actual border.
That'd be the whole code:
from PIL import Image, ImageDraw, ImageOps
# Open image, enforce RGB with alpha channel
img = Image.open('path/to/your/image.png').convert('RGBA')
w, h = img.size
# Set up edge margin to look for dominant color
me = 3
# Set up border margin to be added in dominant color
mb = 30
# On an image copy, set non edge pixels to (0, 0, 0, 0)
img_copy = img.copy()
draw = ImageDraw.Draw(img_copy)
draw.rectangle((me, me, w - (me + 1), h - (me + 1)), (0, 0, 0, 0))
# Count colors, first entry most likely is color used to overwrite pixels
n_colors = sorted(img_copy.getcolors(2 * me * (w + h) + 1), reverse=True)
dom_color = n_colors[0][1] if n_colors[0][1] != (0, 0, 0, 0) else n_colors[1][1]
# Add border
img = ImageOps.expand(img, mb, dom_color).convert('RGB')
# Save image
img.save('with_border.png')
That'd be the result for your example:
And, that's some output for another image:
It's up to you to decide, whether there are several dominant colors, which you want to mix or average. You'd need to inspect the n_colors appropriately on the several counts for that. That's quite a lot of work, which is left out here.
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System information
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Platform: Windows-10-10.0.16299-SP0
Python: 3.9.1
PyCharm: 2021.1
Pillow: 8.2.0
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I'm looking to composite one image on top of another, but the location and size will be determined by a green box in the original image.
Is this something I can do w/ jimp?
Thanks!
--
Here's what I'm hoping to do:
--
Edit: It looks like OpenCV is a great place to start. After some digging, I'm thinking these are the correct steps:
Detect if there is a square with a specific color in the image
Capture the coordinates & dimensions of that square
Create the new image by compositing face.jpg on top of the original image
I'm not quite sure the best path to approach step 1.
I am using the imagemagick module with Nodejs
im = require('imagemagick');
The imagemagick module uses the imagemagick command line tools.
I use the convert method to crop an image
im.convert([image_path, '-crop', '200x150', '-gravity', 'center', target_path],
function(err, stdout){}
);
This results in two images. The one with the cropped image area - the second with the image garbage i tried to get rid of.
How can i force imagemagick to output one image file only?
Per the imagemagick documentation for cropping, which is admittedly a little obtuse (emphasis added):
The width and height of the geometry argument give the size of the image that remains after cropping, and x and y in the offset (if present) gives the location of the top left corner of the cropped image with respect to the original image.
...
If the x and y offsets are present, a single image is generated, consisting of the pixels from the cropping region.
...
If the x and y offsets are omitted, a set of tiles of the specified geometry, covering the entire input image, is generated.
... so, you just need to specify your x and y offsets as part of your geometry argument, like so: 200x150-100-75
Notice that I've specified -100 and -75 for the upper left corner of your crop region, this is because you set your gravity to center, but it appears that imagemagick tries to intelligently determine the appropriate distance target based on your gravity, and I don't see exactly how it behaves when you choose center. So you may have to play around with this one a bit, or you could omit the gravity and use the actual offset from the top left corner of your original image.
I had to use the +delete parameter to remove the last image from the image sequence.
im.convert([image_file.path, '-crop', geometry, '+delete', thumb_path], ...
Following Quote from this source:
http://www.cambridgeincolour.com/tutorials/image-projections.htm
Equirectangular image projections map the latitude and longitude
coordinates of a spherical globe directly onto horizontal and vertical
coordinates of a grid, where this grid is roughly twice as wide as it
is tall.
I have a 13312 px width and 6656 pixel height Panorama picture. It's a equirectangular projection of a room and have a 2:1 ratio.
I use following formular to calculate the xPosition:
var xPosition = ( panorama.width / 360 ) * azimuth
Azimuth = Phi = Heading = Angle to the left or right
How do I project this now on a 1366x768px browser screen?
I think my results are wrong, because it's not on the point where it should be.. it could be because the sphere has a distortion on the left and right:
Is there any formular to calculate the position with attention to the distortion and scale it to fit on the browser screen? I looked many (MANY) sources to find a solution for this, but they always just say that equirectangular are just lat and long.. they don't consider the distortion.
Last question: To find a special solution, I tryed to put a plane on the circle and expanded the line which shows the alpha angle. I though with Phytagoras I could find the position.. but this didn't worked either.. maybe I did something wrong? Is this the way even possible or am I doing it wrong?
edit
THIS is what I'm actually looking for: http://othree.github.io/360-panorama/three-2d/
The black grid in the background. What is the name of this? For what do I have to google or look for? When you start the 2D Panorama, if you want to get the coordinations of the top right corner of the window, what do you have to do?
The whole calculation problem was about to create a Google Streetview similiar view from a 2:1 equirectangular image. We already found a solution for this with a great help from Martin Matysiak (https://github.com/marmat | Google).
It's been a while so I can't give a direct answer to what the main solution is, but I can provide a URL to an AddOn Martin wrote for adding the custom Markers that we actually were trying to make.
You can follow https://github.com/marmat/google-maps-api-addons and look for yourself. In the end it helped a lot to solve the main problem and let us continue with our main Framework for Google Business Tours.
If you follow the link in the threejs demo you included, it would take you to the source code.
particularly look at:
https://github.com/mrdoob/three.js/blob/dev/examples/webgl_panorama_equirectangular.html
and
https://github.com/mrdoob/three.js/blob/dev/src/geometries/SphereBufferGeometry.js
not sure if there is distortion though. The distortion comes from the fact that the texture is mapped to the sphere, and the sphere is rendered in 3D (openGL).
I have to reduce the dimensions of a JPG image to exactly 350 * 450 pixels.Please tell how do I do so.My JPG image is of dimensions 392*520 pixels.
UPDATE:
Ok, thanks for the answers . Checking them now.
Thanks.
You can't do that with a linear operation, the ratio of the image will be skewed.
There is nearly 3 % absolute difference, I don't know how much that is to the human perception.
350/392 = 0.8929
450/520 = 0.8654
I would recommend scaling it to 89% and then cut the rest of the image to get your 350 * 450. This way you would keep the ratios intact.
You can use paint. Go to the resize button (on top of the rotate button) and click the radio box that says pixels. Choose the amount of pixels you want.
There are many online resize tools, e.g.: http://www.picresize.com/
Upload your image, enter its new dimensions and download it again.