I have an indexed image (2-D, not rgb), and i use imagesc to display the image. This function gives a range from blue to red, which can be set by colormap, and can be viewed by colorbar.
Now, I want to change the background, that is usually 0's or NaN's, to white or black, but that it will not affect or change the range of the colormap/colorbar. I've tried converting all the image to 3-D rgb, but this prevent the option of changing the contrast, or the clims, like in imagesc.
So, there is other way to do that?
EDIT:
#Shai's solution was good, but it caused other problem-
When I have an image with a range of values and the background is NaN's, and I display the image in a specific contrast (by imagesc(img, [-1,1]) for example), I get range of colors between -1 to 1, and i get white in the background (because i put [1 1 1] in the first entry of the colormap), but also all the values under the minimum of contrast (i.e., <-1 in the example) also get the white color instead of the bottom dark blue in the colorbar scale.
Any idea for that...?
Many Thanks.
Consider putting [1, 1, 1] as the first entry of your colormap to get the background as white.
Related
I have a problem with a binarization method.
I have images with text, which I want to binarize.
I want the text ends up being white, but there are images with the text darker than the background and there are images with text less dark than the background.
I want to binarize images like these, but I want the text in white color in the binarized images.
By the way, I am binarizing images with this code. This code is good for images with text darker than the background but it isn't good for text less darker than the background. I think I need a method to know if the text is more or less dark than the background for to invert or no invert the binarization.
umb = graythresh(originalImage);
binaryImage =(~im2bw(originalImage,umb));
How can I do it?
Thanks a lot for the help
there are 2 possible solutions which I had in mind:
solution1:
generate a grayscale image using rgb2gray function.
generate a histogram from the grayscale image, and ignore the transparent pixels. you can use imhist function.
check what is the histogram maximal value. if the value is high - the background is probably light and the text should be darker than the background. in this case - take the negative image (for example, by using imcomplement), and then binarize it. otherwise - you can binarize it as is.
solution 2:
the solution asserts that the image is simple enough, i.e. doesn't have a lot of connected components other than the letters.
binarize the input image.
divide the image into connected components using bwconncomp function.
for each connected component find it's representative value, it can either be 0 or 1
check what is the most common represantative value. if it is 1 - the letters are dark. in this case take the negative image and than binarize. otherwise - binarize the input image as is.
good luck!
I wanna to color a sprite/icon with a transparent background and with shadows. I tried to shift the hue to all pixels but it looks not so natural and I have problems with the black and the white colors in an image. If an image tend to be black shifting the hue do not change the black in red or another color even shifting by 360 degrees.
Tried to color addicting and subtracting color and even in that case the black and the white tend to be colored or disappears at all.
Maybe should I put an image on the icon to achieve the coloring effect ?
Any suggestions on how to proceed.
I lost.
You've been asking a lot about this hue shifting thing, so I figured I'd try to work out an example: http://jsfiddle.net/EMujN/3/
Here's another that uses an actual icon: http://jsfiddle.net/EMujN/4/
There's a lot in there. There's a huge data URL which you can ignore unless you want to replace it. Here's the relevant part where we modify HSL.
//SHIFT H HERE
var hMod = .3;
hsl[0]=(hsl[0]+hMod)%1;
//MODIFY S HERE
var sMod = .6;
hsl[1]=Math.max(0,Math.min(1,
hsl[1]+sMod
));
//MODIFY L HERE
var lMod = 0;
hsl[2]=Math.max(0,Math.min(1,
hsl[2]+lMod
));
I've converted to HSL because it's a lot easier to accomplish what you want in that color space than RGB.
Without getting any more complex, you have three variables you can tune: how much to add to either Hue, Saturation, or Lightness. I have the lightness variable set to 0 because any higher and you will see some nasty JPEG artifacts (if you can find a decent .png that would be better, but I went with the first CC night image I could find).
I think the hue shift (yellow to green) looks pretty good though and I have maxed out the saturation, so even a normally white light appears bright purple. Like I said in my comment, you will need to increase the lightness and saturation if you want to colorize patches of black and white. Hopefully, you can figure out what you need from this example.
image used: http://commons.wikimedia.org/wiki/File:Amman_(Jordan)_at_night.jpg
I found a better solution by myself which can solve the problem with the black and white.
So basically the solution can be solved in multiple steps. Here I will define the steps. Later I'll provide some working code:
Get the image
Calculate the predominant color, averaging the image pixels or simply providing an input RGB value which is the predominant that your eye can catch.
If the predominant tends to be black or white, or both, the image has to be recolored with an addictive or subtractive method, addictive if black, subtractive if white. So basically all RGB pixels should be attenuated or sharpened until RED. I think that the best solution should be RED, because RED is first in the HUE scale, and this can help when we will hue-shift the pixels.
To have a unique algorithm which can work with different kind of images, not only black predominant or white, ideally the input the non-black and non-white predominant images should be pre-hueshifted manually, using photoshop or with another algorithm in a way that the new predominant color results to be RED too
After that the Hue shifting coloring is straighforward. We know that the predominant color is RED for all the images, and we'll shift the HUE values with a difference between the HSV value of the desired color and the HSV of the predominant color (RED).
Game over. We have a pretty universal way to color different images with hue shifting in a natural way.
Another question could be how to authomatically pre-shift the input images which predominant color is not black or white.
But this is another question.
Why this coloring method could be considered natural. Simply consider one thing. Generally the non dominant black or white colors are part of the shadows and light which gives a 3D feel to the images. On the other hand if my shoes are 100% black and i will tint them with some colors, they will no more be black. Color the dominant black cannot be achieved simply shifting the HSV parameters but other steps should be performed. The steps are the above described.
I have more then 1 week reading about selective color change of an image. It meand selcting a color from a color picker and then select a part of image in which I want to change the color and apply the changing of color form original color to color of color picker.
E.g. if I select a blue color in color picker and I also select a red part in the image I should be able to change red color to blue color in all the image.
Another example. If I have an image with red apples and oranges and if I select an apple on the image and a blue color in the color picket, then all apples should be changing the color from red to blue.
I have some ideas but of course I need something more concrete on how to do this
Thank you for reading
As a starting point, consider clustering the colors of your image. If you don't know how many clusters you want, then you will need methods to determine whether to merge or not two given clusters. For the moment, let us suppose that we know that number. For example, given the following image at left, I mapped its colors to 3 clusters, which have the mean colors as shown in the middle, and representing each cluster by its mean color gives the figure at right.
With the output at right, now what you need is a method to replace colors. Suppose the user clicks (a single point) somewhere in your image, then you know the positions in the original image that you will need to modify. For the next image, the user (me) clicked on a point that is contained by the "orange" cluster. Then he clicked on some blue hue. From that, you make a mask representing the points in the "orange" cluster and play with that. I considered a simple gaussian filter followed by a flat dilation 3x5. Then you replace the hues in the original image according to the produced mask (after the low pass filtering, the values on it are also considered as a alpha value for compositing the images).
Not perfect at all, but you could have a better clustering than me and also a much-less-primitive color replacement method. I intentionally skipped the details about clustering method, color space, and others, because I used only basic k-means on RGB without any pre-processing of the input. So you can consider the results above as a baseline for anything else you can do.
Given the image, a selected color, and a target new color - you can't do much that isn't ugly. You also need a range, some amount of variation in color, so you can say one pixel's color is "close enough" while another is clearly "different".
First step of processing: You create a mask image, which is grayscale and varying from 0.0 to 1.0 (or from zero to some maximum value we'll treat as 1.0), and the same size as the input image. For each input pixel, test if its color is sufficiently near the selected color. If it's "the same" or "close enough" put 1.0 in the mask. If it's different, put 0.0. If is sorta borderline, put an in-between value. Exactly how to do this depends on the details of the image.
This might work best in LAB space, and testing for sameness according to the angle of the A,B coordinates relative to their origin.
Once you have the mask, put it aside. Now color-transform the whole image. This might be best done in HSV space. Don't touch the V channel. Add a constant to S, modulo 360deg (or mod 256, if S is stored as bytes) and multiply S by a constant chosen so that the coordinates in HSV corresponding to the selected color is moved to the HSV coordinates for the target color. Convert the transformed S and H, with the unchanged L, back to RGB.
Finally, use the mask to blend the original image with the color-transformed one. Apply this to each channel - red, green, blue:
output = (1-mask)*original + mask*transformed
If you're doing it all in byte arrays, 0 is 0.0 and 255 is 1.0, and be careful of overflow and signed/unsigned problems.
How can I select a mask of irregular shape in an image in Matlab without using the imfreehand but automatically? The background of the image is black and I would like to select the entire image as mask just without the black background (everything without the black background).
try this:
threshold=0; % or a different value if needed
mask=image>threshold;
given that the background is truly black, i.e. pixels values are 0, the set threshold to zero. Otherwise, select a value that captures the background (there are ways to do that automatically)
in Matlab,
if I do:
output = false(5, 5);
imshow(output);
it will show me a black square instead of a white binary square image. Is there any reason to this? How can I output a white binary square?
The reason is that false is mapped to 0, and true is mapped to 1.
Also, when showing images, higher number is shown by higher intensity. White has more intensity than black.
Another way to think about it, is that usually you have 256 values - 0-255. 0 is totally black and 255 is totally white. Now, imagine that you do a quantization to two colors. It is now obvious that 0 should be black.
In order to show white square, use
output = true(5,5)
You could use imcomplement
imshow(imcomplement(false(5, 5)))
or modify the default color mapping (quoting from imshow's documentation)
imshow(X,map)
displays the indexed image X with the colormap map. A color map matrix may have any number of rows, but it must have exactly 3 columns. Each row is interpreted as a color, with the first element specifying the intensity of red light, the second green, and the third blue. Color intensity can be specified on the interval 0.0 to 1.0.
You could also change the figure's colormap to customize how MATLAB maps values to colors:
BW = [false,true;true,false];
imshow(BW)
set(gcf, 'Colormap',[1,1,1;0,0,0])