I would like to understand a bit better the behavior of rotate_map_pixel() in this particular case. Below I try to provide a minimal toy example:
map = hp.ma(hp.read_map(mapFile))
map.mask = np.logical_not(hp.read_map(maskFile))
rot = hp.Rotator(coord=['G','C'])
map = rot.rotate_map_pixel(map)
mask = hp.read_map(maskFile)
mask = rot.rotate_map_pixel(mask)
This produces a slightly different map.mask and mask. By "different", I mean
skycut = np.where(map.mask[obj_pix]==False))
and
skycut = np.where(mask[obj_pix]==1.)
or
skycut = np.where(np.isclose(mask[obj_pix],1.))
all produce different skycut. I'd guess this is due to the interpolation procedure?
If I wish to use the mask for selection purpose, what should I do?
If you display the 2 masks, you can see that in one case the mask is a float, in the other case it is a boolean.
Now one case healpy fills the map with UNSEEN and then interpolation is handled by HEALPix C++. I don't know how internally HEALPix handles that.
In the other case we pass a map of 0 and 1 and HEALPix does the interpolation, but we don't trigger any special case of handling UNSEEN values.
See https://zonca.dev/2021/06/healpy-rotate-mask.html a full notebook investigating that.
I find out experimentally that if I round the mask with a number very close to 1 I find the same mask.
Related
I have data that change in size and want to display them in the same window. The command
void ImageResize( BasicImage im, Number num_dim, Number... )
seems like a potential fit, but the syntax is not clear at all.
Let's say I have 512x5 data set and now it needs to be 367x5.
The , Number...) indicates that this command takes a different number of parameters, all of them interpreted as number parameters. Commands which do this, usually use one of their other parameters to specify how many such parameters follow.
A typical example for this is also the SliceN command.
In this particular case, the command not only allows you to change the size of the dimensions in the image, but also the number of dimensions. It is a very useful command to f.e. change a 2D image into a 3D stack or the like.
The command ImageResize( BasicImage im, Number num_dim, Number... ) does several things:
It replaces im in-place, so the meta-data, display and window remains the same
It adjusts the dimension calibration when the dimension size is changed. Here, the assumption is, that the field-of-view before and
after the resize is the same. (The command can be used to easily scale
images as shown in the example below.)
All values of the image im are set to zero. ( If you need to keep the values, you need to act on an image clone!)
Example 1: Resizing image with bilinar interpolation
image before := GetFrontImage()
number sx, sy
before.GetSize(sx,sy)
number factor = 1.3
image after := before.ImageClone()
after.ImageResize( 2, factor*sx, factor*sy ) // Adjusts the empty container with meta-data
after = warp(before, icol/factor, irow/factor ) // interpolate data
after.ShowImage()
Example 2: Extend 2D image into 3D stack
number sx = 100
number sy = 100
image img := RealImage("2D",4,sx,sy)
img = iradius* Random()
img.ShowImage()
OKDialog("Now into a stack...")
number sz = 10
img.ImageResize(3,sx,sy,sz) // All values are zero now!
img = iradius * Random()
So I have the code to import a stack of images, but I am getting an error: Subscripted assignment dimension mismatch.
myPath = 'E:\folder name\'; %'
fileNames = dir(fullfile(myPath, '*.tif'));
width = 1400;
height = 1050;
nbImages = length(fileNames);
C=uint8(zeros(width, height, nbImages));
for i=1:length(fileNames)
C(:,:,i)=imread(cat(2,'E:\folder name\',fileNames(i).name));
i
end
I understand that the error is originating from the for loop, but I don't know of any other way to fill in an empty matrix with images.
Your images must not be all the same size. You can handle this by using explicit assignment for the first two dimensions. This will zero-pad any images which are smaller than the rest.
im = imread(...);
C(1:size(im, 1), 1:size(im, 2), i) = im;
Also, there is a good chance that your images have multiple color channels (the third dimension), so you'll likely want to concatenate along the fourth dimension rather than the third.
C(:,:,:,i) = imread(...)
Obviously it all depends what you want to do with the images, but in general, if you want a "stack" of images (or a "stack" of anything, really), then it sounds like you should be collecting them as a cell array instead.
Also, the correct way to create safe filenames is using the fullfile command
e.g.
C = cell(1, length(nbImages));
for i = 1 : length (fileNames)
C{i} = imread (fullfile ('E:','folder name', fileNames(i).name));
end
If you really want to concatenate to a 3D matrix from your cell array, assuming you have checked this is possible, you can do this very easily using comma-separated-list generator syntax:
My3DMatrix = cat(3, C{:});
I recently asked how to convert Float32 or Uint8 arrays into images in the Images package. I got an answer for the Float32 case, but am still having trouble figuring out how to save a Uint8 array.
As an example, let's create a random Uint8 array using the traditional Matlab scheme where the dimensions are (m,n,3):
array = rand(Uint8, 50, 50, 3);
img = convert(Image, array);
Using the same approach as works for the Float32 case,
imwrite(img, "out.png")
fails with message
ERROR: method 'mapinfo' has no method matching mapinfo(::Type{ImageMagick}, ::Image{Uint8, 3, Image{Uint8, 3, Array{Uint8, 3}}}).
I checked the documentation, and it says
If data encodes color information along one of the dimensions of the array (as opposed to using a ColorValue array, from the Color.jl package), be sure to specify the "colordim" and "colorspace" in properties.
However, inspecting the img object previously created shows that it has colordim = 3 and colorspace = RGB already set up, so this can't be the problem.
I then searched the documentation for all instances of MapInfo. In core.md there is one occurrence:
scalei: a property that controls default contrast scaling upon display. This should be a MapInfo value, to be used for setting the contrast upon display. In the absence of this property, the range 0 to 1 will be used.
But there was no information on what exactly a MapInfo object is, so I looked further, and in function_reference.md it says:
Here is how to directly construct the major concrete MapInfo types:
MapNone(T), indicating that the only form of scaling is conversion to type T. This is not very safe, as values "wrap around": for example, converting 258 to a Uint8 results in 0x02, which would look dimmer than 255 = 0xff.
...
and some other examples. So I tried to specify scalei = MapNone(Uint8) as follows:
img2 = Image(img, colordim = 3, colorspace = "RGB", scalei = MapNone(Uint8));
imwrite(img, "out.png")
but got the same error again.
How do you encode Uint8 image data using Images in Julia?
You can convert back and forth between arrays of primitive types such as UInt8 and arrays of color types. These conversions are achieved in a unified way via two functions: colorview and channelview.
Example
Convert array of UInt8 to array of RGB:
arr = rand(UInt8, 3, 50, 50)
img = colorview(RGB, arr / 255)
Convert back to channel view:
channelview(img)
Notes
In this example the RGB color type requires that the entries of the array live in [0,1] as floating point. I manually converted UInt8 to Float64 using an explicit division by 255. There is probably a more generic way of achieving this result with reinterpret or some other function in Images.jl
The colorview and channelview functions assume that the channel dimension is the first dimension of the array. You can use permutedims in case your channels live in a different dimension, or use some function in Images.jl (maybe reinterpretc?) to do it efficiently without memory copies.
I want to get only leaf from an image.
The background is a normal white paper(A4) and there is some shadow.
I apply some method (structure element,edge detection using filter) but I cannot find the general way which can apply all the image.
these are examples.
Are there better methods for this problem??
thank you
another example.
and the result I got is
By using
hsv_I = rgb2hsv(I);
Is = hsv_I(:,:,2);
Is_d = imdilate(Is,strel('diamond',4));
Is_e = imerode(Is,strel('diamond',2));
Is_de = imerode(Is_d,strel('disk',2));
Is_def = imfill(Is_de,'holes');
Is_defe = imerode(Is_def,strel('disk',5));
Then Is_defe is a mask to segment
But the method that i did is very specific. I cannot use this in general.
If you have the Image Processing Toolbox, you could do as follows:
The code below first estimates the threshold with the function graythresh, thresholds the image and fills holes with the imfill function. Suppose I is a cell containing your RGB images:
for k=1:length(I)
t=graythresh(rgb2gray(I{k}));
BW{k}=imfill(~im2bw(I{k}, t), 'holes');
subplot(length(I),1,k), imshow(BW{k});
end
I am trying to overlay an activation map over a baseline vasculature image but I keep getting the same error below:
X and Y must have the same size and class or Y must be a scalar double.
I resized each to 400x400 so I thought it would work but no dice. Is there something I am missing? It is fairly straight forward for a GUI I am working on. Any help would be appreciated.
a=imread ('Vasculature.tif');
b = imresize (a, [400,400]);
c=imread ('activation.tif');
d= imresize (c, [400,400]);
e=imadd (b,d);
Could it be the bit depth or dpi?
I think one of your images is RGB (size(...,3)==3) and the other is grayscale (size(...,3)==1). Say the vasculature image a is grayscale and the activation image c is RGB. To convert a to RGB to match c, use ind2rgb, then add.
aRGB = ind2rgb(a,gray(256)); % assuming uint8
Alternatively, you could do aRGB = repmat(a,[1 1 3]);.
Or to put the activation image into grayscale:
cGray = rgb2gray(c);
Also, according to the documentation for imadd the two images must be:
nonsparse numeric arrays with the same size and class
To get the uint8 and uint16 images to match use the im2uint8 or im2uint16 functions to convert. Alternatively, just rescale and cast (e.g. b_uint8 = uint8(double(b)*255/65535);).
Note that in some versions of MATLAB there is a bug with displaying 16-bit images. The fix depends on whether the image is RGB or gray scale, and the platform (Windows vs. Linux). If you run into problems displaying 16-bit images, use imshow, which has the fix, or use the following code for integer data type images following image or imagesc:
function fixint16disp(img)
if any(strcmp(class(img),{'int16','uint16'}))
if size(img,3)==1,
colormap(gray(65535)); end
if ispc,
set(gcf,'Renderer','zbuffer'); end
end
chappjc's answers is just fine, I want to add a more general answer to the question how to solve the error message
X and Y must have the same size and class or Y must be a scalar double
General solving strategy
At which line does the error occur
Try to understand the error message
a. "... must have the same size ...":
Check the sizes of the input.
Try to understand the meaning of your code for the given (type of) input parameters. Is the error message reasonable?
What do you want to achieve?
Useful command: size A: returns the size of A
b. "... must have the same class ...":
Check the data types of the input arguments.
Which common data type is reasonable?
Convert it to the chosen data type.
Usefull command: whos A: returns all the meta information of A, i.e. size, data type, ...
Implement the solution: your favorite search engine and the matlab documentation are your best friend.
Be happy: you solved your problem and learned something new.
A simple code :
a=imread ('image1.jpg');
b=imresize (a, [400,400]);
subplot(3,1,1), imshow(b), title('image 1');
c=imread ('image2.jpg');
d= imresize (c, [400,400]);
subplot(3,1,2), imshow(d), title('image 2');
[x1, y1] = size(b) %height and wedth of 1st image
[x2, y2] = size(d) %height and wedth of 2nd image
for i = 1: x1
for j = 1: y1
im3(i, j)= b(i, j)+d(i, j);
end
end
subplot(3,1,3), imshow (im3), title('Resultant Image');