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I am trying to implement convolution by hand in Julia. I'm not too familiar with image processing or Julia, so maybe I'm biting more than I can chew.
Anyway, when I apply this method with a 3*3 edge filter edge = [0 -1 0; -1 4 -1; 0 -1 0] as convolve(img, edge), I am getting an error saying that my values are exceeding the allowed values for the RGBA type.
Code
function convolve(img::Matrix{<:Any}, kernel)
(half_kernel_w, half_kernel_h) = size(kernel) .÷ 2
(width, height) = size(img)
cpy_im = copy(img)
for row ∈ 1+half_kernel_h:height-half_kernel_h
for col ∈ 1+half_kernel_w:width-half_kernel_w
from_row, to_row = row .+ (-half_kernel_h, half_kernel_h)
from_col, to_col = col .+ (-half_kernel_h, half_kernel_h)
cpy_im[row, col] = sum((kernel .* RGB.(img[from_row:to_row, from_col:to_col])))
end
end
cpy_im
end
Error (original)
ArgumentError: element type FixedPointNumbers.N0f8 is an 8-bit type representing 256 values from 0.0 to 1.0, but the values (-0.0039215684f0, -0.007843137f0, -0.007843137f0, 1.0f0) do not lie within this range.
See the READMEs for FixedPointNumbers and ColorTypes for more information.
I am able to identify a simple case where such error may occur (a white pixel surrounded by all black pixels or vice-versa). I tried "fixing" this by attempting to follow the advice here from another stackoverflow question, but I get more errors to the effect of Math on colors is deliberately undefined in ColorTypes, but see the ColorVectorSpace package..
Code attempting to apply solution from the other SO question
function convolve(img::Matrix{<:Any}, kernel)
(half_kernel_w, half_kernel_h) = size(kernel) .÷ 2
(width, height) = size(img)
cpy_im = copy(img)
for row ∈ 1+half_kernel_h:height-half_kernel_h
for col ∈ 1+half_kernel_w:width-half_kernel_w
from_row, to_row = row .+ [-half_kernel_h, half_kernel_h]
from_col, to_col = col .+ [-half_kernel_h, half_kernel_h]
cpy_im[row, col] = sum((kernel .* RGB.(img[from_row:to_row, from_col:to_col] ./ 2 .+ 128)))
end
end
cpy_im
end
Corresponding error
MethodError: no method matching +(::ColorTypes.RGBA{Float32}, ::Int64)
Math on colors is deliberately undefined in ColorTypes, but see the ColorVectorSpace package.
Closest candidates are:
+(::Any, ::Any, !Matched::Any, !Matched::Any...) at operators.jl:591
+(!Matched::T, ::T) where T<:Union{Int128, Int16, Int32, Int64, Int8, UInt128, UInt16, UInt32, UInt64, UInt8} at int.jl:87
+(!Matched::ChainRulesCore.AbstractThunk, ::Any) at ~/.julia/packages/ChainRulesCore/a4mIA/src/tangent_arithmetic.jl:122
Now, I can try using convert etc., but when I look at the big picture, I start to wonder what the idiomatic way of solving this problem in Julia is. And that is my question. If you had to implement convolution by hand from scratch, what would be a good way to do so?
EDIT:
Here is an implementation that works, though it may not be idiomatic
function convolve(img::Matrix{<:Any}, kernel)
(half_kernel_h, half_kernel_w) = size(kernel) .÷ 2
(height, width) = size(img)
cpy_im = copy(img)
# println(Dict("width" => width, "height" => height, "half_kernel_w" => half_kernel_w, "half_kernel_h" => half_kernel_h, "row range" => 1+half_kernel_h:(height-half_kernel_h), "col range" => 1+half_kernel_w:(width-half_kernel_w)))
for row ∈ 1+half_kernel_h:(height-half_kernel_h)
for col ∈ 1+half_kernel_w:(width-half_kernel_w)
from_row, to_row = row .+ (-half_kernel_h, half_kernel_h)
from_col, to_col = col .+ (-half_kernel_w, half_kernel_w)
vals = Dict()
for method ∈ [red, green, blue, alpha]
x = sum((kernel .* method.(img[from_row:to_row, from_col:to_col])))
if x > 1
x = 1
elseif x < 0
x = 0
end
vals[method] = x
end
cpy_im[row, col] = RGBA(vals[red], vals[green], vals[blue], vals[alpha])
end
end
cpy_im
end
First of all, the error
Math on colors is deliberately undefined in ColorTypes, but see the ColorVectorSpace package.
should direct you to read the docs of the ColorVectorSpace package, where you will learn that using ColorVectorSpace will now enable math on RGB types. (The absence of default support it deliberate, because the way the image-processing community treats RGB is colorimetrically wrong. But everyone has agreed not to care, hence the ColorVectorSpace package.)
Second,
ArgumentError: element type FixedPointNumbers.N0f8 is an 8-bit type representing 256 values from 0.0 to 1.0, but the values (-0.0039215684f0, -0.007843137f0, -0.007843137f0, 1.0f0) do not lie within this range.
indicates that you're trying to write negative entries with an element type, N0f8, that can't support such values. Instead of cpy_im = copy(img), consider something like cpy_im = [float(c) for c in img] which will guarantee a floating-point representation that can support negative values.
Third, I would recommend avoiding steps like RGB.(img...) when nothing about your function otherwise addresses whether images are numeric, grayscale, or color. Fundamentally the only operations you need are scalar multiplication and addition, and it's better to write your algorithm generically leveraging only those two properties.
Tim Holy's answer above is correct - keep things simple and avoid relying on third-party packages when you don't need to.
I might point out that another option you may not have considered is to use a different algorithm. What you are implementing is the naive method, whereas many convolution routines using different algorithms for different sizes, such as im2col and Winograd (you can look these two up, I have a website that covers the idea behind both here).
The im2col routine might be worth doing as essentially you can break the routine in several pieces:
Unroll all 'regions' of the image to do a dot-product with the filter/kernel on, and stack them together into a single matrix.
Do a matrix-multiply with the unrolled input and filter/kernel.
Roll the output back into the correct shape.
It might be more complicated overall, but each part is simpler, so you may find this easier to do. A matrix multiply routine is definitely quite easy to implement. For 1x1 (single-pixel) convolutions where the image and filter have the same ordering (i.e. NCHW images and FCHW filter) the first and last steps are trivial as essentially no rolling/unrolling is necessary.
A final word of advice - start simpler and add in the code to handle edge-cases, convolutions are definitely fiddly to work with.
Hope this helps!
I'm trying to obtain a numerical solution to the following integral:
1
The correct answer is -0.324 + 0.382i but as seen below I am not getting a numerical answer and would appreciate help with the Maxima syntax.
2
Perhaps related to why I am not getting a numerical output are two specific questions:
I read that e and i in Maxima need to be preceded by % in input but should these also appear as %e and %i as seen in the Maxima output?
Why is dy missing at the end of the integral in the Maxima output?
Thank you!
Looks to me like your input is okay, however, the function to compute approximations to integrals is named quad_qags. (There are actually several related functions. See ?? quad_ for more info.) Also, a wrinkle here is that the integrand is a complex-valued function (of a real variable), and quad_qags can only work on real-valued integrands, so we'll have to work around it. Here's how I would arrange it.
myintegrand: exp(%i*(1 + %i*y))/(1 + %i*y + 1/(1 + %i*y));
result_realpart: quad_qags (realpart (myintegrand), y, 0, 6);
result_imagpart: quad_qags (imagpart (myintegrand), y, 0, 6);
result: result_realpart[1] + %i*result_imagpart[1];
I get 0.3243496676292901*%i + 0.3820529930785175 as the final result. That's a little different from what you said; maybe a minus sign went missing? or there's a missing or extra factor of %i?
A quick approximation
0.1 * lsum (x, x, float (rectform (makelist (ev (myintegrand, y = k/10), k, 0, 60))));
seems to show the result from quad_qags is reasonable.
Given input signal x (e.g. a voltage, sampled thousand times per second couple of minutes long), I'd like to calculate e.g.
/ this is not q
y[3] = -3*x[0] - x[1] + x[2] + 3*x[3]
y[4] = -3*x[1] - x[2] + x[3] + 3*x[4]
. . .
I'm aiming for variable window length and weight coefficients. How can I do it in q? I'm aware of mavg and signal processing in q and moving sum qidiom
In the DSP world it's called applying filter kernel by doing convolution. Weight coefficients define the kernel, which makes a high- or low-pass filter. The example above calculates the slope from last four points, placing the straight line via least squares method.
Something like this would work for parameterisable coefficients:
q)x:10+sums -1+1000?2f
q)f:{sum x*til[count x]xprev\:y}
q)f[3 1 -1 -3] x
0n 0n 0n -2.385585 1.423811 2.771659 2.065391 -0.951051 -1.323334 -0.8614857 ..
Specific cases can be made a bit faster (running 0 xprev is not the best thing)
q)g:{prev[deltas x]+3*x-3 xprev x}
q)g[x]~f[3 1 -1 -3]x
1b
q)\t:100000 f[3 1 1 -3] x
4612
q)\t:100000 g x
1791
There's a kx white paper of signal processing in q if this area interests you: https://code.kx.com/q/wp/signal-processing/
This may be a bit old but I thought I'd weigh in. There is a paper I wrote last year on signal processing that may be of some value. Working purely within KDB, dependent on the signal sizes you are using, you will see much better performance with a FFT based convolution between the kernel/window and the signal.
However, I've only written up a simple radix-2 FFT, although in my github repo I do have the untested work for a more flexible Bluestein algorithm which will allow for more variable signal length. https://github.com/callumjbiggs/q-signals/blob/master/signal.q
If you wish to go down the path of performing a full manual convolution by a moving sum, then the best method would be to break it up into blocks equal to the kernel/window size (which was based on some work Arthur W did many years ago)
q)vec:10000?100.0
q)weights:30?1.0
q)wsize:count weights
q)(weights$(((wsize-1)#0.0),vec)til[wsize]+) each til count v
32.5931 75.54583 100.4159 124.0514 105.3138 117.532 179.2236 200.5387 232.168.
If your input list not big then you could use the technique mentioned here:
https://code.kx.com/q/cookbook/programming-idioms/#how-do-i-apply-a-function-to-a-sequence-sliding-window
That uses 'scan' adverb. As that process creates multiple lists which might be inefficient for big lists.
Other solution using scan is:
q)f:{sum y*next\[z;x]} / x-input list, y-weights, z-window size-1
q)f[x;-3 -1 1 3;3]
This function also creates multiple lists so again might not be very efficient for big lists.
Other option is to use indices to fetch target items from the input list and perform the calculation. This will operate only on input list.
q) f:{[l;w;i]sum w*l i+til 4} / w- weight, l- input list, i-current index
q) f[x;-3 -1 1 3]#'til count x
This is a very basic function. You can add more variables to it as per your requirements.
In GeoGebra, is there a way to define a function of two variables which is the pointwise minimum of two functions.
Like h(x, y):= min(x² + y², x + y).
(The GeoGebra Min command does something different.)
I could work around by means of the abs function, whichis available, using min(a, b) = (a + b - |a - b|) / 2, but this is not very convenient (actually I need to take the minimum of more than two functions).
You could use a Conditional Function to create a piecewise function that is equal to f(x) if f(x) < g(x) and g(x) otherwise. The definition of this is:
If(f(x) < g(x), f, g)
Here's an example of this in action.
One option: Plot a point A on the x-axis. Compute F=min({f(x(A)),g(x(A)),h(x(A))}). Plot B=(x(A),F). Create Locus(B,A).
If you want the minimum of two functions f(x) and g(x) then you can define (f(x) + g(x) - abs(f(x) - g(x)))/2
Hello i am trying to solve a algebric equation in maxima, the equation has alpha, delta and psi as variable. I want the alpha in equation to be solved in terms of psi and delta. I tried using the solve command but i am getting alpha in terms of alpha.
Here is the equation to solve Equation to solve
And this is the output from maxima
.
This is the code i am trying -->
solve([(sqrt(-4*alpha*delta*psi-4*delta*psi+alpha^2*delta^2)/(delta^2+delta)-(alpha*delta/(delta^2+delta)))/2-sqrt(4*alpha^2*delta^2+6*alpha*delta^2+3*delta^2+2*alpha^2*delta+4*alpha*delta+2*delta)/(3*delta^2+2*delta)+alpha*delta/(3*delta^2+2*delta)=0],alpha);
Thank you
The problem with that algebraic equation is that involves square roots,or radicals and normal polynomial, and that that type of equation is not easy to solve, take a look at this equation:
(%i30) solve(x=sqrt(x+6),x);
(%o30) x = sqrt{x+6}
So Maxima doesn't return any value, but for example other software Mathematica does.
Let's square both sides of equation and try to solve it
(%i31) solve(x^2=x+6,x);
(%o31) x=3 , x=-2
we get two solutions, let's try with first equation:
3 = sqrt(3+6) => 3 = sqrt(9) => 3 = 3
-2 = sqrt(-2+6) => -2 = sqrt(4) => -2 = 2 ??????
So the second solution is not valid,
maxima solve program in Macsyma/Maxima generally avoids methods that
produce false solutions, like "square both sides". It may still
make errors based on dividing by expressions that appear to be
non-zero, but actually ARE zero, and maybe some other similar
situations.
from this mailing list
In your case I will factor the equation to get a simplified version but with those free variable this will be difficult so, try to assume some values for psi and delta:
(%i26) solve(factor((sqrt(-4*alpha*delta*psi-4*delta*psi+alpha^2*delta^2)/(delta^2+delta)-(alpha*delta/(delta^2+delta)))/2-sqrt(4*alpha^2*delta^2+6*alpha*delta^2+3*delta^2+2*alpha^2*delta+4*alpha*delta+2*delta)/(3*delta^2+2*delta)+alpha*delta/(3*delta^2+2*delta))=0,alpha);
(\%o26) \left[ \alpha=\ifrac{\left(3\,\delta+2\right)\,\isqrt{\left(-4\,\alpha-4\right)\,\delta\,\psi+\alpha^2\,\delta^2}+\left(-2\,\delta-2\right)\,\isqrt{\left(4\,\alpha^2+6\,\alpha+3\right)\,\delta^2+\left(2\,\alpha^2+4\,\alpha+2\right)\,\delta}}{\delta^2} \right]
Expand your Equation and try to remove square roots or some assumptions:
Equation : (sqrt(-4*alpha*delta*psi-4*delta*psi+alpha^2*delta^2)/(delta^2+delta)-(alpha*delta/(delta^2+delta)))/2-sqrt(4*alpha^2*delta^2+6*alpha*delta^2+3*delta^2+2*alpha^2*delta+4*alpha*delta+2*delta)/(3*delta^2+2*delta)+alpha*delta/(3*delta^2+2*delta);