I'm having some serious issues fitting an exponential function (Beer-Lambert law) to my data. The optimization toolset function that I'm using produces terrible fits:
function [ Coefficients ] = fitting_new( Modified_Spectrum_Data,trajectory )
x_axis = trajectory;
fun = #(x,x_axis) (x(1)*exp((-x(2))*x_axis));
start = [Modified_Spectrum_Data(1) 0.05];
nlm = nlinfit(x_axis,Modified_Spectrum_Data,fun,start,opts);
Coefficients = nlm;
end
Data:
Modified_Spectrum_Data = [1.11111111111111, 1.08784976353957, 1.06352170731165, 1.04099672033640, 1.02649723285838, 1.00423806910703, 0.994116452961827, 0.975928861361604, 0.963081773802984, 0.953191520906905, 0.940636278551651, 0.930360007604054, 0.922259178548511, 0.916659345499171, 0.909149956799775, 0.901241601559703, 0.895375741449218, 0.893308346234150, 0.887985459843162, 0.884657500398024, 0.883852990694089, 0.877158499678129, 0.874817832833850, 0.875428444059047, 0.873170360623947, 0.871461252768665, 0.867913776631497, 0.866459074988087, 0.863819528471106, 0.863228815347816 ,0.864369045426273 ,0.860602502500599, 0.862653463581049, 0.861169231463016, 0.858658616425390, 0.864588421841755, 0.858668693409622, 0.857993365648639]
trajectory = [0.0043, 0.9996, 2.0007, 2.9994, 3.9996, 4.9994, 5.9981, 6.9978, 7.9997, 8.9992, 10.0007, 10.9993, 11.9994, 12.9992, 14.0001, 14.9968, 15.9972, 16.9996, 17.9996, 18.999, 19.9992, 20.9996, 21.9994, 23.0003, 23.9992, 24.999, 25.9987, 26.9986, 27.999, 28.9991, 29.999, 30.9987, 31.9976, 32.9979, 33.9983, 34.9988, 35.999, 36.9991]
I've tried using multiple different fitting functions and messing around with the options, but they don't seem to make too big of a difference. Additionally, I've tried changing the initial guess, but again that doesn't really make a difference.
Excel seems to be able to fit the data perfectly fine, but I have 900 rows of data I want to fit so doing it in Excel is not possible.
Any help would be greatly appreciated, thank you.
You'll want to use the cftool. Your data looks to follow a power law. Then choose 'Modified Spectrum Data' as your x axis and 'Trajectory' as your y. Select 'Power' from the drop down menu towards the top of the GUI.
Modified_Spectrum_Data = [1.11111111111111, 1.08784976353957, 1.06352170731165, 1.04099672033640, 1.02649723285838, 1.00423806910703, 0.994116452961827, 0.975928861361604, 0.963081773802984, 0.953191520906905, 0.940636278551651, 0.930360007604054, 0.922259178548511, 0.916659345499171, 0.909149956799775, 0.901241601559703, 0.895375741449218, 0.893308346234150, 0.887985459843162, 0.884657500398024, 0.883852990694089, 0.877158499678129, 0.874817832833850, 0.875428444059047, 0.873170360623947, 0.871461252768665, 0.867913776631497, 0.866459074988087, 0.863819528471106, 0.863228815347816 ,0.864369045426273 ,0.860602502500599, 0.862653463581049, 0.861169231463016, 0.858658616425390, 0.864588421841755, 0.858668693409622, 0.857993365648639]
trajectory = [0.0043, 0.9996, 2.0007, 2.9994, 3.9996, 4.9994, 5.9981, 6.9978, 7.9997, 8.9992, 10.0007, 10.9993, 11.9994, 12.9992, 14.0001, 14.9968, 15.9972, 16.9996, 17.9996, 18.999, 19.9992, 20.9996, 21.9994, 23.0003, 23.9992, 24.999, 25.9987, 26.9986, 27.999, 28.9991, 29.999, 30.9987, 31.9976, 32.9979, 33.9983, 34.9988, 35.999, 36.9991]
cftool
Screenshot:
For more information on the curve fitting (cftool), see: https://www.mathworks.com/help/curvefit/curvefitting-app.html
Here's the thing: I want to modify (and then return) a matrix of integers that is given in the parameters of the function. The funcion average (of the class MatrixMotionBlur) gives the average between the own pixel, upper, down and left pixels. Follows the following formula:
result(x, y) = (M1(x, y)+M1(x-1, y)+M1(x, y-1)+M1(x, y+1)) / 4
This is the code i've implemented so far
MatrixMotionBlur - Average function
MotionBlurSingleThread - run
The objetive here is to apply "average" method to alter the matrix value and return that matrix. The thing is the program gives me error when I to insert the value on the matrix.
Any ideas how to do this ?
The functional way
val updatedData = data.map{ outter =>
outter(i).map{ inner =>
mx.average(i.j)
}
}
Pay attention that Seq is immutable collection type and you can't just modify it, you can create new, modified collection only.
By the way, why you iterate starting 1, but not 0. Are you sure you want it?
I have the following code that generates a matrix of 15 blocks that will then be used in a Montecarlo approach as multiple starting points. How can I get the same exact result in a smarter way?
assume that J=15*100 are the total simulation and paramNum the number of parameters
[10^-10*ones(paramNum,round(J/15)) 10^-9*ones(paramNum,round(J/15)) 10^-8*ones(paramNum,round(J/15)) 10^-7*ones(paramNum,round(J/15)) 10^-6*ones(paramNum,round(J/15)) 10^-5*ones(paramNum,round(J/15)) rand*10^-5*ones(paramNum,round(J/15)) 10^-4*ones(paramNum,round(J/15)) rand*10^-4*ones(paramNum,round(J/15)) 10^-3*ones(paramNum,round(J/15)) 10^-2*ones(paramNum,round(J/15)) 10^-1*ones(paramNum,round(J/15)) 10^-abs(randn/2)*ones(paramNum,round(J/15))];
you could do
v = 10.^[-10:-5 rand*10^-5 -4:-1 10^-abs(randn/2)];
repmat(repelem(v, 1, round(J/15)), paramNum) .* ...
repmat(ones(paramNum,round(J/15)), numel(v))
Or mimic the repmat/repelem functionality with a for loop. The first is shorter, the later is more understandable.
By the way... it's less than 15 blocks...
I'm looking for a function or solution to the following:
For the chart in SQL Reporting i need to multiply values from a Column A. For summation i would use =SUM(COLUMN_A) for the chart. But what can i use for multiplication - i was not able to find a solution so far?
Currently i am calculating the value of the stacked column as following:
=ROUND(SUM(Fields!Value_Is.Value)/SUM(Fields!StartValue.Value),3)
Instead of SUM i need something to multiply the values.
Something like that:
=ROUND(MULTIPLY(Fields!Value_Is.Value)/MULTIPLY(Fields!StartValue.Value),3)
EDIT #1
Okay tried to get this thing running.
The expression for the chart looks like this:
=Exp(Sum(Log(IIf(Fields!Menge_Ist.Value = 0, 10^-306, Fields!Menge_Ist.Value)))) / Exp(Sum(Log(IIf(Fields!Startmenge.Value = 0, 10^-306, Fields!Startmenge.Value))))
If i calculate my 'needs' manually i have to get the following result:
In my SQL Report i get the following result:
To make it easier, these are the raw values:
and you have the possibility to group the chart by CW, CQ or CY
(The values from the first pictures are aggregated Sum values from the raw values by FertStufe)
EDIT #2
Tried your expression, which results in this:
Just to make it clear:
The values in the column
=Value_IS / Start_Value
in the first picture are multiplied against each other
0,9947 x 1,0000 x 0,59401 = 0,58573
Diffusion Calenderweek 44 Sums
Startvalue: 1900,00 Value Is: 1890,00 == yield:0,99474
Waffer unbestrahlt Calenderweek 44 Sums
Startvalue: 620,00 Value Is: 620,00 == yield 1,0000
Pellet Calenderweek 44 Sums
Startvalue: 271,00 Value Is: 160,00 == yield 0,59041
yield Diffusion x yield Wafer x yield Pellet = needed Value in chart = 0,58730
EDIT #3
The raw values look like this:
The chart ist grouped - like in the image - on these fields
CY (Calendar year), CM (Calendar month), CW (Calendar week)
You can download the data as xls here:
https://www.dropbox.com/s/g0yrzo3330adgem/2013-01-17_data.xls
The expression i use (copy / past from the edit window)
=Exp(Sum(Log(Fields!Menge_Ist.Value / Fields!Startmenge.Value)))
I've exported the whole report result to excel, you can get it here:
https://www.dropbox.com/s/uogdh9ac2onuqh6/2013-01-17_report.xls
it's actually a workaround. But I am pretty sure is the only solution for this infamous problem :D
This is how I did:
Exp(∑(Log(X))), so what you should do is:
Exp(Sum(Log(Fields!YourField.Value)))
Who said math was worth nothing? =D
EDIT:
Corrected the formula.
By the way, it's tested.
Addressing Ian's concern:
Exp(Sum(Log(IIf(Fields!YourField.Value = 0, 10^-306, Fields!YourField.Value))))
The idea is change 0 with a very small number. Just an idea.
EDIT:
Based on your updated question this is what you should do:
Exp(Sum(Log(Fields!Value_IS.Value / Fields!Start_Value.Value)))
I just tested the above code and got the result you hoped for.