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how to enter equation into matrix column and refer to adjacent column.

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Hi all. I am a Matlab beginner trying to find full width half max. Below is my code. I start with my .csv file, identify x- and y-values, fit a gaussian curve, and find half max. Next, I added x-values at evenly spaced small increments to a column in a matrix (defined as A). I want the second column (B) to find the y-values at the given x-values using the equation found with the gaussian fit (in other words, the x-value at a given row will be used in the equation to find the y-value in the adjacent column). I am stuck on the for loop here. Then, I think I can use the "find" command to find my 2 x-values to find the full width, but this may need some help too. Thank you for your help!
data = load ('Practice.csv');
datax = data(:,2);
datay = data(:,5);
f = fit(datax, datay,'gauss1')
yfitted = feval(f,datax);
hold on
[ypk] = findpeaks(yfitted)
HalfMax = 0.5*ypk
Matrix = zeros(2,1000);
A = Matrix(1,:);
for A =1:1000;
x1= 0;
x2= 1.1;
n = 1000;
A = linspace(x1, x2, n);
B = Matrix(2,:); %%this is where I need help with a for loop please!
index1 = find(B <= HalfMax, 1, 'first');
index2 = find(B >= HalfMax, 1, 'last')
fwhm = index2-index1
  3 Kommentare
Emily Pendleton
Emily Pendleton am 15 Dez. 2017
With the first column in the matrix (1000 rows), I am just trying to give values to x at small, evenly spaced increment. Here, my first 5 rows of A (x-values) are: 0, 0.0011, 0.0022, 0.0033, 0.0044, 0.0055. This continues to my last 5 rows: 1.0956, 1.0967, 1.0978, 1.0989, 1.1000. I'm sorry if this code is redundant and the loop is not needed. I am attempting to have evenly spaced increases in x-value to find FWHM; is there a better way to increment x-values? Even if my code is repetitive, does my question make sense?
Walter Roberson
Walter Roberson am 15 Dez. 2017
Perhaps you just want
A = linspace(0, 1.1, 1000);
Matrix(1,:) = A;

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Akzeptierte Antwort

Matthew am 15 Dez. 2017
If I understand what you're attempting to do, you can replace the 'A' for loop with one line
A = linspace(0, 1.1, 1000);
B can then be calculated as
B= feval(f,A);
  2 Kommentare
Emily Pendleton
Emily Pendleton am 15 Dez. 2017
Yes. That worked for finding the correspoinding y-values. This line is now wrong and returns a value of 1.
index1 = find(B <= HalfMax, 1, 'first')
I'm just trying to locate the row at which B is <= Halfmax. I can then use that location to find the value of A in the same row. I will do this for when B <= Halfmax too and find the difference between the x-values. Any suggestions on how to get this done?
Matthew am 15 Dez. 2017
Bearbeitet: Matthew am 15 Dez. 2017
You probably want to reverse the sign of your inequality.
index1 = find(B >= HalfMax, 1, 'first')
The current code will most likely always return 1 because the first point of B is the first point less than the HalfMax.

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