Writing data to table with for loops

I currently have the data displayed as a printed equation in the Command window, but I would rather have it set up in a table with individual variable names. I simplified the code to just the parts that would affect the table, but I'm not sure how to get it to actually print the values how I want. The variable names for the table would be Output (from spurs(m,n)), m_value, LO_value, n_value, and RF_value
for LO = 11:2:21
lo = [0 1 2 3 4 5].*LO
for RF = 2:0.1:18
rf = [1; 2; 3; 4; 5]*RF
rf = rf(:)
spurs = [(lo(1)-rf) (lo(2)-rf) (lo(3)-rf) (lo(4)-rf) (lo(5)-rf) lo(6)-rf]
for m = [1 2 3 4 5]
for n = [1 2 3 4 5 6]
if spurs(m,n) > 3 && spurs(m,n) < 5.1
Print = [num2str(spurs(m,n)), ' = ', num2str(m), '*', num2str(LO), '-', num2str(n), '*', num2str(RF)];
disp(Print)
end
end
end
end
end

3 comentarios

Jan
Jan el 11 de Jun. de 2019
How is "table" defined here? Do you mean a table object? Then what should be the contents? Currently you display char vectors only, but should the table contain numerical vectors or symbolic equations? Or do you mean an uitable object? Or a tabular display with automatic tab stops in the command window?
Please explain exactly, what you want to achieve.
Danielle Rivera
Danielle Rivera el 11 de Jun. de 2019
It would be a table object where for each spur(m,n) output in the set range, the spur(m,n) number, the m value, the n value, the RF value, and the LO value would be shown in the same row (columns for each of those variables) as numerical values not char/string values. Ideally I would have a table for each LO value, where the LO is still included in the table but a new table is made every time the LO value changes (6 tables instead of just 1).
This line:
rf = rf(:)
does nothing since you've made sure to create a column vector in the previous line.
This line:
spurs = [(lo(1)-rf) (lo(2)-rf) (lo(3)-rf) (lo(4)-rf) (lo(5)-rf) lo(6)-rf]
is simply:
spurs = lo - rf; %if on R2016b or later
%spurs = bsxfun(@minus, lo, rf); %on earlier versions
Note that none of the loops are needed, you could just create a 4D matrix (m x n x RF x LO) in one go:
LO = 11:2:21;
RF = 2:0.1:18;
n = 0:5;
m = 1:5;
[mm, nn, rf, lo] = ndgrid(m, n, RF, LO);
spurs = nn .* lo - mm .* rf;

Iniciar sesión para comentar.

 Respuesta aceptada

Guillaume
Guillaume el 11 de Jun. de 2019
Editada: Guillaume el 11 de Jun. de 2019
As commented you don't need loops to construct all your spurs:
LO = 11:2:21;
RF = 2:0.1:18;
n = 0:5;
m = 1:5;
[mm, nn, rf, lo] = ndgrid(m, n, RF, LO);
spurs = nn .* lo - mm .* rf;
It's then trivial to create a table from that:
tokeep = spurs > 3 & spurs < 5.1
result = table(mm(tokeep), nn(tokeep), rf(tokeep), lo(tokeep), spurs(tokeep), 'VariableNames', {'m', 'n', 'RF', 'LO', 'spur'})
There is no point in splitting that table into several tables according to LO. It will only complicate further processing.

3 comentarios

Danielle Rivera
Danielle Rivera el 11 de Jun. de 2019
This is helpful, however I do need the for loop for the LO because I am plotting a new graph for every LO value after the spur table creation. Also, I think it would still be useful to break up the table by LO because it resulted in 1255 rows
You probably don't need a loop for your plotting. It probably can be done all at once for each LO. For example to plot RF vs m for each LO:
figure; hold on;
rowfun(@(m, rf, lo) plot(m, rf, 'DisplayName', sprintf('LO = %d', lo(1))), result, 'GroupingVariables', 'LO', 'InputVariables', {'m', 'RF', 'LO'});
legend('show'); xlabel('m'); ylabel('RF');
it's probably not what you're plotting, so what are you plotting exactly?
I only get 631 rows but whatever the height why does it matter? Again, splitting it makes processing harder not easier.
Danielle Rivera
Danielle Rivera el 11 de Jun. de 2019
I'm making 3d stem plots with RF(x-axis), spurs(y-axis), and power(z-axis, values from a matrix). The table is intended to help with filter and mixer design and indicate isolation necessities and provide notice of undesired values that show up in the output range (3-5)

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Operators and Elementary Operations en Centro de ayuda y File Exchange.

Preguntada:

el 10 de Jun. de 2019

Comentada:

el 11 de Jun. de 2019

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by