help me with this for loop please

1 Ansicht (letzte 30 Tage)
soloby
soloby am 23 Jun. 2015
Kommentiert: soloby am 25 Jun. 2015
x = -10:0.1:10;
y = 0:0.005:1;
minNmax1 = 0:.005:1;
upper_1 = 0:.005:1;
y2 = 0:.005:1;
f1 = trapmf(x,[-2 0 0 2].');
minNmax1 = zeros(length(x), 2);
for k = 1 :length(y)
minNmax1(k,:) = [min(x),max(x)];
end
I'm trying to get [min(x),max(x)] for 201 values of y but this will only give me 1 answer [-10 10] which i assume is the first iteration?
help!
  6 Kommentare
soloby
soloby am 23 Jun. 2015
Bearbeitet: soloby am 23 Jun. 2015
my function f1 ranges from 0 to 1. if you plot(x,f1) it can be seen.
I need to split my f1 values (0 to 1) into 201 pieces and find the upper and lower x values for each iterations of y. Think of fuzzy logic, it is similar to how it is solved.
Easy concept, but I'm afraid a lot harder to execute.
so my f1 is 0 to 1
I want to divide it up into equal numbers (201 in my case)
at each iteration, I want to find the x values Please someone help!
soloby
soloby am 23 Jun. 2015
Check my picture please, I need to find those starred points for all the red lines (201 of them)

Melden Sie sich an, um zu kommentieren.

Antworten (1)

Muhannad
Muhannad am 24 Jun. 2015
will min(x) always be the same number as x is not changing? you can use repmat(min(x), 201,1)) if that is the case
  1 Kommentar
soloby
soloby am 25 Jun. 2015
min(x) will be the same for each iterations of y but I need it for all 201 y's.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by