Function using for loop

3 views (last 30 days)
Sayanta
Sayanta on 31 May 2012
Hi All,
I want to write a function using a for loop but the function will do the same functionality. The function is shown below
How can I do that any tips ?
Here is function
f unction [state1,state2] = stat(x)
% for previous state
if (x(1) <= 5)
state1 = 1;
end
if (x(1) > 5 && x(1) <= 10)
state1 = 2;
end
if (x(1) > 10 && x(1) <= 15)
state1 = 3;
end
if (x(1) > 15 && x(1) <= 20)
state1 = 4;
end
if (x(1) > 20 && x(1) <= 25)
state1 = 5;
end
if (x(1) > 25 && x(1) <= 30)
state1 = 6;
end
% for current state
if (x(2) <= 5)
state2 = 1;
end
if (x(2) > 5 && x(2) <= 10)
state2 = 2;
end
if (x(2) >10 && x(2) <= 15)
state2 = 3;
end
if (x(2) > 15 && x(2) <= 20)
state2 = 4;
end
if (x(2) > 20 && x(2) <= 25)
state2 = 5;
end
if (x(2) > 25 && x(2) <= 30)
state2 = 6;
end
Many Thanks in advance
Sayanta
  2 Comments
Oleg Komarov
Oleg Komarov on 31 May 2012
you can actually avoid a for loop and use histc()
Oleg Komarov
Oleg Komarov on 31 May 2012
From my perspective this user has asked the same question for the 6th time, got an answer and did not interact with the contributors to get to the bottom of the problem. At this point I suggest to invest time with other users of the forum unless Sayanta starts giving some feedback.

Sign in to comment.

Answers (2)

Andrei Bobrov
Andrei Bobrov on 31 May 2012
x = unifrnd(3,36,20,1);
[state2,state2 ] = histc(x,[-inf,5:5:30] + eps(100));
or
k = [-30:5:-5 inf]
[j1,j1] = histc(-x,k )
n = numel(k)
state2 = rem(abs(j1-n),n)
  1 Comment
Walter Roberson
Walter Roberson on 31 May 2012
As a usage note: andrei has used these mechanisms of adding eps() or negating the values involved, in order to compensate for the fact that histc() defines its bins as a <= x < b whereas your request was for a < x <= b . If your boundaries can be switched so that (for example) _exactly_ 25 is state 6 instead of state 5, then histc() can be used directly with no fiddling.

Sign in to comment.


Geoff
Geoff on 31 May 2012
There's a pattern here...
state1 = max( ceil(x(1) / 5), 1 );
state2 = max( ceil(x(2) / 5), 1 );
Just as a point on style... if you want to write a bunch of if statements for a continuous range, do this:
if x(2) <= 5
state2 = 1;
elseif x(2) <= 10
state2 = 2;
elseif x(2) <= 15
state2 = 3;
elseif % etc etc
% etc etc
else
state2 = NaN; % What happens if out of range?
end
  1 Comment
Geoff
Geoff on 31 May 2012
Note: you don't need the 'max' if x(?) is always greater than zero.

Sign in to comment.

Categories

Find more on Loops and Conditional Statements in Help Center and File Exchange

Community Treasure Hunt

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

Start Hunting!