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Need help with a While loop

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Nicolo Zaza
Nicolo Zaza 2012-7-19
So I'm trying to write a rainflow counting program and I'm having trouble with the repetitions
while i <= numel(PV) ;
i = i +1;
N = N+ 1;
PV(N);
switch N
case 3
X = abs(PV(N) - PV(N-1))
Y = abs(PV(N-1) - PV(N-2))
case 4
X = abs(PV(N) - PV(N-1))
Y = abs(PV(N-1) - PV(N-2))
end
end
Instead of doing cases all the way to 24 there must be an easier way but I suck at matlab and would appreciate any help. Thanks
  1 个评论
Nicolo Zaza
Nicolo Zaza 2012-7-19
What Im actually trying to do is to re arrange a previous code to make work with some other scripts I developed but it is in fortran and I'm having a hard time translating it.
DIMENSION E(50)
N=0
1 N=N+1
CALL DATA(E(N),K)
IF(K == 1) STOP
2 IF(N < 3) GO TO 1
X=ABS((E(N)-E(N-1))
Y=ABS(E(N-1)-E(N-2))
3 IF(X.LT.Y) GO TO 1
4 RANGE=Y
XMEAN=(E(N-1)+E(N-2))/2
N=N-2
E(N)=E(N+2)
GO TO 2
END

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采纳的回答

Kevin Claytor
Kevin Claytor 2012-7-19
编辑：Kevin Claytor 2012-7-19
You may want to have a look at this file to see how they implement the rainflow method; http://www.mathworks.com/matlabcentral/fileexchange/3026
Now I'm just going to take a stab at my interpretation and re-writing of your FORTRAN code (added matlab-style comments to the end);
DIMENSION E(50) % Not sure what this does; same as E=[1:50]?
N=0 % Initalize counters
1 N=N+1
CALL DATA(E(N),K) % K = DATA(E(N))
IF(K == 1) STOP % Throw an error if the first point is 1?
2 IF(N < 3) GO TO 1 % Can't do algorithm for N < 3, increment N
X=ABS((E(N)-E(N-1)) % Compute X and Y
Y=ABS(E(N-1)-E(N-2))
3 IF(X.LT.Y) GO TO 1 % Continue if X > Y - this is the 'flow part'
4 RANGE=Y % Set some other parameters
XMEAN=(E(N-1)+E(N-2))/2
N=N-2 % Go back to see if this is still max
E(N)=E(N+2) % Update counter with prev max at location N
GO TO 2
END
So to me, and I'm likely wrong, as I've never done rainflow counting before and just read the wiki page on it now in about 15 secs...
DATA = [...blargh...];
LDATA = length(DATA);
E = [1:LDATA]; % A linear index of where our flows start from?
N = 3; % no sense starting before the algorithm can kick in
while N < LDATA
X = abs(DATA(N) - DATA(N-1));
Y = abs(DATA(N-1) - DATA(N-2));
if x > Y
RANGE = Y;
XMEAN = (DATA(N-1) + DATA(N-2))/2;
E(N-2) = E(N);
N = N-2;
else
N = N+1;
end
end
Maybe this helps?
  1 个评论
Nicolo Zaza
Nicolo Zaza 2012-7-19
This helped far more than what I was expecting.. Had to do some changes but it gave me a great idea.. Thanks a lot man.

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更多回答（4 个）

Walter Roberson
Walter Roberson 2012-7-19
Why have a switch at all? Why not just
if N >= 3
X = abs(PV(N) - PV(N-1))
Y = abs(PV(N-1) - PV(N-2))
end
  2 个评论
Andrei Bobrov
Andrei Bobrov 2012-7-19
if N >= 3 && N <= 4
X = abs(PV(N) - PV(N-1))
Y = abs(PV(N-1) - PV(N-2))
end
Albert Yam
Albert Yam 2012-7-19
Following from Nicolo's response about keeping X,Y values.
for N = 3:length(PV)
if N >= 3 && N <= 4
X(N-2) = abs(PV(N) - PV(N-1))
Y(N-2) = abs(PV(N-1) - PV(N-2))
end
end
but at this point.. Walter is right .. why loop?
X = PV(3:end);
Y = PV(3:end);
X(1) = abs(PV(3) - PV(2));
X(2) = abs(PV(2) - PV(1));
Y(1) = abs(PV(3) - PV(2));
Y(2) = abs(PV(2) - PV(1));

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Matt Kindig
Matt Kindig 2012-7-19
编辑：Matt Kindig 2012-7-19
I reformatted the code for easier reading. A few questions:
  1. 1. what are X and Y for? They don't appear to be used anywhere in the loop, and are overwritten each time.
  2. 2. For N=3 or N=4, X and Y are defined in the same way.
  3. 3. Are i and N initialized prior to the loop?
  4. 4. What exactly is this code supposed to do? If we understand your intended output more clearly, we can probably recommend a more direct approach.
Your code:
while i <= numel(PV)
i = i +1;
N = N+ 1;
PV(N);
switch N
case 3
X = abs(PV(N) - PV(N-1))
Y = abs(PV(N-1) - PV(N-2))
case 4
X = abs(PV(N) - PV(N-1))
Y = abs(PV(N-1) - PV(N-2))
end
end
  1 个评论
Nicolo Zaza
Nicolo Zaza 2012-7-19
编辑：Jan 2012-7-19
This is just a part of the code Im trying to write, I actually have a fortran-like version of it if it would be of any help. The thing is that case 3 and 4 are not the only ones. What Im trying to do is for the loop to continue until theres no more values
PV = [ 80 -40 60 0 60 0 80 -100 60 20 60 20 80 -100 40 -80 0 -20 ...
0 -20 40 -80 80 -40];
SA = length(PV);
N = 0;
i = 1;
while i <= numel(PV) ;
i = i +1;
N = N+ 1;
PV(N);
switch N
case 3
X = abs(PV(N) - PV(N-1))
Y = abs(PV(N-1) - PV(N-2))
case 4
X = abs(PV(N) - PV(N-1))
Y = abs(PV(N-1) - PV(N-2))
end
end
This is what i have until now, The program is suppose to array the values from max to mins in a way defined by Rainflow Counting Algorithms(not sure if you've heard of it.) I could provide the fortran code if it would be of any help.
[EDITED, code formatted, Jan] Please learn how to format the code by your own: empty line above and below the code, mark code, hit "{} code" button or insert two spaces in front of each line manually. It is not so hard.

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Andrei Bobrov
Andrei Bobrov 2012-7-19
编辑：Andrei Bobrov 2012-7-19
switch N
case {3,4}
X = abs(PV(N) - PV(N-1));
Y = abs(PV(N-1) - PV(N-2));
end
OR
k = abs(diff(PV(1:4)));
xy = flipud(k(hankel(1:2,2:3)))
X = xy(1,N-2);
Y = xy(2,N-2);
Nicolo Zaza
Nicolo Zaza 2012-7-19
Well yeah I already checked the Rainflow Method your sending me a link to but it doesn't seem to work properly. The problem I have is that I can't get it counting. I tried your code but it kept going and only displayed 1 answer for each element. It totally looks much better than mine. I need to get a bunch of X and Y's for 24 elements i should get around 12 of those(and for every other variable, ex: xmean) that's where I get lost. How can I keep it going and outputting results until there is an index with no info. And need all the results, like a vector. Also why cant an algorithm be done for {If N<3 , N = N + 1,end} something like that? Thanks
  1 个评论
Albert Yam
Albert Yam 2012-7-19
The reason you only get 1 set of X, Y, is because you tell Matlab to over write it. This is the first instance of you mentioning that you need to keep previous X, Y values.
for N=1:length(PV)
if N<3
%stuff
end
end

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