Sorting some data

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Lena Heffern
Lena Heffern on 29 Jul 2011
I'm having trouble sorting out some data:
2 % incident#
4 % x (0-15)
59.783
18 % y (16-31)
59.543
3 % incident#
5 % x (0-15)
33.89
6 % x (0-15)
22.462
18 % y (16-31)
58.528
2 % incident#
7 % x (0-15)
59.407
18 % y (16-31)
59.465
2 % incident#
6 % x (0-15)
44.616
17 % y (16-31)
39.891
2 % incident#
19 % y (16-31)
59.886
6 % x (0-15)
59.796
3 % incident#
2 % x (0-15)
13.022
19 % y (16-31)
59.995
3 % x (0-15)
46.00
I'm trying to sort out x's and y's with their corresponding times (under each x or y value). However, some of the times need to add up to ~59, but some of them don't need to add up (x's w/ x's, y's w/ y's depending on incident#). I'm not sure how to sort the data because of the incident#'s being different and having a different number of values after. Plus, the x's and y's aren't always in order. Is there some sort of Boolean operation using integers I can use? In other words, identify when there's 2 integers in sequence (aka the 2, 4, 59.783...) so I can identify when to sort out values?
Sorry if this is extremely confusing, but any help is greatly appreciated.
edit: I'm really trying to avoid having to do a ton of switch/case and if/else statements.
edit: so far I have the following code, but its still not quite working...
i=1;sx=1;sy=1;
while i < length(data)+1
if (data(i+1,1) == 0:1:31) % is the second line of data a channel number?
n=fix(data(i,:)); % tells how many incident values there are
for j=1:1:n % iterates in groups depending on number of incidents
% is the channel an x or a y?
if (data(i+j,:) < 16 && data(i+j,:) > -1)
x(sx,1) = [data(i+j,:),data(1+i+j,:)];
i=i+1; sx=sx+1;
else (data(i+j,:) > 15 && data(i+j,:) > 32)
y(sy,1) = [data(i+j,:),data(1+i+j,:)];
i=i+1; sy=sy+1;
end
end
end
end

Answers (3)

Walter Roberson
Walter Roberson on 29 Jul 2011
strfind([false; v(:) == fix(v(:))].', [0 1 1]) - 1
This will return the indices of the beginning of each run of 2 or more integers.
  5 Comments
Fangjun Jiang
Fangjun Jiang on 30 Jul 2011
I don't get it. If I want to find 2 consecutive 1s, it's not going to give me the correct result.
>> strfind(true(1,6),[1 1])
ans =
1 2 3 4 5
>> strfind([false,true(1,6)],[0 1 1])
ans =
1
Walter Roberson
Walter Roberson on 30 Jul 2011
The code is to find the *beginning* of each run of 2 or more integers.

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Fangjun Jiang
Fangjun Jiang on 29 Jul 2011
Your data doesn't seem to be consistent. Did you have some typo? I would assume your data should be a repetition of the following 5 lines:
2 % incident#
4 % x (0-15)
59.783
18 % y (16-31)
59.543
If that's the case, you could use reshape() to re-share your data and then it's easy to sort. I modified your data a little bit and got the following result. Is it supposed to be that way?
>> B=reshape(A,5,[])
B =
2.0000 3.0000 2.0000 2.0000 2.0000 3.0000
4.0000 5.0000 7.0000 6.0000 19.0000 2.0000
59.7830 33.8900 59.4070 44.6160 59.8860 13.0220
18.0000 18.0000 18.0000 17.0000 6.0000 19.0000
59.5430 58.5280 59.4650 39.8910 59.7960 59.9950
  3 Comments
Lena Heffern
Lena Heffern on 29 Jul 2011
The incident# tells how many groups of 2 numbers there are...
int
int
float
int
float
this would be a 2 #incident
int
int
float
int
float
int
float
would be a 3 #incident
yes, my data is just that annoying
Fangjun Jiang
Fangjun Jiang on 29 Jul 2011
Okay, then Walter's solution should help!

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Lena Heffern
Lena Heffern on 1 Aug 2011
Almost got it... I just need to figure out how to get rid of my error:
??? Attempted to access data(336607,:); index out of bounds because size(data)=[336606,1].
Code:
i=1;sx=1;sy=1;j=1;xs=zeros((length(data)),(1));ys=zeros((length(data)),(1));m=1;r=0;k=1;
while i+j < length(data)
if ceil(data(i+1,:)) == floor(data(i+1,:)) % is the second line of data a channel number?
n=fix(data(i,:)); % tells how many incident values there are
r=0;
for j=1:2*n % iterates in groups depending on number of incidents, n
% is the channel an x or a y?
if ceil(data(i+j,:)) == floor(data(i+j,:))
if (data(i+j,1) < 16 && data(i+j,1) > -1) % its an x
xo(sx,:) = [data(i+j,:),data(j+i+1,:)];
if ceil(xo(sx,2)) ~= floor(xo(sx,2))
x(sx,:) = [data(i+j,:),data(j+i+1,:)];
xs(m,:)= xs(m,:)+ data(j+i+1,:);
sx=sx+1; j=j+1;
i=i+1;
else
j=j+1;
end
else
y(sy,:) = [data(i+j,:),data(j+i+1,:)];
ys(m,:)= ys(m,:)+ data(j+i+1,:);
j=j+1;
sy=sy+1;
end
else
j=j+1;
end
end
i=i+1; m=m+1;
else
i=i+1;
end
i=i+1;
end

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