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# how to change data

Asked by Tian Lin on 6 Jan 2013

i have a matrix like this: [1 0 0 1 1 1 1 0 1 1 0]

i need to change to: [1 0 0 4 4 4 4 0 2 2 0]

how to make a loop?thanks

## 1 Comment

Image Analyst on 6 Jan 2013

Tian, as you can see there are a number of different methods to do that. But I've never done that. It seems like a strange thing to want. Why do you want this output array? What are you going to do with it after you get it? (It's possible you don't really need it, you just think you do.)

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Answer by Azzi Abdelmalek on 6 Jan 2013
Edited by Azzi Abdelmalek on 6 Jan 2013

Improve my previous code speed three times

```x=[1 0 0 1 1 1 1 0 1 1 0]
a=find(x);
b=[ 1 diff(a)];
b(b==1)=0;
b(b~=0)=1;
idx=zeros(numel(a),2);
e=1;
c=0;
d=[];
for k=1:numel(a)
e=e+b(k);
c=c*not(b(k))+1;
d(c)=a(k);
idx(e,:)=[d(1) c];
end
for k=1:e
x(idx(k,1):idx(k,1)+idx(k,2)-1)=idx(k,2);
end
```

Tian Lin on 14 Jan 2013

Thanks for help! If I have a matrix like this [1 0 0 1 1 0; 0 1 0 1 1 0; ... 0 1 1 0 0 1], that means there are many rows，how to change this:b=[ 1 diff(a)]?

Azzi Abdelmalek on 14 Jan 2013
```y=[1 0 0 1 1 1 1 0 1 1 0;0 0 0 1 1 0 1 0   1 1 0]
for k1=1:size(y,1)
x=y(k1,:);
a=find(x);
b=[ 1 diff(a)];
b(b==1)=0;
b(b~=0)=1;
idx=zeros(numel(a),2);
e=1;
c=0;
d=[];
for k=1:numel(a)
e=e+b(k);
c=c*not(b(k))+1;
d(c)=a(k);
idx(e,:)=[d(1) c];
end
for k=1:e
x(idx(k,1):idx(k,1)+idx(k,2)-1)=idx(k,2);
end
out(k1,:)=x;
end
```
Tian Lin on 15 Jan 2013

Thank you very much

Answer by Jan Simon on 6 Jan 2013

And finally an improved loop method which is about twice as fast as the vectorized method:

```function a = RunLength_IgnoZero_loop2(a)
len = length(a);
c   = a(1);
ini = 1;
b   = zeros(size(a));
for ii = 2:len
if a(ii) ~= c
if c == 0
ini = ii;
else
b(ini) = ii  - ini;
b(ii)  = ini - ii;
end
c = a(ii);
end
end
```
```if c ~= 0
b(ini) = len - ini + 1;
end
a = cumsum(b);
```

Image Analyst on 6 Jan 2013

My method does not allow more than 65563 separate regions, like the ~160,000 your code produces, so I ran it for 1e5 elements instead. Then I figure we'd just multiply by 10 to compare times. The regionprops() & intlut() takes 0.12 seconds for 1e5 elements, and presumably 1.2 seconds if it were able to process more than 65536 regions.

Jan Simon on 6 Jan 2013
```x = double(rand(1, 1e6) > 0.8);
tic; for ii = 1:100, y = FCN(x); end; toc
```
```% RunLength_IgnoZero_loop2:
0.40317 seconds
% Image Analyst's regionprops:
45.40 seconds
```

(Matlab R2009a/64/Win7/Core2Duo) I assume 0.12 sec means 1 iteration and you are using a modern machine which is 4 times faster than my older processor.

Image Analyst on 6 Jan 2013

OK, I didn't notice at first that you were doing the same thing 100 times.

Answer by Roger Stafford on 6 Jan 2013

Let x be the original row vector of 1's and 0's.

``` n = length(x);
d = diff([0,x,0]);
f1 = find(d(1:n)>0);
f2 = find(d(2:n+1)<0)+1;
y = zeros(1,n+1);
y([f1,f2]) = [f2-f1,f1-f2];
y = cumsum(y(1:n));```

Answer by Azzi Abdelmalek on 6 Jan 2013
Edited by Azzi Abdelmalek on 6 Jan 2013
```clear
x=[1 0 0 1 1 1 1 0 1 1 0]
e=0,c=0,d=[]
for k=1:numel(x)
if x(k)
e=e+not(c)
c=c+1
d=[d k]
idx(e,:)=[d(1) c]
else
c=0
d=[]
end
end
for k=1:size(idx,1)
x(idx(k,1):idx(k,1)+idx(k,2)-1)=idx(k,2)
end
```

## 1 Comment

Jan Simon on 6 Jan 2013

Pre-allocating idx to the maximum possible size increases the speed.

Answer by Image Analyst on 6 Jan 2013

Very simple. No loop needed. You just reassign it:

```m = [1 0 0 1 1 1 1 0 1 1 0]
% Now make it into what you want:
m = [1 0 0 4 4 4 4 0 2 2 0]
```

If you have some other algorithm then let's hear it. For example, leave the first element alone but take the next contiguous stretch of 1's and multiply them by 4, and take the next stretch and multiply them by 2. I couldn't figure out what algorithm you were applying, and you didn't say, and didn't say how general you needed this to be (for example can m have values other than 0 and 1, or can it be other lengths, or can it be 2D or 3D?).

Azzi Abdelmalek on 6 Jan 2013

I guess 4 is the number of consecutive 1 in the array, then 2 is the number of consecutive 1, and so on

Image Analyst on 6 Jan 2013

Oh, thanks Azzi, I didn't notice that. In that case, you can use bwlabel, regionprops, and intlut:

```m = logical([1 0 0 1 1 1 1 0 1 1 0])
% Group into connected regions.
labeledArray = bwlabel(m)
% Measure the area of all regions.
measurements = regionprops(labeledArray, 'Area');
areas = [measurements.Area]
numberOfAreas = length(areas);
```
```% Assign each connected area with its area.
% Make a look up table to map each region number into the area of that region.
lookUpTable = uint16([0 areas, zeros(1,65536-numberOfAreas-1)]);
% Do the actual mapping.
output = intlut(uint16(labeledArray), lookUpTable)
```

(Requires the Image Processing Toolbox.) This should also work with 2D arrays.

Answer by Jan Simon on 6 Jan 2013
Edited by Jan Simon on 6 Jan 2013

An inplace method, which changes the input vector on the fly without storing an index list - this muight be an advantage for large data sets:

```function a = RunLength_IgnoZero_loop1(a)
```
```len = length(a);
c   = a(1);
ini = 1;
for ii = 2:len
if a(ii) ~= c
if c == 0
ini = ii;
else
a(ini:ii-1) = ii - ini;
end
c = a(ii);
end
end
```
```% care about last segement:
if c ~= 0
a(ini:len) = len - ini + 1;
end
```

Answer by Jan Simon on 6 Jan 2013
Edited by Jan Simon on 6 Jan 2013

For the test data x = double(rand(1, 1e6) > 0.8) this vectorized method is 9 times faster than my loop approach:

```function a = RunLength_IgnoZero_Vec(a)
```
```pos   = [false, a > 0, false];
start = strfind(pos, [false, true]);
stop  = strfind(pos, [true, false]) -1;
run   = stop - start + 1;
```
```b         = zeros(size(a));
b(start)  = run;
b(stop+1) = -run;
if length(b) == length(a)
a = cumsum(b);
else
a = cumsum(b(1:length(b) - 1));
end
```