how to change data
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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.)
Accepted Answer
Azzi Abdelmalek
on 6 Jan 2013
Edited: 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
3 Comments
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
More Answers (6)
Jan
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);
4 Comments
Jan
on 6 Jan 2013
Edited: Jan
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.
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));
0 Comments
Azzi Abdelmalek
on 6 Jan 2013
Edited: 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
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?).
2 Comments
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
Edited: 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.
Jan
on 6 Jan 2013
Edited: Jan
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
0 Comments
Jan
on 6 Jan 2013
Edited: Jan
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
0 Comments
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