how to change data

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

[Image Analyst]

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.)

Answer 1

[Azzi Abdelmalek]

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]

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]

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]

Thank you very much

Answer 2

[Jan]

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);

[Jan]

Some timings:

x = double(rand(1, 1e6) > 0.8);
tic; for ii = 1:100, y = RunLength_IgnoZero(x); end; toc
% Azzi's method without pre-allocation:
21000 seconds (I ran with loop with 1 iteration only)
% Azzi's method with pre-allocation: idx = zeros(numel(a), 2);
241.4 seconds
% RunLength_IgnoZero_loop1:
47.02 seconds
% Jose-Luis' bwconncomp:
25.07 seconds
% RunLength_IgnoZero_Vec:
7.109 seconds
% RunLength_IgnoZero_loop2:
3.687 seconds

[Image Analyst]

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]

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]

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

Answer 3

[Roger Stafford]

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 4

[Azzi Abdelmalek]

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

[Jan]

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

Answer 5

[Image Analyst]

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]

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]

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 6

[Jan]

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 7

[Jan]

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