How to find the index of the first element which is greater than a particular value in a matrix for all the columns.

Suppose I have a 300 by 300 matrix, containg the values from 0 to 200. For each column, I want to find the position of the first element which is greater than 50. how do I do it. I have tried to write a code but can't do it any further.

no_row=size(c,1);
no_col=size(c,2);
for i=1:no_col
    for j=1:no_row
        if c(j,i)>=50
        end
    end
end

Answer 1

[Stephen23]

N = 5;
M = randi(9,5,7)

M = 5×7

9 3 7 2 8 1 2 1 7 6 4 4 2 9 7 1 6 1 7 6 2 7 3 5 1 1 6 2 6 8 2 9 1 1 7

Method one: CUMSUM and FIND:

X = M>N;
X = X & cumsum(X,1)<2;
[R,~] = find(X)

R = 7×1

1 2 1 5 1 3 2

Method two: FIND and ACCUMARRAY:

[Rx,Cx] = find(M>N);
R = accumarray(Cx,Rx,[],@min)

R = 7×1

1 2 1 5 1 3 2

Method three**: NUM2CELL and CELLFUN (fails if there is no such element in any column):

F = @(v)find(v>N,1,'first');
R = cellfun(F,num2cell(M,1))

R = 1×7

1 2 1 5 1 3 2

[DARSHAN KUMAR BISWAS]

Sorry but I can't understand what you just wrote. It would be really helpful if you could just explain what you are writing.

p.s. I am new to matlab.

[Stephen23]

Forget about loops. Loops are what people write in low-level languages. MATLAB is a high-level languages based around matrices. Learn how to process entire matrices (arrays/vectors..). If you do not understand what code is doing, then:

• read the documentation for all of the functions and operators
• print out all of the intermediate results.

doing both of those will go a long way to understanding what the code is doing. Lets look at the first method in my answer, which is neat and efficient (note that if any column does not contain such a value it will not throw an error, just return fewer indices in the output):

N = 5;
M = randi(9,5,7)

M = 5×7

7 3 9 4 4 8 2 8 3 6 8 5 7 5 1 7 6 2 8 9 4 9 9 3 6 8 2 2 1 7 9 3 1 4 6

X = M>N % logical indices of all values > 50

X = 5×7 logical array
   1   0   1   0   0   1   0
   1   0   1   1   0   1   0
   0   1   1   0   1   1   0
   1   1   0   1   1   0   0
   0   1   1   0   0   0   1

Y = cumsum(X,1) % count 1's in each column

Y = 5×7

1 0 1 0 0 1 0 2 0 2 1 0 2 0 2 1 3 1 1 3 0 3 2 3 2 2 3 0 3 3 4 2 2 3 1

X = X & Y<2 % logical index of first value > 50 in each column 

X = 5×7 logical array
   1   0   1   0   0   1   0
   0   0   0   1   0   0   0
   0   1   0   0   1   0   0
   0   0   0   0   0   0   0
   0   0   0   0   0   0   1

[R,C] = find(X) % row of each value > 50 in each column

R = 7×1

1 3 1 2 3 1 5

C = 7×1

1 2 3 4 5 6 7

When you use loops you break your data into pieces and repeat tasks on the pieces. When you think in terms of matrices you apply those tasks to all of the data at once. Much neater!

[Jan]

@Stephen23: "Forget about loops. Loops are what people write in low-level languages." - except if they need fast code ;-)

s1 = 300;
s2 = 300;
M  = randi([0, 200], s1, s2);
N  = 50;

tic
for rep = 1:1e4    % Run this repeatedly for accurate timings
   R = nan(s2, 1);
   for k = 1:s2
      v = find(M(:, k) > N, 1);
      if ~isempty(v)
         R(k) = v;
      end
   end
end
toc

Elapsed time is 3.994913 seconds.

tic
for rep = 1:1e4
   R = nan(s2, 1);
   for k = 1:s2
      for kk = 1:s1
        if M(kk, k) > N
           R(k) = kk;
           break;
        end
      end  
   end
end
toc

Elapsed time is 0.051487 seconds.

tic
for rep = 1:1e4
   MM = M > N;
   MM = MM & cumsum(MM,1) < 2;
   [R,~] = find(MM);
end
toc

Elapsed time is 2.443751 seconds.

tic
for rep = 1:1e4
   [Rx,Cx] = find(M>N);
   R = accumarray(Cx,Rx,[],@min);
end
toc

Elapsed time is 8.819041 seconds.

tic
for rep = 1:1e4
   F = @(v)find(v>N,1,'first');
   R = cellfun(F,num2cell(M,1));
end
toc

Elapsed time is 15.418575 seconds.

On my local R2018b/i7 I get these timings:

% Elapsed time is 1.851131 seconds.    Loop find
% Elapsed time is 0.051859 seconds.    Loop loop !!!
% Elapsed time is 2.296044 seconds.    cumsum
% Elapsed time is 16.072823 seconds.   accumarray
% Elapsed time is 25.542278 seconds.   cellfun

Why does the loop/find version takes the double time here in the forum?

[Stephen23]

" except if they need fast code"

Which is only some of the time. In most cases, getting the correct result has a higher priority.

[Jan]

@Stephen23: Of course the methods shown above produce the same result. The dull loop/loop method is 305 times faster than the cellfun approach and does not fail on missing values. I am impressed by this speed, although I actually use Matlab to avoid such low level programming.

I assume, it is much more difficult for a beginner to write a working accumarray appraoch than to implement the nested loop appraoch correctly. So speed and valid code are arguments for the low level method. Unfortunately.

Answer 2

[Image Analyst]

Try this. It's a loop but only 300 iterations for it's fast, completing in 0.001 seconds:

% Prepare sample image because user forgot to upload the data.
grayImage = 200 * rand(300);
subplot(1, 2, 1);
imshow(grayImage, []);
[rows, columns] = size(grayImage);

% Now scan across columns finding the first row where the value exceeds 200.
tic
topRows = nan(1, columns);
for col = 1 : columns
    t = find(grayImage(:, col) > 50, 1, "first");
    if ~isempty(t)
        topRows(col) = t;
    end
end
toc % Takes 0.001 seconds for me

% Plot the location of the top rows.
subplot(1, 2, 2);
plot(topRows, 'r.', 'MarkerSize', 15);
grid on;
xlabel('Column');
ylabel('First row where value > 200')

[Jan]

I've implemented an equivalent loop, which runs in 1.8e-4 s on my local i7/R2018b, but here in the forum it needs 3.7e-4s (see my comment). To my surprise a nested loop instead of find(x >= 50, 1) is 36 times faster. Could you please run the code provided there and post the timings?

I've learned and teached Matlab for so long, and now the JIT got such powerful, that Matlab's loops gets near to C-code. Sigh. See also: Mandelbrot with dull loops and two reals instead of a complex number

[Image Analyst]

It runs in 0.0046 here in the forum using MATLAB online. I've brought this up before. You'd think the Answers server would use a very powerful computer so why is it like 4 or 5 times slower than my laptop? They basically said that they expect that, and it was not surprising, and that's just the way it is.

[Jan]

Thanks, @Image Analyst, I've found the source of the different speeds: The online interpreter in the forum seems to let the JIT work more powerful in functions than in scripts:

s1 = 300;
s2 = 300;
M  = randi([0, 200], s1, s2);
N  = 50;

tic
for rep = 1:1e4
    R = nan(s2, 1);
    for k = 1:s2
        v = find(M(:, k) > N, 1);
        if ~isempty(v)
            R(k) = v;
        end
    end
end
toc  % Code in the main script:

Elapsed time is 3.724234 seconds.

tic;
R = fcn(M, s1, s2, N);
toc  % Identical code inside a function:

Elapsed time is 1.392424 seconds.

function R = fcn(M, s1, s2, N)
for rep = 1:1e4
    R = nan(s2, 1);
    for k = 1:s2
        v = find(M(:, k) > N, 1);
        if ~isempty(v)
            R(k) = v;
        end
    end
end
end

I'm surprised that this nested loop is 35 times faster (even inside a script):

R = nan(s2, 1);
for k = 1:s2
    for kk = 1:s1
        if M(kk, k) > N
            R(k) = kk;
            break;
        end
    end  
end

Answer 3

[Mathan]

[no_row,no_col] = size(c);
Index_final = [];
 
for ii = 1:no_col
    Index =[];
    for jj=1:no_row
        if c(jj,ii)>=50
            Index = [Index jj];  % Index contains all the elements greater than 50 in that particular column
        end
    end
    if (sum(Index)>0)
        Index_final = [Index_final Index(1)]; % Index_final should contain all the required elements in your case.
    end
end

[DARSHAN KUMAR BISWAS]

I am getting this in the command window when I write your code.

Index exceeds the number of array elements. Index must not exceed 0.

Error in ykf (line 12)

Index_final = [Index_final Index(1)]; % Index_final should contain all the required elements in your case.

[Mathan]

can you give me your c matrix? Perhaps attach it as a .mat file here.

[Mathan]

I have now modified the code so as to include an if statement so that you will get the indices of the first element which is greater than 50 from all those columns which have at least one element that is greater than 50. Previously, you were getting the error because you were having some columns that did not have any element greater than 50.

[Jan]

@Mathan: Avoid iteratively growing arrays, because tzhey are very expensive. Although this is hardly a problem for a tiny array as a [300x300] matrix, but it is worth to use methods, which run for large problems also.

x = [];
for k = 1:1e6
    x = [x, k];
end

This creates a new array in each iteration and copies the former values. Although the final result uses 8 MB only (8 byte per element), the Matlab has to allocate and copy sum(1:1e6)*8 Byte = 4 TeraByte ! This is a massive waste of processing time. The solution is easy:

x = zeros(1, 1e6);   % Pre-allocate!!!
for k = 1:1e6
    x(k) = k;
end

Now the new value is inserted in the existing array.

Instead of proceeding to compare the elements, you can use break after the first match.

[Mathan]

Thanks Jan - makes sense!

Answer 4

[Jan]

s1 = 300;
s2 = 300;
C  = randi([0, 200], s1, s2);   % Test data
N  = 50;                        % Search

R = nan(s2, 1);        % Pre-allocate the output array
for i2 = 1:s2          % Loop over columns
  for i1 = 1:s1        % Loop over rows
    if C(i1, i2) >= N  % Compare the element
       R(i2) = i1;     % Store in output
       break;          % Stop "for i1" loop
    end
  end  
end

This is a loop approach equivalent to C or even Basic. Actually Matlab offers more elegant command, see Stephen23's answer. But these "vectorized" methods are slower in modern Matlab versions, because they have to create temporary arrays. The CPU is much faster than the RAM (especially if the data sizes exceed the 1st and 2nd level caches). So this ugly code is much faster than the other suggestions.

This is the case for larger arrays also: I've tested it with 30'000x30'000 matrices.

Answer 5

[DARSHAN KUMAR BISWAS]

A = [29 33 35 25 33 32 12;3 37 10 22 19 30 9;12 33 19 23 41 11 12;18 23 32 16 27 16 22;42 28 40 26 18 24 16;1 15 5 26 47 12 47;3 38 47 41 44 43 22; 9 10 39 40 28 10 10]

k=zeros(1,7);

for i=1:7

p=A(:,i);

for j=1:8

if p(j)>=40

k(i)=j

plot(7-k)

break

end

end

end

[DARSHAN KUMAR BISWAS]

I tried and tried and wrote this one. constructive criticism is always welcome.

[Image Analyst]

Yeah, lots of room for improvement in that code.

1. I'd split A onto separate lines for readability, breaking after the semicolons.
2. To be more general and robust, I'd get the number of rows and columns of A using size, and use those sizes in the "for" lines and in the call to zeros().
3. You need "hold on" after plot(), plus you need an (x,y) pair.
4. No need to use a for loop and break when you can simply use find

See my Answer above to see better coding practice.