Please help me store data from iterations

How can I store the values of cell_count and tot_hrs1 from each iteration and then plot them against each other.

n = [-1:32];
%n=32;
%n=input('counts, n= ')
cell_count=1;
tot_hrs1=0;
for i=0:n
  r=0:7;
  t=3+4*r;
  %r=1:(70/8);
  %multiples_8=8.*r;
 if i~=t(1)
    cell_count(i)=cell_count(i)*0.90+cell_count(i);
 %end
 elseif i==t(1)
    cell_count(i)=cell_count(i)*0.90+cell_count(i)-1;
 end

 if n < 0
    tot_hrs1=0;
 elseif n >= 0
     tot_hrs1=tot_hrs1+2;
 end    

end
tot_hrs1
format('shortE')
cell_count(i)

Thanks

Answer 1

[Image Analyst]

Make tot_hrs1 an array, just like you did for cell_count.

tot_hrs1(i)

[James]

I did that and it is only storing the first iteration, not the next values. Also do you think my counter i=0:n is correct as n is an array and I don't think that it is stepping through the array n.

Thanks

[Image Analyst]

No, it will probably only take the first one, which may not even enter the loop. Don't use i as the loop iterator - i is the imaginary variable - use something else, like k. You could do

for k = n

in which case k will take on all values in n, but then you can't do something like tot_hrs1(k) because indexes must be integers starting at 1 - no 0 or -1 allowed.

Finally, try using descriptive variable names, for example the descriptive "totalHours" rather than the cryptic "tot_hrs1." It will make it easier for people to follow and understand your code.

[James]

Thanks for your comments

Can you see my answer below. This is actually what I am trying to do.

Any suggestions?

Thanks

Answer 2

[Amit Davidi]

Hello James,

If you your loop to run n times, use:

n = 32;
for i = 1 : n
...
end

An array in Matlab start from the index 1, so your loop need to start from 1 (not 0). Also, variable like r and t, could be calculated once before the loop. It's unnecessary and wastes time. To further improve running time, pre-allocate memory to arrays (assuming you know their size). I've tried to rewrite your code, although I'm not sure I understood the purpose of your code:

n = 4;
cell_count = ones(1,n);
tot_hrs1 = zeros(1,n);
r = 0 : 7;
t = 3 + 4 * r;

for i = 1 : n

    if i ~= t(1)
        cell_count(i) = cell_count(1) * 0.90 + cell_count(1);
    elseif i==t(1)
        cell_count(i) = cell_count(1) * 0.90 + cell_count(1) - 1;
    end

    if n < 0
        tot_hrs1(i) = 0;
    elseif n >= 0
        tot_hrs1(i) = tot_hrs1(1) + 2;
    end
end

tot_hrs1
cell_count

plot(tot_hrs1, cell_count, 'o-')

PS - a cosmetic remark, if I may: I would recommend you to use more spaces in your codes, both in between lines (as I did above), it'll make your codes much more readable.

Good luck, Amit

[James]

Thanks for your comments.

Can you see my answer below. That is what I am trying to do.

with your method it is not actually counting.

Thanks

Answer 3

[James]

I am wanting to do the following: Calculate the total hours to reach a cell count of 1e-9 The parameters: Each cell divides 1 every 2 hours Each cell has a 10% chance of dying each time it divides The rate of cell loss is 1 every 8 hours

Time Cell Count 
 x0      1              Begin with 1 cell 
 x2   x0*0.90+x0 
 x4   x2*0.90+x2 
 x6   x4*0.90+x4 
 x8   x6*0.90+x6-1      Loss of a cell after 8 hours

Then plot the total hours vs the cell count

Thanks

[Image Analyst]

How can you have less than 1 cell? A cell is either there and alive, or it's dead. How can you have 10e-9 of a cell? It looks like your cell count is cellCount(k) = (k-1)*1.9 except for iteration 5 (time x8). And it doesn't seem like a valid Monte Carlo approach since you're not using the rand() function at all to get a random number that gets the possibility that the cell dies. Something like

if rand(1) < 0.1

% Cell dies

else

% Cell lives

end

[James]

So I have one cancer tumor of a particular size. Inside of that tumor there are 1e-9 cancer cells. There are some calculations that I did to figure out how many cancer cells are in a tumor of 1cm size if the cancer cells are 10micron size. This is not that important thought to my actually Matlab question. I want to find out how many hours it takes to reach 1e-9 cancer cells. So starting out at time 0 the cell count is 1. Then in 2 hours you have a division, but at every division there is a 10% chance of the cell dying. Then at every 8 hours one of those cells dies. This is just a hypothetical case.

Thanks

[Image Analyst]

Do you mean 1e+9 instead of 1e-9?

[James]

yes sorry about that.

Answer 4

[Image Analyst]

James, Since you made an attempt at it, I improved (actually totally rewrote it) for you. Try this:

clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
workspace;  % Make sure the workspace panel is showing.
fontSize = 20;
% Determine some starting number of cells.
% It must be greater than 1 since 10 percent die
% on the first iteration.
cell_count=10;
totalHours=0;
index = 1;
while cell_count(index) < 1e+9
    % Increment the generation index.
    index = index + 1
    % Determine and save the hours since we started.
    totalHours(index) = totalHours(index-1) + 2;

      % 10 percent of the cells dies at this iteration
      % before they can divide.
      cell_count(index) = int32(0.9 * cell_count(index-1));
      if cell_count(index) < 1
          % No more cells left so bail out.
          break;
      end
      % Multiply the remaining cells by two.
      % Now cell_count(index) is the remaining number of cells
      % so we don't use the -1 anymore.
      % I guess in biology multiply and divide mean the same thing.
      cell_count(index) = cell_count(index) * 2
  end
  totalHours
  cell_count

% Plot linearly.
subplot(1,2,1);
plot(totalHours, cell_count);
grid on;
xlabel('Time [hours]', 'FontSize', fontSize);
ylabel('Cell Count', 'FontSize', fontSize);
title('Cell Count vs. Time', 'FontSize', fontSize);

% Plot on a semilog plot
subplot(1,2, 2);
semilogy(totalHours, cell_count);
grid on;
xlabel('Time [hours]', 'FontSize', fontSize);
ylabel('Cell Count', 'FontSize', fontSize);
title('Cell Count vs. Time', 'FontSize', fontSize);

% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]); % Maximize figure.
set(gcf,'name','Demo by ImageAnalyst','numbertitle','off')

[James]

Thanks for your suggestions. I'm not sure what your cell_count=10; in the beginning means. The starting cell_count has to be 1.

I know that it takes a total of 66 hrs to reach a cell count of 1.4595e+9

I am having a hard time getting this with your code. Any insight?

Thanks

[James]

So at 0 hours you start with 1 cell

Then at 2 hours you have 1.9 cells (which is your previous cell count*1.9)

Then at 4 hours you have 3.61 cells (which is your previous cell count*1.9)

Then at 6 hours you have 6.859 cells (which is your previous cell count*1.9)

Then at 8 hours you have 12.0321 cells (which is your previous cell count*1.9 - 1)

This should just repeat until you reach 66 hours and get 1.4595e+9 cell count.

Thanks

[James]

Also, what is int32?

Thanks