so, i have tried to Plot all of values of (h,rex)
but its plot just the last values of them.
i think must use array but i dont know how to do that. i hope to help me !
clc
Pr=1.05;
lam=0.167;
l=2;
rex=100;
while (rex<2900)
disp('rex');
disp(rex);
h=((0.664*Pr^(1/3)*lam)/l)*rex^(1/2) ;
disp('h');
disp(h);
rex=rex+100;
end
plot(rex,h);
You are not storing your intermediate values, hence rex and h are overwritten each time and only the values from the last iteration are available. Solve this by storing the values in an array:
Pr=1.05;
lam=0.167;
l=2;
rex=100;
rex_array = [];
h_array = [];
while (rex<2900)
h=((0.664*Pr^(1/3)*lam)/l)*rex^(1/2) ;
rex=rex+100;
% store values in array (appending)
rex_array(end+1) = rex;
h_array(end+1) = h;
end
figure(1); clf;
plot(rex_array,h_array);
This is not recommended though, since rex_array and h_array are increasing in size with every iteration. So instead you use a for loop.
rex = 100:100:2900; % or to give same result as the while loop: 100:100:2800
h = zeros(size(rex)); % initialize h to be a zero vector with the same size as rex
% loop over rex
for k = 1:numel(rex)
% store each value in h(k):
h(k)=((0.664*Pr^(1/3)*lam)/l)*rex(k)^(1/2) ;
end
figure(1); clf;
plot(rex,h);
The best option though is to avoid the loop completely, and use vectorization:
rex = 100:100:2900; % or to give same result as the while loop: 100:100:2800
h=((0.664*Pr^(1/3)*lam)/l)*rex.^(1/2);
By using .^ each element in rex will be squared element wise.
Related
Suppose we are running an infinite for loop in MATLAB, and we want to store the iterative values in a vector. How can we declare the vector without knowing the size of it?
z=??
for i=1:inf
z(i,1)=i;
if(condition)%%condition is met then break out of the loop
break;
end;
end;
Please note first that this is bad practise, and you should preallocate where possible.
That being said, using the end keyword is the best option for extending arrays by a single element:
z = [];
for ii = 1:x
z(end+1, 1) = ii; % Index to the (end+1)th position, extending the array
end
You can also concatenate results from previous iterations, this tends to be slower since you have the assignment variable on both sides of the equals operator
z = [];
for ii = 1:x
z = [z; ii];
end
Sadar commented that directly indexing out of bounds (as other answers are suggesting) is depreciated by MathWorks, I'm not sure on a source for this.
If your condition computation is separate from the output computation, you could get the required size first
k = 0;
while ~condition
condition = true; % evaluate the condition here
k = k + 1;
end
z = zeros( k, 1 ); % now we can pre-allocate
for ii = 1:k
z(ii) = ii; % assign values
end
Depending on your use case you might not know the actual number of iterations and therefore vector elements, but you might know the maximum possible number of iterations. As said before, resizing a vector in each loop iteration could be a real performance bottleneck, you might consider something like this:
maxNumIterations = 12345;
myVector = zeros(maxNumIterations, 1);
for n = 1:maxNumIterations
myVector(n) = someFunctionReturningTheDesiredValue(n);
if(condition)
vecLength = n;
break;
end
end
% Resize the vector to the length that has actually been filled
myVector = myVector(1:vecLength);
By the way, I'd give you the advice to NOT getting used to use i as an index in Matlab programs as this will mask the imaginary unit i. I ran into some nasty bugs in complex calculations inside loops by doing so, so I would advise to just take n or any other letter of your choice as your go-to loop index variable name even if you are not dealing with complex values in your functions ;)
You can just declare an empty matrix with
z = []
This will create a 0x0 matrix which will resize when you write data to it.
In your case it will grow to a vector ix1.
Keep in mind that this is much slower than initializing your vector beforehand with the zeros(dim,dim) function.
So if there is any way to figure out the max value of i you should initialize it withz = zeros(i,1)
cheers,
Simon
You can initialize z to be an empty array, it'll expand automatically during looping ...something like:
z = [];
for i = 1:Inf
z(i) = i;
if (condition)
break;
end
end
However this looks nasty (and throws a warning: Warning: FOR loop index is too large. Truncating to 9223372036854775807), I would do here a while (true) or the condition itself and increment manually.
z = [];
i = 0;
while !condition
i=i+1;
z[i]=i;
end
And/or if your example is really what you need at the end, replace the re-creation of the array with something like:
while !condition
i=i+1;
end
z = 1:i;
As mentioned in various times in this thread the resizing of an array is very processing intensive, and could take a lot of time.
If processing time is not an issue:
Then something like #Wolfie mentioned would be good enough. In each iteration the array length will be increased and that is that:
z = [];
for ii = 1:x
%z = [z; ii];
z(end+1) = ii % Best way
end
If processing time is an issue:
If the processing time is a large factor, and you want it to run as smooth as possible, then you need to preallocating.If you have a rough idea of the maximum number of iterations that will run then you can use #PluginPenguin's suggestion. But there could still be a change of hitting that preset limit, which will break (or severely slow down) the program.
My suggestion:
If your loop is running infinitely until you stop it, you could do occasional resizing. Essentially extending the size as you go, but only doing it once in a while. For example every 100 loops:
z = zeros(100,1);
for i=1:inf
z(i,1)=i;
fprintf("%d,\t%d\n",i,length(z)); % See it working
if i+1 >= length(z) %The array as run out of space
%z = [z; zeros(100,1)]; % Extend this array (note the semi-colon)
z((length(z)+100),1) = 0; % Seems twice as fast as the commented method
end
if(condition)%%condition is met then break out of the loop
break;
end;
end
This means that the loop can run forever, the array will increase with it, but only every once in a while. This means that the processing time hit will be minimal.
Edit:
As #Cris kindly mentioned MATLAB already does what I proposed internally. This makes two of my comments completely wrong. So the best will be to follow what #Wolfie and #Cris said with:
z(end+1) = i
Hope this helps!
I wrote this matlab code in order to concatenate the results of the integration of all the columns of a matrix extracted form a multi matrix array.
"datimf" is a matrix composed by 100 matrices, each of 224*640, vertically concatenated.
In the first loop i select every single matrix.
In the second loop i integrate every single column of the selected matrix
obtaining a row of 640 elements.
The third loop must concatenate vertically all the lines previously calculated.
Anyway i got always a problem with the third loop. Where is the error?
singleframe = zeros(224,640);
int_frame_all = zeros(1,640);
conc = zeros(100,640);
for i=0:224:(22400-224)
for j = 1:640
for k = 1:100
singleframe(:,:) = datimf([i+1:(i+223)+1],:);
int_frame_all(:,j) = trapz(singleframe(:,j));
conc(:,k) = vertcat(int_frame_all);
end
end
end
An alternate way to do this without using any explicit loops (edited in response to rayryeng's comment below. It's also worth noting that using cellfun may not be more efficient than explicitly looping.):
nmats = 100;
nrows = 224;
ncols = 640;
datimf = rand(nmats*nrows, ncols);
% convert to an nmats x 1 cell array containing each matrix
cellOfMats = mat2cell(datimf, ones(1, nmats)*nrows, ncols);
% Apply trapz to the contents of each cell
cellOfIntegrals = cellfun(#trapz, cellOfMats, 'UniformOutput', false);
% concatenate the results
conc = cat(1, cellOfIntegrals{:});
Taking inspiration from user2305193's answer, here's an even better "loop-free" solution, based on reshaping the matrix and applying trapz along the appropriate dimension:
datReshaped = reshape(datimf, nrows, nmats, ncols);
solution = squeeze(trapz(datReshaped, 1));
% verify solutions are equivalent:
all(solution(:) == conc(:)) % ans = true
I think I understand what you want. The third loop is unnecessary as both the inner and outer loops are 100 elements long. Also the way you have it you are assigning singleframe lots more times than necessary since it does not depend on the inner loops j or k. You were also trying to add int_frame_all to conc before int_frame_all was finished being populated.
On top of that the j loop isn't required either since trapz can operate on the entire matrix at once anyway.
I think this is closer to what you intended:
datimf = rand(224*100,640);
singleframe = zeros(224,640);
int_frame_all = zeros(1,640);
conc = zeros(100,640);
for i=1:100
idx = (i-1)*224+1;
singleframe(:,:) = datimf(idx:idx+223,:);
% for j = 1:640
% int_frame_all(:,j) = trapz(singleframe(:,j));
% end
% The loop is uncessary as trapz can operate on the entire matrix at once.
int_frame_all = trapz(singleframe,1);
%I think this is what you really want...
conc(i,:) = int_frame_all;
end
It looks like you're processing frames in a video.
The most efficent approach in my experience would be to reshape datimf to be 3-dimensional. This can easily be achieved with the reshape command.
something along the line of vid=reshape(datimf,224,640,[]); should get you far in this regard, where the 3rd dimension is time. vid(:,:,1) then would display the first frame of the video.
l_0=1.5;
l_1=1.6;
Lambda_min=2*(1+1)*l_0;
Lambda_max=2*(1+1)*l_1;
n_0=linspace(2,2.11,10);
n_1=linspace(2.30,2.50,10);
for i=1:10
for j=1:10
for k=1:10
l(i) = Lambda_min * ( Lambda_max/Lambda_min)^(i/10)
sum=sum(l)
d_0(:,j)= l(i)/((n_0(i)/n_1(i)+1))
d_1(:,k)= (n_0(i)/n_1(i))*d_0(:,j)
end
end
end
First of all; I want to find values of l(i) which is a vector, then take the sum of that vector. second, for d_0(:,j) I want to create a matrix so I can plot it later, that takes different values from l(i),n_0,n_1 each time. If I take the values for n_0 and n_1 and put in the for loop I will get index error because it should be logic or integer number.
My matrix is overwritten and do not know how to avoid it. Note, I want in d_0 and d_1 n_0 and n_1 to take values from linspace. for example in the first iteration n_0= 2 n_1= 2.30 then second iteration take the next value in linspace.
I tried to see the value of n_0(i) and does it give me 10 iterations. It gives me more that that overwritten.
Try:
l_0=1.5;
l_1=1.6;
Lambda_min = 4*l_0;
Lambda_max = 4*l_1;
n_0 = linspace(2,2.11,10) % don't add semicolon so you can check this is giving 10 values
n_1 = linspace(2.30,2.50,10) %
for i=1:10
l(i) = Lambda_min * ( Lambda_max/Lambda_min)^(i/10) % should give you 10 values
end
d_0= l./((n_0./n_1+1)); % This will only give you a vector, not a matrix.
d_1= (n_0./n_1).*d_0;
Lsum = sum(l); % should give you one value
Hi for my code I would like to know how to best save my variable column. column is 733x1. Ideally I would like to have
column1(y)=column, but I obtain the error:
Conversion to cell from logical is not possible.
in the inner loop. I find it difficult to access these stored values in overlap.
for i = 1:7
for y = 1:ydim % ydim = 436
%execute code %code produces different 'column' on each iteration
column1{y} = column; %'column' size 733x1 %altogether 436 sets of 'column'
end
overlap{i} = column1; %iterates 7 times.
end
Ideally I want overlap to store 7 variables saved that are (733x436).
Thanks.
I'm assuming column is calculated using a procedure where each column is dependent on the latter. If not, then there are very likely improvements that can be made to this:
column = zeros(733, 1); % Might not need this. Depends on you code.
all_columns = zeros(xdim, ydim); % Pre-allocate memory (always do this)
% Note that the first dimension is usually called x,
% and the second called y in MATLAB
overlap = cell(7, 1);
overlap(:) = {zeros(xdim, ydim)}; % Pre-allocate memory
for ii = 1:numel(overlap) % numel is better than length
for jj = 1:ydim % ii and jj are better than i and j
% several_lines_of_code_to_calculate_column
column = something;
all_columns(:, jj) = column;
end
overlap{ii} = all_columns;
end
You can access the variables in overlap like this: overlap{1}(1,1);. This will get the first element in the first cell. overlap{2} will get the entire matrix in the second cell.
You specified that you wanted 7 variables. Your code implies that you know that cells are better than assigning it to different variables (var1, var2 ...). Good! The solution with different variables is bad bad bad.
Instead of using a cell array, you could instead use a 3D-array. This might make processing later on faster, if you can vectorize stuff for instance.
This will be:
column = zeros(733, 1); % Might not need this. Depends on you code.
overlap = zeros(xdim, ydim, 7) % Pre-allocate memory for 3D-matrix
for ii = 1:7
for jj = 1:ydim
% several_lines_of_code_to_calculate_column
column = something;
all_column(:, jj, ii) = column;
end
end
I am trying to concatenate several structs. What I take from each struct depends on a function that requires a for loop. Here is my simplified array:
t = 1;
for t = 1:5 %this isn't the for loop I am asking about
a(t).data = t^2; %it just creates a simple struct with 5 data entries
end
Here I am doing concatenation manually:
A = [a(1:2).data a(1:3).data a(1:4).data a(1:5).data] %concatenation function
As you can see, the range (1:2), (1:3), (1:4), and (1:5) can be looped, which I attempt to do like this:
t = 2;
A = [for t = 2:5
a(1:t).data
end]
This results in an error "Illegal use of reserved keyword "for"."
How can I do a for loop within the concatenate function? Can I do loops within other functions in Matlab? Is there another way to do it, other than copy/pasting the line and changing 1 number manually?
You were close to getting it right! This will do what you want.
A = []; %% note: no need to initialize t, the for-loop takes care of that
for t = 2:5
A = [A a(1:t).data]
end
This seems strange though...you are concatenating the same elements over and over...in this example, you get the result:
A =
1 4 1 4 9 1 4 9 16 1 4 9 16 25
If what you really need is just the .data elements concatenated into a single array, then that is very simple:
A = [a.data]
A couple of notes about this: why are the brackets necessary? Because the expressions
a.data, a(1:t).data
don't return all the numbers in a single array, like many functions do. They return a separate answer for each element of the structure array. You can test this like so:
>> [b,c,d,e,f] = a.data
b =
1
c =
4
d =
9
e =
16
f =
25
Five different answers there. But MATLAB gives you a cheat -- the square brackets! Put an expression like a.data inside square brackets, and all of a sudden those separate answers are compressed into a single array. It's magic!
Another note: for very large arrays, the for-loop version here will be very slow. It would be better to allocate the memory for A ahead of time. In the for-loop here, MATLAB is dynamically resizing the array each time through, and that can be very slow if your for-loop has 1 million iterations. If it's less than 1000 or so, you won't notice it at all.
Finally, the reason that HBHB could not run your struct creating code at the top is that it doesn't work unless a is already defined in your workspace. If you initialize a like this:
%% t = 1; %% by the way, you don't need this, the t value is overwritten by the loop below
a = []; %% always initialize!
for t = 1:5 %this isn't the for loop I am asking about
a(t).data = t^2; %it just creates a simple struct with 5 data entries
end
then it runs for anyone the first time.
As an appendix to gariepy's answer:
The matrix concatenation
A = [A k];
as a way of appending to it is actually pretty slow. You end up reassigning N elements every time you concatenate to an N size vector. If all you're doing is adding elements to the end of it, it is better to use the following syntax
A(end+1) = k;
In MATLAB this is optimized such that on average you only need to reassign about 80% of the elements in a matrix. This might not seam much, but for 10k elements this adds up to ~ an order of magnitude of difference in time (at least for me).
Bare in mind that this works only in MATLAB 2012b and higher as described in this thead: Octave/Matlab: Adding new elements to a vector
This is the code I used. tic/toc syntax is not the most accurate method for profiling in MATLAB, but it illustrates the point.
close all; clear all; clc;
t_cnc = []; t_app = [];
N = 1000;
for n = 1:N;
% Concatenate
tic;
A = [];
for k = 1:n;
A = [A k];
end
t_cnc(end+1) = toc;
% Append
tic;
A = [];
for k = 1:n;
A(end+1) = k;
end
t_app(end+1) = toc;
end
t_cnc = t_cnc*1000; t_app = t_app*1000; % Convert to ms
% Fit a straight line on a log scale
P1 = polyfit(log(1:N),log(t_cnc),1); P_cnc = #(x) exp(P1(2)).*x.^P1(1);
P2 = polyfit(log(1:N),log(t_app),1); P_app = #(x) exp(P2(2)).*x.^P2(1);
% Plot and save
loglog(1:N,t_cnc,'.',1:N,P_cnc(1:N),'k--',...
1:N,t_app,'.',1:N,P_app(1:N),'k--');
grid on;
xlabel('log(N)');
ylabel('log(Elapsed time / ms)');
title('Concatenate vs. Append in MATLAB 2014b');
legend('A = [A k]',['O(N^{',num2str(P1(1)),'})'],...
'A(end+1) = k',['O(N^{',num2str(P2(1)),'})'],...
'Location','northwest');
saveas(gcf,'Cnc_vs_App_test.png');