I have two for loops in MATLAB.
One of the for loops leads to different variables being inserted into the model, which are 43 and then I have 5 horizons.
So I estimate the model 215 times.
My problem is I want to store this in 215x5 matrix, the reason I have x5 is that I am estimating 5 variables, 4 are fixed and the other one comes from the for loop.
I have tried to do this in two ways,
Firstly, I create a variable called out,
out=zeros(215,5);
The first for loop is,
for i=[1,2,3,4,5];
The second for loop is,
for ii=18:60;
The 18:60 is how I define my variables using XLS read, e.g. they are inserted into the model as (data:,ii).
I have tried to store the data in two ways, I want to store OLS which contains the five estimates
First,
out(i,:)=OLS;
This method creates a 5 x 5 matrix, with the estimates for one of the (18:60), at each of the horizons.
Second,
out(ii,:)=OLS;
This stores the variables for each of the variables (18:60), at just one horizon.
I want to have a matrix which stores all of the estimates OLS, at each of the horizons, for each of my (18:60).
Minimal example
clear;
for i=[1,2,3,4,5];
K=i;
for ii=18:60
x=[1,2,3,i,ii];
out(i,:)=x;
end
end
So the variable out will store 1 2 3 5 60
I want the variable out to store all of the combinations
i.e.
1 2 3 1 1
1 2 3 1 2
...
1 2 3 5 60
Thanks
The simplest solution is to use a 3D matrix:
for jj=[1,2,3,4,5];
K=jj;
for ii=18:60
x=[1,2,3,jj,ii];
out(ii-17,jj,:)=x;
end
end
If you now reshape the out matrix you get the same result as the first block in etmuse's answer:
out = reshape(out,[],size(out,3));
(Note I replaced i by jj. i and ii are too similar to use both, it leads to confusion. It is better to use different letters for loop indices. Also, i is OK to use, but it also is the built-in imaginary number sqrt(-1). So I prefer to use ii over i.)
As you've discovered, using just one of your loop variables to index the output results in most of the results being overwritten, leaving only the results from the final iteration of the relevant loop.
There are 2 ways to create an indexing variable.
1- You can use an independent variable, initialised before the loops and incremented at the end of the internal loop.
kk=1;
for i=1:5
for ii=18:60
%calculate OLC
out(kk,:)=OLC;
kk = kk+1;
end
end
2- Use a calculation of i and ii
kk = i + 5*(ii-18)
(Use in loop as before, without increment)
Related
Parallel Looping.
Hi there, I am having trouble figuring out what kind of loop I should be using for a section of my code.
my issue it that I would like to run the program with multiple initial conditions.for example lets say I have a matrix q where its dimensions are 4 by 2. q(:,1) is the first set of initial conditions and q(:,2) is the second set of initial conditions.
what I want is the recursive for loop (lines 11-22) to take the independent initial conditions q(:,1) and q(:,2) and simultaneously run them through the for loop independently.
Here is some mock code (unfortunately I cannot use the actual code due to an NDA):
1 p=ones(3,2); %initial condition
2 q=2.*ones(4,2); %initial condition
3 r=3.*ones(3,2); %initial condition
4 k=1; %here i want it to iterate for k=1:2 in parallel (i think)
5 pp=p(:,k);
6 qq=q(:,k);
7 rr=r(:,k);
8 p=p(:,k);
9 q=q(:,k);
10 r=r(:,k); %for loop below requires vector form.
11 for t=0:.5:60
12 v=[p;q;r]; %initial conditions for ode45
13 tspan=[t,t+.25,t+.5];
14 [T,Y]=ode45('somefunc',tspan,v,...);
15 v=Y(3,:);
16 p=v(1:3);
17 q=v(4:7);
18 r=v(8:10);
%building up a matrix with all the data from the iterated/recursive ode45
19 pp=cat(2,pp,p);
20 qq=cat(2,qq,q);
21 rr=cat(2,rr,r);
22 end
The code above evaluates for only k=1 or k=2. it does not iterate k=1:2.
what I want is to end up with a multi-dimensional array for pp, qq and rr for the two (at some point n) possible initial conditions. ie pp(:,:,1) would be the matrix with the first initial conditions and pp(:,:,2) would be the second initial conditions. same goes for qq and rr.
This is why I think I need a parallel for loop instead? so that the initial conditions in lines 1 through 3 can be called again rather than the redefined values of my variables due to the for loop in lines 11-22.
I am also confused on how to satisfy parfor's requirement of independent iterations while having a nested for loop with dependent iterations within.
Please let me know if there is anything else you would like me to explain further or clarify.
Thank you for your time and help!
I have a 5-by-200 matrix where the i:50:200, i=1:50 are related to each other, so for example the matrix columns 1,51,101,151 are related to each other, and columns 49,99,149,199 are also related to each other.
I want to use a for-loop to create another matrix that re-sorts the previous matrix based on this relationship.
My code is
values=zeros(5,200);
for j=1:50
for m=1:4:200
a=factor_mat(:,j:50:200)
values(:,m)=a
end
end
However, the code does not work.
Here's what's happening. Let's say we're on the first iteration of the outer loop, so j == 1. This effectively gives you:
j = 1;
for m=1:4:200
a=factor_mat(:,j:50:200)
values(:,m)=a;
end
So you're creating the same submatrix for a (j doesn't change) 50 times and storing it at different places in the values matrix. This isn't really what you want to do.
To create each 4-column submatrix once and store them in 50 different places, you need to use j to tell you which of the 50 you're currently processing:
for j=1:50
a=factor_mat(:,j:50:200);
m=j*4; %// This gives us the **end** of the current range
values(:,m-3:m)=a;
end
I've used a little trick here, because the indices of Matlab arrays start at 1 rather than 0. I've calculated the index of the last column we want to insert. For the first group, this is column 4. Since j == 1, j * 4 == 4. Then I subtract 3 to find the first column index.
That will fix the problem you have with your loops. But loops aren't very Matlab-ish. They used to be very slow; now they're adequate. But they're still not the cool way to do things.
To do this without loops, you can use reshape and permute:
a=reshape(factor_mat,[],50,4);
b=permute(a,[1,3,2]);
values=reshape(b,[],200);
I'm struggling with one of my matlab assignments. I want to create 10 different models. Each of them is based on the same original array of dimensions 1x100 m_est. Then with for loop I am choosing 5 random values from the original model and want to add the same random value to each of them. The cycle repeats 10 times chosing different values each time and adding different random number. Here is a part of my code:
steps=10;
for s=1:steps
for i=1:1:5
rl(s,i)=m_est(randi(numel(m_est)));
rl_nr(s,i)=find(rl(s,i)==m_est);
a=-1;
b=1;
r(s)=(b-a)*rand(1,1)+a;
end
pert_layers(s,:)=rl(s,:)+r(s);
M=repmat(m_est',s,1);
end
for k=steps
for m=1:1:5
M_pert=M;
M_pert(1:k,rl_nr(k,1:m))=pert_layers(1:k,1:m);
end
end
In matrix M I am storing 10 initial models and want to replace the random numbers with indices from rl_nr matrix into those stored in pert_layers matrix. However, the last loop responsible for assigning values from pert_layers to rl_nr indices does not work properly.
Does anyone know how to solve this?
Best regards
Your code uses a lot of loops and in this particular circumstance, it's quite inefficient. It's better if you actually vectorize your code. As such, let me go through your problem description one point at a time and let's code up each part (if applicable):
I want to create 10 different models. Each of them is based on the same original array of dimensions 1x100 m_est.
I'm interpreting this as you having an array m_est of 100 elements, and with this array, you wish to create 10 different "models", where each model is 5 elements sampled from m_est. rl will store these values from m_est while rl_nr will store the indices / locations of where these values originated from. Also, for each model, you wish to add a random value to every element that is part of this model.
Then with for loop I am choosing 5 random values from the original model and want to add the same random value to each of them.
Instead of doing this with a for loop, generate all of your random indices in one go. Since you have 10 steps, and we wish to sample 5 points per step, you have 10*5 = 50 points in total. As such, why don't you use randperm instead? randperm is exactly what you're looking for, and we can use this to generate unique random indices so that we can ultimately use this to sample from m_est. randperm generates a vector from 1 to N but returns a random permutation of these elements. This way, you only get numbers enumerated from 1 to N exactly once and we will ensure no repeats. As such, simply use randperm to generate 50 elements, then reshape this array into a matrix of size 10 x 5, where the number of rows tells you the number of steps you want, while the number of columns is the total number of points per model. Therefore, do something like this:
num_steps = 10;
num_points_model = 5;
ind = randperm(numel(m_est));
ind = ind(1:num_steps*num_points_model);
rl_nr = reshape(ind, num_steps, num_points_model);
rl = m_est(rl_nr);
The first two lines are pretty straight forward. We are just declaring the total number of steps you want to take, as well as the total number of points per model. Next, what we will do is generate a random permutation of length 100, where elements are enumerated from 1 to 100, but they are in random order. You'll notice that this random vector uses only a value within the range of 1 to 100 exactly once. Because you only want to get 50 points in total, simply subset this vector so that we only get the first 50 random indices generated from randperm. These random indices get stored in ind.
Next, we simply reshape ind into a 10 x 5 matrix to get rl_nr. rl_nr will contain those indices that will be used to select those entries from m_est which is of size 10 x 5. Finally, rl will be a matrix of the same size as rl_nr, but it will contain the actual random values sampled from m_est. These random values correspond to those indices generated from rl_nr.
Now, the final step would be to add the same random number to each model. You can certainly use repmat to replicate a random column vector of 10 elements long, and duplicate them 5 times so that we have 5 columns then add this matrix together with rl.... so something like:
a = -1;
b = 1;
r = (b-a)*rand(num_steps, 1) + a;
r = repmat(r, 1, num_points_model);
M_pert = rl + r;
Now M_pert is the final result you want, where we take each model that is stored in rl and add the same random value to each corresponding model in the matrix. However, if I can suggest something more efficient, I would suggest you use bsxfun instead, which does this replication under the hood. Essentially, the above code would be replaced with:
a = -1;
b = 1;
r = (b-a)*rand(num_steps, 1) + a;
M_pert = bsxfun(#plus, rl, r);
Much easier to read, and less code. M_pert will contain your models in each row, with the same random value added to each particular model.
The cycle repeats 10 times chosing different values each time and adding different random number.
Already done in the above steps.
I hope you didn't find it an imposition to completely rewrite your code so that it's more vectorized, but I think this was a great opportunity to show you some of the more advanced functions that MATLAB has to offer, as well as more efficient ways to generate your random values, rather than looping and generating the values one at a time.
Hopefully this will get you started. Good luck!
Could somebody explain the following code snippet? I have no background in computer science or programming and just recently became aware of Matlab. I understand the preallocation part from data=ceil(rand(7,5)*10)... to ...N*(N-1)/2).
I need to understand every aspect of how matlab processes the code from kk=0 to the end. Also, the reasons why the code is codified in that manner. There's no need to explain the function of: bsxfun(#minus), just how it operates in the scheme of the code.
data=ceil(rand(7,5)*10);
N = size(data,2);
b=cell(N-1,1);
c=NaN(size(data,1),N*(N-1)/2);
kk=0;
for ii=1:N-1
b{ii} = bsxfun(#minus,data(:,ii),data(:,ii+1:end));
c(:,kk+(1:N-ii)) = bsxfun(#minus,data(:,ii),data(:,ii+1:end));
kk=kk+N-ii;
end
Start at zero
kk=0;
Loop with ii going from 1 up to N-1 incrementing by 1 every iteration. Type 1:10 in the command line of matlab and you'll see that it outputs 1 2 3 4 5 6 7 8 9 10. Thuis colon operator is a very important operator to understand in matlab.
for ii=1:N-1
b{ii} = ... this just stores a matrix in the next element of the cell vector b. Cell arrays can hold anything in each of their elements, this is necessary as in this case each iteration is creating a matrix with one fewer column than the previous iteration.
data(:,ii) --> just get the iith column of the matrix data (: means get all the rows)
data(:, ii + 1:end) means get a subset of the matrix data consisting of all the rows but only of columns that appear after column ii
bsxfun(#minus, data(:,ii), data(:,ii+1:end)) --> for each column in the matrix data(:, ii+1:end), subtract the single column data(:,ii)
b{ii} = bsxfun(#minus,data(:,ii),data(:,ii+1:end));
%This does the same thing as the line above but instead of storing the resulting matrix of the loop in a separate cell of a cell array, this is appending the original array with the new matrix. Note that the new matrix will have the same number of rows each time but one fewer column, so this appends as new columns.
%c(:,kk + (1:N-ii)) = .... --> So 1:(N-ii) produces the numbers 1 up to the number of columns in the result of this iteration. In matlab, you can index an array using another array. So for example try this in the command line of matlab: a = [0 0 0 0 0]; a([1 3 5]) = 1. The result you should see is a = 1 0 1 0 1. but you can also extend a matrix like this so for example now type a(6) = 2. The result: a = 1 0 1 0 1 2. So by using c(:, 1:N-ii) we are indexing all the rows of c and also the right number of columns (in order). Adding the kk is just offsetting it so that we do not overwrite our previous results.
c(:,kk+(1:N-ii)) = bsxfun(#minus,data(:,ii),data(:,ii+1:end));
Now we just increment kk by the number of new columns we added so that in the next iteration, c is appended at the end.
kk=kk+N-ii;
end;
I suggest that you put a breakpoint in this code and step through it line by line and look at how the variables change in matlab. To do this click on the little dashed line next to k=0; in the mfile, you will see a red dot appear there, and then run the code. The code will only execute as far as the dot, you are now in debug mode. If you hover over a variable in debug mode matlab will show its contents in a tool tip. For a really big variable check it out in the workspace. Now step through the code line by line and use my explanations above to make sure you understand how each line is changing each variable. For more complex lines like b{ii} = bsxfun(#minus,data(:,ii),data(:,ii+1:end)); you should highlight code snippets and ruin these in the command line to see what each part is doing so for example run data(:,ii) to see what that does and then try data(:,ii+1:end)) or even just ii+1:end (well in that case it wont work, replace end with size(data, 2)). Debugging is the best way to understand code that confuses you.
bsxfun(#minus,A,B)
is almost the same as
A-B
The difference is that the bsxfun version will handle inputs of different size: In each dimension (“direction,” if you find it easier to think about that way), if one of the inputs is scalar and the other one a vector, the scalar one will simply be repeated sufficiently often.
http://www.mathworks.com/help/techdoc/ref/bsxfun.html
I'm a newbie to Matlab and just stumped how to do a simple task that can be easily performed in excel. I'm simply trying to get the percent change between cells in a matrix. I would like to create a for loop for this task. The data is setup in the following format:
DAY1 DAY2 DAY3...DAY 100
SUBJECT RESULTS
I could only perform getting the percent change between two data points. How would I conduct it if across multiple days and multiple subjects? And please provide explanation
Thanks a bunch
FOR EXAMPLE, FOR DAY 1 SUBJECT1(RESULT=1), SUBJECT2(RESULT=4), SUBJECT3(RESULT=5), DAY 2 SUBJECT1(RESULT=2), SUBJECT2(RESULT=8), SUBJECT3(RESULT=10), DAY 3 SUBJECT1(RESULT=1), SUBJECT2(RESULT=4), SUBJECT3(RESULT=5).
I WANT THE PERCENT CHANGE SO OUTPUT WILL BE DAY 2 SUBJECT1(RESULT=100%), SUBJECT2(RESULT=100%), SUBJECT3(RESULT=100%). DAY3 SUBJECT1(RESULT=50%), SUBJECT2(RESULT=50%), SUBJECT3(RESULT=50%)
updated:
Hi thanks for responding guys. sorry for the confusion. zebediah49 is pretty close to what I'm looking for. My data is for example a 10 x 10 double. I merely wanted to get the percentage change from column to column. For example, if I want the percentage change from rows 1 through 10 on all columns (from columns 2:10). I would like the code to function for any matrix dimension (e.g., 1000 x 1000 double) zebediah49 could you explain the code you posted? thanks
updated2:
zebediah49,
(data(1:end,100)- data(1:end,99))./data(1:end,99)
output=[data(:,2:end)-data(:,1:end-1)]./data(:,1:end-1)*100;
Observing the code above, How would I go about modifying it so that column 100 is used as the index against all of the other columns(1-99)? If I change the code to the following:
(data(1:end,100)- data(1:end,:))./data(1:end,:)
matlab is unable because of exceeding matrix dimensions. How would I go about implementing that?
UPDATE 3
zebediah49,
Worked perfectly!!! Originally I created a new variable for the index and repmat the index to match the matrices which was not a good idea. It took forever to replicate when dealing with large numbers.
Thanks for you contribution once again.
Thanks Chris for your contribution too!!! I was looking more on how to address and manipulate arrays within a matrix.
It's matlab; you don't actually want a loop.
output=input(2:end,:)./input(1:end-1,:)*100;
will probably do roughly what you want. Since you didn't give anything about your matlab structure, you may have to change index order, etc. in order to make it work.
If it's not obvious, that line defines output as a matrix consisting of the input matrix, divided by the input matrix shifted right by one element. The ./ operator is important, because it means that you will divide each element by its corresponding one, as opposed to doing matrix division.
EDIT: further explanation was requested:
I assumed you wanted % change of the form 1->1->2->3->1 to be 100%, 200%, 150%, 33%.
The other form can be obtained by subtracting 100%.
input(2:end,:) will grab a sub-matrix, where the first row is cut off. (I put the time along the first dimension... if you want it the other way it would be input(:,2:end).
Matlab is 1-indexed, and lets you use the special value end to refer to the las element.
Thus, end-1 is the second-last.
The point here is that element (i) of this matrix is element (i+1) of the original.
input(1:end-1,:), like the above, will also grab a sub-matrix, except that that it's missing the last column.
I then divide element (i) by element (i+1). Because of how I picked out the sub-matrices, they now line up.
As a semi-graphical demonstration, using my above numbers:
input: [1 1 2 3 1]
input(2,end): [1 2 3 1]
input(1,end-1): [1 1 2 3]
When I do the division, it's first/first, second/second, etc.
input(2:end,:)./input(1:end-1,:):
[1 2 3 1 ]
./ [1 1 2 3 ]
---------------------
== [1.0 2.0 1.5 0.3]
The extra index set to (:) means that it will do that procedure across all of the other dimension.
EDIT2: Revised question: How do I exclude a row, and keep it as an index.
You say you tried something to the effect of (data(1:end,100)- data(1:end,:))./data(1:end,:). Matlab will not like this, because the element-by-element operators need them to be the same size. If you wanted it to only work on the 100th column, setting the second index to be 100 instead of : would do that.
I would, instead, suggest setting the first to be the index, and the rest to be data.
Thus, the data is processed by cutting off the first:
output=[data(2:end,2:end)-data(2:end,1:end-1)]./data(2:end,1:end-1)*100;
OR, (if you neglect the start, matlab assumes 1; neglect the end and it assumes end, making (:) shorthand for (1:end).
output=[data(2:,2:end)-data(2:,1:end-1)]./data(2:,1:end-1)*100;
However, you will probably still want the indices back, in which case you will need to append that subarray back:
output=[data(1,1:end-1) data(2:,2:end)-data(2:,1:end-1)]./data(2:,1:end-1)*100];
This is probably not how you should be doing it though-- keep data in one matrix, and time or whatever else in a separate array. That makes it much easier to do stuff like this to data, without having to worry about excluding time. It's especially nice when graphing.
Oh, and one more thing:
(data(:,2:end)-data(:,1:end-1))./data(:,1:end-1)*100;
is identically equivalent to
data(:,2:end)./data(:,1:end-1)*100-100;
Assuming zebediah49 guessed right in the comment above and you want
1 4 5
2 8 10
1 4 5
to turn into
1 1 1
-.5 -.5 -.5
then try this:
data = [1,4,5; 2,8,10; 1,4,5];
changes_absolute = diff(data);
changes_absolute./data(1:end-1,:)
ans =
1.0000 1.0000 1.0000
-0.5000 -0.5000 -0.5000
You don't need the intermediate variable, you can directly write diff(data)./data(1:end,:). I just thought the above might be easier to read. Getting from that result to percentage numbers is left as an exercise to the reader. :-)
Oh, and if you really want 50%, not -50%, just use abs around the final line.