Okay so this sounds easy but no matter how many times I have tried I still cannot get it to plot correctly. I need only 3 lines on the same graph however still have an issue with it.
iO = 2.0e-6;
k = 1.38e-23;
q = 1.602e-19;
for temp_f = [75 100 125]
T = ((5/9)*temp_f-32)+273.15;
vd = -1.0:0.01:0.6;
Id = iO*(exp((q*vd)/(k*T))-1);
plot(vd,Id,'r',vd,Id,'y',vd,Id,'g');
legend('amps at 75 F', 'amps at 100 F','amps at 125 F');
end;
ylabel('Amps');
xlabel('Volts');
title('Current through diode');
Now I know the plot function that is currently in their isn't working and that some kind of variable needs setup like (vd,Id1,'r',vd,Id2,'y',vd,Id3,'g'); however I really can't grasp the concept of changing it and am seeking help.
You can use the "hold on" function to make it so each plot command plots on the same window as the last.
It would be better to skip the for loop and just do this all in one step though.
iO = 2.0e-6;
k = 1.38e-23;
q = 1.602e-19;
temp_f = [75 100 125];
T = ((5/9)*temp_f-32)+273.15;
vd = -1.0:0.01:0.6;
% Convert this 1xlength(vd) vector to a 3xlength(vd) vector by copying it down two rows.
vd = repmat(vd,3,1);
% Convert this 1x3 array to a 3x1 array.
T=T';
% and then copy it accross to length(vd) so each row is all the same value from the original T
T=repmat(T,1,length(vd));
%Now we can calculate Id all at once.
Id = iO*(exp((q*vd)./(k*T))-1);
%Then plot each row of the Id matrix as a seperate line. Id(1,:) means 1st row, all columns.
plot(vd,Id(1,:),'r',vd,Id(2,:),'y',vd,Id(3,:),'g');
ylabel('Amps');
xlabel('Volts');
title('Current through diode');
And that should get what you want.
Related
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.
total_Route = zeros(4,4);
tmp = evalin('base', 't'); % i initialise t in the Workspace with the value 1
if(tmp==5)
tmp=1;
end
total_Route(tmp,1) = Distance_Traveled_CM;
total_Route(tmp,2) = Hauptantrieb_Verbrauchte_Energie_CM;
total_Route(tmp,3) = Nebenaggregate_Verbrauch_Real_CM;
total_Route(tmp,4) = t;
Total_Distance_Traveled_CM = sum(total_Route(:,1));
set(handles.edit3, 'string',Total_Distance_Traveled_CM);
Total_Hauptantrieb_Verbrauchte_Energie_CM = sum(total_Route(:,2));
set(handles.edit4, 'string',Total_Hauptantrieb_Verbrauchte_Energie_CM);
Total_Nebenaggregate_Verbrauch_Real_CM = sum(total_Route(:,3));
set(handles.edit5, 'string',Total_Nebenaggregate_Verbrauch_Real_CM);
%% Index
set(handles.edit15, 'string',tmp);
assignin('base', 't', tmp + 1); % with this line i can increment "t" after each pass
guidata(hObject,handles);
Sorry that I did not explain my problem well.
#Sardar_Usama I want to run the loop only once but t should be incremented after each time I click on my Button.
# Sembei Norimaki end is at the end of my codes, have forgotten to write it in my question
#Patrik & #Dennis Jaheruddin let me explain my problem again
I created a Matrix with 4×4 Elements with the Goal to save the results of each my Variable (Total_Distance_Traveled_CM, Total_Hauptantrieb_Verbrauchte_Energie_CM etc...) after each Simulation in the element of my Matrix (See image below).
I want by pressing a button (on my GUI) to get always the sum of each Column.
Example
The first pass: t = 1--> Distance_Traveled(1,1) is 900 the GUI will take through clicking on the Button, the sum of the first column (which is 900+0+0+0) and write it in a static test.
The second pass t = 2--> Distance_traveled(2,1) is 800 the GUI will take the sum of the first column (which is 900+800+0+0) and write it in a static test and the same thing should happen with the other column.
This should continue until t = 4 i.e. until it does the same thing for each column, then it should reset.
I hope, I have explained my problem better this time and I apologize for my bad English.
I appreciate any help.
Based on your code fragment the for loop is only called once.
However, the contents of the for loop are ran for four times. (first for i=1 then for 1=2 etc..)
If you only want to run one of these options the solution is very simple:
i = 1
yourLoopContent
If i is always 0 the first time, and you always want to run it for the current i, it would also be simple:
yourLoopContent
i = i+1;
However if i may not be set properly the first time, things get messy. This is because i is by default defined as the square root of minus 1.
Therefore I would recommend you to use a different letter like t instead. Then you could do this:
if ~exists(t)
t=0;
end
yourLoopContent %Everywhere using t instead of i
t = t+1;
In general you may want to avoid i as an index to stay clear of complex number issues.
I'm not sure if I got your question correctly, but it seems to me that what you look for is a cumulative sum. This can be done either buy summing on 1:t or by using cumsum. I'm not sure why you use a loop, but if this is only for the summing then cumsum can replace that.
Here is some example in your code:
total_Route = zeros(4,4);
% I commented below what is not part of the question
for t = 1:4
total_Route(t,:) = [Distance_Traveled_CM,
Hauptantrieb_Verbrauchte_Energie_CM,
Nebenaggregate_Verbrauch_Real_CM,
t];
% the following line compute the comulative sum from the top of each
% column to every element in it, so cs_total_Route(3,2) is like
% sum(total_Route(1:3,2)):
cs_total_Route = cumsum(total_Route);
Total_Distance_Traveled_CM = cs_total_Route(t,1); % OR sum(total_Route(1:t,1))
% set(handles.edit3, 'string',Total_Distance_Traveled_CM);
Total_Hauptantrieb_Verbrauchte_Energie_CM = cs_total_Route(t,2); % OR sum(total_Route(1:t,2))
% set(handles.edit4, 'string',Total_Hauptantrieb_Verbrauchte_Energie_CM);
Total_Nebenaggregate_Verbrauch_Real_CM = cs_total_Route(t,3); % OR sum(total_Route(1:t,3))
% set(handles.edit5, 'string',Total_Nebenaggregate_Verbrauch_Real_CM);
% set(handles.edit15, 'string',t);
end
And here is a quick look on what cumsum does (with some random numbers for total_Route):
total_Route =
671 4.6012 1.0662 1
840 3.6475 0.58918 2
354 8.6056 2.1313 3
893 4.1362 2.0118 4
cs_total_Route =
671 4.6012 1.0662 1
1511 8.2487 1.6554 3
1865 16.854 3.7867 6
2758 20.991 5.7985 10
Is this what you looked for?
I am trying to find the max value and its location. Following is the example of the programme,
fname = dir('*.mat');
nfiles = length(fname);
vals = cell(nfiles,1);
phen = cell(nfiles,1);
for i = 1:nfiles
vals{i} = load(fname(i).name);
phen{i} = (vals{i}.phen);
[M, position] = max(phen{i},[],3);
clear vals
end
After the program is executed, all the position is showing 1. There are total 15 files and M is taking the values of the last file.
How to overcome this prpoblem? Any help will be appreciated
I am not sure I understand your question.
However, at every iteration you are computing the max value and position and overwriting them in the next iteration (i.e. not storing them anywhere). So at the end of the loop M and position would correspond to the last entry phen{nfiles}.
Each time you run through your for loop, you are overwriting M with the max from the most recently loaded phen from the dimension of 3. Since your data is only two dimensional, you probably should be using a dimension of 1 or 2 instead of 3. Because you are using 3, max is returning 1 to position. Fix the dimension issue and position should then be the correct value.
What you could do is make M and position the size of nfiles. So instead of
[M, position] = max(phen{i},[],3);
do
%create M and positions arrays here
%ex. M(nfiles) = 0; or a smaller value if your values are negative
%do the same for positions
[M(i), positions(i)] = max(phen{i},[],1); %1 or 2 correction here here!
then after your for loop
...
end
[maxM, maxMposition] = max(M);
position = positions(maxMposition);
I'm looking for a way to update certain elements in a vector [nx113] for every full rotation of my system.
%% # Iterate through timesteps
for tt = 1:nTimeSteps
% # Initialise ink on transfer roller
rollers(2).ink = [zeros(1,98),ones(1,5),zeros(1,113)];
% # Rotate all rollers
for ii = 1:N
rollers(ii).ink(:) = ...
circshift(rollers(ii).ink(:),rollers(ii).rotDirection);
end
% # Update all roller-connections
for ii = 1:N
for jj = 1:nBins(ii)
if(rollers(ii).connections(jj) ~= 0)
index1 = rollers(ii).connections(jj);
index2 = find(ii == rollers(index1).connections);
ink1 = rollers(ii).ink(jj);
ink2 = rollers(index1).ink(index2);
rollers(ii).ink(jj) = (ink1+ink2)/2;
rollers(index1).ink(index2) = (ink1+ink2)/2;
end
end
end
% # Calculate average amount of ink on each roller
for ii = 1:N
averageAmountOfInk(tt,ii) = mean(rollers(ii).ink);
end
rollers(18).TakeOff = averageAmountOfInk*0.6;
end
the vector rollers(2).ink is the vector i'd like to update. currently the vector is populated only once so i have ones from row 98:103. I would like this range of elements to be populated for each 'rotation' of my system not just the first time.
The reason - I'm trying to show ink being added intermittently from only a small section of the roller surface, hence the need for only five cells to be populated.
i thought that if i iterated from 1 to the number of timesteps, in steps size nBins-Max in the loop:
for tt = 1:nBins_max:nTimeSteps
this doesn't seem to be what i'm after.
I'm also hoping to remove ink from the system at the end. for every revolution i would like to be able to remove a percentage of ink on each rotation so it does not stay in the system (as if it was being printed onto a sheet and taken away).
Hopefully someone can understand this and perhaps offer some advice on how to proceed on either or both of my issues.
Your explanation doesn't quite match your code (or vice-versa if you prefer) so I'm not entirely sure what you want to do, but the following may help you towards a solution or towards expressing your problem more clearly.
The vector rollers(2).ink has 1 row and 216 columns, so an operation such as rollers(2).ink(98:103) = something is not updating rows 98 through to 103. Note also that element 98 of that vector is initialised to 0, it's not included in the elements which are initialised to 1.
You write that you want to update a range of the elements in that vector, then write a loop statement for tt = 1:nBins_max:nTimeSteps which strides over a vector of time steps. Surely you want to write something like rollers(2).ink(99:103) = new_values.
As for removing ink from the rollers at every rotation, you could just execute a line such as rollers(2).ink = rollers(2).ink * 0.975 every rotation; obviously you'll want to replace the removal rate of 2.5% every rotation that I have chosen with whatever is right for your simulation.
What is the best way to do random sample with replacement from dataset? I am using 316 * 34 as my dataset. I want to segment the data into three buckets but with replacement. Should I use the randperm because I need to make sure I keep the index intact where that index would be handy in identifying the label data. I am new to matlab I saw there are couple of random sample methods but they didn't look like its doing what I am looking for, its strange to think that something like doesn't exist in matlab, but I did the follwoing:
My issue is when I do this row_idx = round(rand(1)*316) sometimes I get zero, that leads to two questions
what should I do to avoid zeor?
What is the best way to do the random sample with replacement.
shuffle_X = X(randperm(size(X,1)),:);
lengthOf_shuffle_X = length(shuffle_X)
number_of_rows_per_bucket = round(lengthOf_shuffle_X / 3)
bucket_cell = cell(3,1)
bag_matrix = []
for k = 1:length(bucket_cell)
for i = 1:number_of_rows_per_bucket
row_idx = round(rand(1)*316)
bag_matrix(i,:) = shuffle_X(row_idx,:)
end
bucket_cell{k} = bag_matrix
end
I could do following:
if row_idx == 0
row_idx = round(rand(1)*316)
assuming random number will never give two zeros values in two consecutive rounds.
randi is a good way to get integer indices for sampling with replacement. Assuming you want to fill three buckets with an equal number of samples, then you can write
data = rand(316,34); %# create some dummy data
number_of_data = size(data,1);
number_of_rows_per_bucket = 50;
bucket_cell = cell(1,3);
idx = randi([1,number_of_data],[number_of_rows_per_bucket,3]);
for iBucket = 1:3
bucket_cell{iBucket} = data(idx(:,iBucket),:);
end
To the question: if you use randperm it will give you a draw order without replacement, since you can draw any item once.
If you use randi it draws you with replacement, that is you draw an item possibly many times.
If you want to "segment" a dataset, that usually means you split the dataset into three distinct sets. For that you use draw without replacement (you don't put the items back; use randperm). If you'd do it with replacement (using randi), it will be incredibly slow, since after some time the chance that you draw an item which you have not before is very low.
(Details in coupon collector ).
If you need a segmentation that is a split, you can just go over the elements and independently decide where to put it. (That is you choose a bucket for each item with replacement -- that is you put any chosen bucket back into the game.)
For that:
% if your data items are vectors say data = [1 1; 2 2; 3 3; 4 4]
num_data = length(data);
bucket_labels = randi(3,[1,num_data]); % draw a bucket label for each item, independently.
for i=1:3
bucket{i} = data(bucket_labels==i,:);
end
%if your data items are scalars say data = [1 2 3 4 5]
num_data = length(data);
bucket_labels = randi(3,[1,num_data]);
for i=1:3
bucket{i} = data(bucket_labels==i);
end
there we go.