I have a following code
x=randi([1 150],1,400);
d=max(x);
m=numel(x);
y=zeros(d,1);
p=zeros(d,d);
for k=1:m-1
y(x(k))=y(x(k))+1;
p(x(k),x(k+1))=p(x(k),x(k+1))+1;
end
p=bsxfun(#rdivide,p,y);
p(isnan(p)) = 0;
j=prod(p(p~=0));
[~,~,idx] = unique(x);
q=prod(hist(idx,1:max(idx))/numel(x));
s=log(j);
l=log(q);
g=s+l
In above code random no. generated from 1 to 150 with sequence length 400. finally I am getting output g,suppose random no. sequence generated by code is
x=[5,96,96,55,55,65,65,65,12,12,6,7,9,9,9,31,31,14,26,26,26,101,101,145,145,...]
so code is given output for whole sequence, but i want output for sliding window of size 10, i.e. o/p for sequence
[5,96,96,55,55,65,65,65,12,12],
then skipping 1st element and adding next element so window size should remain same, i.e.
[96,96,55,55,65,65,65,12,12,6],
then next for
[96,55,55,65,65,65,12,12,6,7],
likewise up to last, how to do it.
One easy way, using for loop, is below code.
x=randi([1 150],1,12)
m=numel(x);
N=10; %Window length
for n=0:m-N;
y=x(1+n:N+n)
end
Full code will be below.
xr=randi([1 150],1,20)
z=numel(xr);
N=10; %Window length
gAll=zeros(1,z-N+1);
for n=0:z-N;
x=xr(1+n:N+n)
d=max(x);
m=numel(x);
y=zeros(d,1);
p=zeros(d,d);
for k=1:m-1
y(x(k))=y(x(k))+1;
p(x(k),x(k+1))=p(x(k),x(k+1))+1;
end
p=bsxfun(#rdivide,p,y);
p(isnan(p)) = 0;
j=prod(p(p~=0));
[~,~,idx] = unique(x);
q=prod(hist(idx,1:max(idx))/numel(x));
s=log(j);
l=log(q);
g=s+l
gAll(n+1)=g;
end
plot(gAll)
Related
I have created this code to generate a 1 set of lottery numbers, but I am trying to make it so that the user can enter how many sets they want (input n), and it will print out as one long matrix of size nX6? I was messing around with a few options from online suggestions, but to no avail. I put the initial for i=1:1:n at the beginning, but I do not know how to store each run into a growing matrix. Right now it still generates just 1 set.
function lottery(n)
for i=1:1:n
xlow=1;
xhigh=69;
m=5;
i=1;
while (i<=m)
lottonum(i)=floor(xlow+rand*(xhigh-xlow+1));
flag=0;
for j=1:i-1
if (lottonum(i)==lottonum(j))
flag=1;
end
end
if flag==0
i=i+1;
end
end
ylow=1;
yhigh=26;
m=1;
lottonum1=floor(ylow+rand*(yhigh-ylow+1));
z = horzcat(lottonum, lottonum1);
end
disp('The lotto numbers picked are')
fprintf('%g ',z)
disp (' ')
The problem is that you are not storing or displaying the newly generated numbers, only the last set. To solve this, initialize z with NaNs or zeros, and later index z to store each set in a row of z, by using z(i,:) = lottonum.
However, you are using i as iterator in the while loop already, so you should use another variable, e.g. k.
You can also set z as an output of the function, so you can use this matrix in some other part of a program.
function z = lottery(n)
% init z
z = NaN(n,6);
for k = 1:n
xlow=1;
xhigh=69;
m=5;
i=1;
while (i<=m)
lottonum(i)=floor(xlow+rand*(xhigh-xlow+1));
flag=0;
for j=1:i-1
if (lottonum(i)==lottonum(j))
flag=1;
end
end
if flag==0
i=i+1;
end
end
ylow=1;
yhigh=26;
lottonum1 = floor(ylow+rand*(yhigh-ylow+1));
z(k,:) = horzcat(lottonum, lottonum1); % put the numbers in a row of z
end
disp('The lotto numbers picked are')
disp(z) % prettier display than fprintf in this case.
disp (' ')
end
The nice answer from rinkert corrected your basic mistakes (like trying to modify your loop iterator i from within the loop => does not work), and answered your question on how to store all your results.
This left you with a working code, however, I'd like to propose to you a different way to look at it.
The porposed architecture is to divide the tasks into separate functions:
One function draw_numbers which can draw N numbers randomly (and does only that)
One function draw_lottery which call the previous function as many times as it needs (your n), collect the results and display them.
draw_lottery
This architecture has the benefit to greatly simplify your main function. It can now be as simple as:
function Draws = draw_lottery(n)
% define your draw parameters
xmin = 1 ; % minimum number drawn
xmax = 69 ; % maximum number drawn
nballs = 5 ; % number of number to draw
% pre allocate results
Draws = zeros( n , nballs) ;
for iDraw=1:1:n
% draw "nballs" numbers
thisDraw = draw_numbers(xmin,xmax,nballs) ;
% add them to the result matrix
Draws(iDraw,:) = thisDraw ;
end
disp('The lotto numbers picked are:')
disp (Draws)
disp (' ')
end
draw_numbers
Instead of using a intricated set of if conditions and several iterators (i/m/k) to branch the program flow, I made the function recursive. It means the function may have to call itself a number of time until a condition is satisfied. In our case the condition is to have a set of nballs unique numbers.
The function:
(1) draws N integer numbers randomly, using randi.
(2) remove duplicate numbers (if any). Using unique.
(3) count how many unique numbers are left Nu
(4a) if Nu = N => exit function
(4b) if Nu < N => Call itself again, sending the existing Nu numbers and asking to draw an additional N-Nu numbers to add to the collection. Then back to step (2).
in code, it looks like that:
function draw = draw_numbers(xmin,xmax,nballs,drawn_set)
% check if we received a partial set
if nargin == 4
% if yes, adjust the number of balls to draw
n2draw = nballs - numel(drawn_set) ;
else
% if not, make a full draw
drawn_set = [] ;
n2draw = nballs ;
end
% draw "nballs" numbers between "xmin" and "xmax"
% and concatenate these new numbers with the partial set
d = [drawn_set , randi([xmin xmax],1,n2draw)] ;
% Remove duplicate
drawn_set = unique(d) ;
% check if we have some more balls to draw
if numel(drawn_set) < nballs
% draw some more balls
draw = draw_numbers(xmin,xmax,nballs,drawn_set) ;
else
% we're good to go, assign output and exit funtion
draw = drawn_set ;
end
end
You can have both functions into the same file if you want.
I encourage you to look at the documentation of a couple of Matlab built-in functions used:
randi
unique
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 write a function transform(A) which when given a matrix A returns a new matrix. The new matrix should be obtained according to following:
if A has more than one row then interchange the first and second row. After this square the elements in the first row.
So far thats what I have written:
function[Anew] = transform(A)
dimension = size(A);
if dimension(1) > 1 %if there is more than 1 row
A([1 2],:)=A([2 1],:);
end
A(1,:,:) = A(1,:,:).^2 %squares elements in the first row
end
I tested my function by invoking it in matlab.
I noticed that because i dont have a semi colon next to A(1,:,:) = A(1,:,:).^2
I still obtain the desired result but not as output of the function.
I obtain A =
instead of Anew =
If i put a semi colon next to A(1,:,:) = A(1,:,:).^2; then I dont get an output at all.
Could you tell me what is wrong and what should I change in my program to obtain output as Anew?
Thank you
To return a value from a function in Matlab, you must directly assign to it.
In your case, you need to assign to Anew at the end of the operation (you could also technically just use that variable all-together).
function [Output1, Output2] = SomeFunction( ... )
% Do Some Work
% Store Output
Output1 = Result1(:,2) %some output
Output2 = Result2(3:end, :) %some other result
end
function[Anew] = transform(A)
dimension = size(A);
Anew = A;
if dimension(1) > 1 %if there is more than 1 row
Anew ([1 2],:)=Anew([2 1],:);
end
Anew(1,:,:) = Anew(1,:,:).^2; %squares elements in the first row
end
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);
Hi have this code and I don't know how to put the output result with every pixel.I think the output code are not well defined.
EDIT:
I'm going to try to explain the code:
% I have an image
imagen1=imread('recor.tif');
imagen2= double(imagen1);
band1= imagen2(:,:,1);
% I preallocate the result (the image size is 64*89*6)
yvan2= zeros(61,89,1);
% For every pixel of the image, I want to get one result (each one is different).
for i = 1 : nfiles
for j = 1 : nrows
for i = 1:numel(band1)
% I'm doing this because I've to multiply the results of this interpolation with that result a2ldb1y= ldcm_1(:,1). This vector has a length of 2151x1 and I need to muliply the result of the interpolation for (101:267) position on the vector, this is the reason because I'm doing the interpolation since 101 to 267 (also because I don't have that values).
interplan= interp1(van1,man2,t2,'spline');
ma(96) = banda1a(i); % I said 96, because I want to do an interpollation
end
van1= [101 96 266]';
mos1= ma(134);
van2= [0 pos1 0];
t= 101:267;
t2= t';
xi= 101:1:267;
xi2=xi';
interplan= interp1(van1,van2,t2,'spline');
% After this, I 'prepare' the vector.
out=zeros(2151,1)
out(101:267) = interplan;
% And then, I do all this operation (are important for the result)
a2ldb1y= ldcm_1(:,1);
a2ldsum_pesos1= sum(a2ldb1y);
a2l7dout1_a= a2ldb1y.*out;
a2l7dout1_b= a2l7dout1_a./a2ldsum_pesos1;
a2l7dout1_c= sum(a2l7dout1_b);
% And the result a2l7dout1_c I want it for every pixel (the results are different because every pixel has a different value...)
**yvan2(:,:,1)= [a2l7dout1_c];**
end
end
Thanks in advance,
I'm shooting in the dark here, but I think you're looking for:
yvan2(i, j, 1)= a2l7dout1_c;
instead of:
yvan2(:,:,1)= [a2l7dout1_c];
and thus your output should be stored in the variable yvan2 after the loops are done.
P.S
Some issues in your code:
Why do you have two loops using the same iteration variable i? Your calculations are probably incorrect since i is being modified by two for loops.
Why do you even need the second loop? Each iteration overruns the value of ma(134) set by the previous iteration. You can just replace the entire loop with:
ma(134) = banda1a(numel(band1))
You shouldn't be using the names i and j for loop variables. They are already reserved for the imaginary unit (that is, sqrt(-1)), so MATLAB needs extra processing time for name resolution. You'd rather use other loop variable names instead, even ii and jj.