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
Related
I am trying to detect high amplitude events and remove them along with rows above and below. I have the following code which does this in part but not fully and I'm not sure where the error is. I have commented out the the audioread function and added randi to allow reproducible results. Code:
%[data, fs] = audioread("noise.wav");
%t1 = linspace(0, (numel(data)-1)/fs, numel(data));
rng(1)
data = randi(10,1000,1);
threshold = 5;
clear_range = 10; %rows/samples
data = clearRange(data, threshold, clear_range);
%t1 = linspace(0, (numel(data)-1)/fs, numel(data));
%plot(t1, data);
plot(data)
function [data] = clearRange(data, threshold, clear_range, compare_column)
% data: matrix of values to clean
% threshold: value to compare values against
% clear_range: number of rows to delete
% compare_column: column to check for value to compare against threshold
if nargin < 4
compare_column = 1;
end
for i = 1:length(data)
if i > length(data)
break
end
if data(i,compare_column) > threshold
data(max(1, i-clear_range):min(length(data), i+clear_range),:) = [];
end
end
end
I think the main problem with your code is that you modify data while looping over it. This means, you delete peaks (or high amplitude events in your words) in rows with an index greater than i, so that they cannot be taken into account in following iterations.
E.g. consider peaks in rows with indices 4 and 6, which should cause that rows up to index 16 are removed (with a value of clear_range equal to 10). However, when i is equal to 4, you remove rows up to index 14. Consequently, you also remove the peak at position 6, so that it is not taken into account in further iterations.
In general, it is easier to rely on MATLAB's matrix/array operations instead of using loops.
Please find below a possible solution with explanations inline.
clc;
% I adjusted inputs to get a minimal example
data = randi(10,30,1);
threshold = 9;
rangeToClear = 1;
columnToCompare = 1;
dataOut = clearRange(data, threshold, rangeToClear, columnToCompare );
disp('In:')
disp( data' );
disp('Out:')
disp( dataOut' ); % Plot for cross-check
function data = clearRange(data, threshold, rangeToClear, columnToCompare)
% rowsWithPeak is 1-D logical array showing where columnToCompare is greater than the threshold
rowsWithPeak = data( :, columnToCompare ) > threshold;
% kernel is a column vector of ones of size Nx1, where N is the number of rows
% that should be removed around a peak
kernel = ones( 2*rangeToClear+1, 1 );
% rowsToRemove is a column vector being greater than one at row indices
% that should be removed from the data. To obtain rowsToRemove,
% we convolute rowsWithPeak with the kernel. The argument 'same' to
% the conv2 function, specifies that rowsToRemove will have the same
% size as rowsWithPeak.
rowsToRemove = conv2( rowsWithPeak, kernel, 'same' );
% rowsToRemoveLogical is a logical array being one, where rowsToRemove is greater than 0.
% Note that, rowsToRemoveLogical = rowsToRemove > 0 would also work here
rowsToRemoveLogical = logical( rowsToRemove);
% Finally, we use rowsToRemoveLogical to mask positions of the rows that should be removed.
data( rowsToRemoveLogical, : ) = [];
end
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.
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
I have a data, which may be simulated in the following way:
N = 10^6;%10^8;
K = 10^4;%10^6;
subs = randi([1 K],N,1);
M = [randn(N,5) subs];
M(M<-1.2) = nan;
In other words, it is a matrix, where the last row is subscripts.
Now I want to calculate nanmean() for each subscript. Also I want to save number of rows for each subscript. I have a 'dummy' code for this:
uniqueSubs = unique(M(:,6));
avM = nan(numel(uniqueSubs),6);
for iSub = 1:numel(uniqueSubs)
tmpM = M(M(:,6)==uniqueSubs(iSub),1:5);
avM(iSub,:) = [nanmean(tmpM,1) size(tmpM,1)];
end
The problem is, that it is too slow. I want it to work for N = 10^8 and K = 10^6 (see commented part in the definition of these variables.
How can I find the mean of the data in a faster way?
This sounds like a perfect job for findgroups and splitapply.
% Find groups in the final column
G = findgroups(M(:,6));
% function to apply per group
fcn = #(group) [mean(group, 1, 'omitnan'), size(group, 1)];
% Use splitapply to apply fcn to each group in M(:,1:5)
result = splitapply(fcn, M(:, 1:5), G);
% Check
assert(isequaln(result, avM));
M = sortrows(M,6); % sort the data per subscript
IDX = diff(M(:,6)); % find where the subscript changes
tmp = find(IDX);
tmp = [0 ;tmp;size(M,1)]; % add start and end of data
for iSub= 2:numel(tmp)
% Calculate the mean over just a single subscript, store in iSub-1
avM2(iSub-1,:) = [nanmean(M(tmp(iSub-1)+1:tmp(iSub),1:5),1) tmp(iSub)-tmp(iSub-1)];tmp(iSub-1)];
end
This is some 60 times faster than your original code on my computer. The speed-up mainly comes from presorting the data and then finding all locations where the subscript changes. That way you do not have to traverse the full array each time to find the correct subscripts, but rather you only check what's necessary each iteration. You thus calculate the mean over ~100 rows, instead of first having to check in 1,000,000 rows whether each row is needed that iteration or not.
Thus: in the original you check numel(uniqueSubs), 10,000 in this case, whether all N, 1,000,000 here, numbers belong to a certain category, which results in 10^12 checks. The proposed code sorts the rows (sorting is NlogN, thus 6,000,000 here), and then loop once over the full array without additional checks.
For completion, here is the original code, along with my version, and it shows the two are the same:
N = 10^6;%10^8;
K = 10^4;%10^6;
subs = randi([1 K],N,1);
M = [randn(N,5) subs];
M(M<-1.2) = nan;
uniqueSubs = unique(M(:,6));
%% zlon's original code
avM = nan(numel(uniqueSubs),7); % add the subscript for comparison later
tic
uniqueSubs = unique(M(:,6));
for iSub = 1:numel(uniqueSubs)
tmpM = M(M(:,6)==uniqueSubs(iSub),1:5);
avM(iSub,:) = [nanmean(tmpM,1) size(tmpM,1) uniqueSubs(iSub)];
end
toc
%%%%% End of zlon's code
avM = sortrows(avM,7); % Sort for comparison
%% Start of Adriaan's code
avM2 = nan(numel(uniqueSubs),6);
tic
M = sortrows(M,6);
IDX = diff(M(:,6));
tmp = find(IDX);
tmp = [0 ;tmp;size(M,1)];
for iSub = 2:numel(tmp)
avM2(iSub-1,:) = [nanmean(M(tmp(iSub-1)+1:tmp(iSub),1:5),1) tmp(iSub)-tmp(iSub-1)];
end
toc %tic/toc should not be used for accurate timing, this is just for order of magnitude
%%%% End of Adriaan's code
all(avM(:,1:6) == avM2) % Do the comparison
% End of script
% Output
Elapsed time is 58.561347 seconds.
Elapsed time is 0.843124 seconds. % ~70 times faster
ans =
1×6 logical array
1 1 1 1 1 1 % i.e. the matrices are equal to one another
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