how to assemble a symbolic series?

Hello everyone, I would like to implement the logic described in the figure below. It is one of the steps necessary for converting a time series into a symbolic series. The image is an explanation of how the logic works, in this case the partition length is l=12. Any help is welcome, thanks for your attention.

Answer 1

[Star Strider]

Try something like this —

b = randi([0 1], 1, 12)

b = 1×12

0 1 0 0 1 1 0 1 0 1 1 1

for k = 1:7
    sb(k,:) = b(k:k+5);
end
sb

sb = 7×6

0 1 0 0 1 1 1 0 0 1 1 0 0 0 1 1 0 1 0 1 1 0 1 0 1 1 0 1 0 1 1 0 1 0 1 1 0 1 0 1 1 1

bc = mat2cell(sb,ones(7,1),6)'

bc = 1×7 cell array
    {[0 1 0 0 1 1]}    {[1 0 0 1 1 0]}    {[0 0 1 1 0 1]}    {[0 1 1 0 1 0]}    {[1 1 0 1 0 1]}    {[1 0 1 0 1 1]}    {[0 1 0 1 1 1]}

That is as close as I can get to the result in the illustration.

.

[Abuh La Ha NedaL]

Thanks for your help. One question, how can I convert 'bc' array to decimal values? God bless you.

[Star Strider]

My pleasure!

Try this —

b = randi([0 1], 1, 12)

b = 1×12

0 0 1 0 0 0 0 1 0 1 0 1

for k = 1:7
    sb(k,:) = b(k:k+5);
end
sb

sb = 7×6

0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 1

bc = mat2cell(sb,ones(7,1),6)'

bc = 1×7 cell array
    {[0 0 1 0 0 0]}    {[0 1 0 0 0 0]}    {[1 0 0 0 0 1]}    {[0 0 0 0 1 0]}    {[0 0 0 1 0 1]}    {[0 0 1 0 1 0]}    {[0 1 0 1 0 1]}

bcd = cellfun(@(x)bin2dec(sprintf('%d',x)),bc, 'Unif',0)

bcd = 1×7 cell array
    {[8]}    {[16]}    {[33]}    {[2]}    {[5]}    {[10]}    {[21]}

.

[Walter Roberson]

Do you need the bit streams or do you just need the decimal outputs? Because there are ways to calculate the decimal without building the cell.

[Abuh La Ha NedaL]

To get the value of my symbolic series you need to group the binary values ​​with L=6, that is, group the sequence of 0 and 1 in subgroups of 6 according to the logic of the figure, and then convert to decimal. If there's another way to do this, that's cool too. I am very grateful for your help.

[Star Strider]

A relatively straightforward way to do this (without using bin2dec) is something like this:

b = randi([0 1], 1, 12)

b = 1×12

1 0 0 0 0 0 1 1 1 0 1 1

for k = 1:7
    sb(k,:) = b(k:k+5);
end
sb

sb = 7×6

1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 1 1 1 0 0 1 1 1 0 0 1 1 1 0 1 1 1 1 0 1 1

v = 2.^(5:-1:0);                            % Powers-Of-2 Vector
sc = sb*v(:)

sc = 7×1

32 1 3 7 14 29 59

sct = sc.'

sct = 1×7

32 1 3 7 14 29 59

With ‘sct’ being the transpose of ‘sc’.

.

[Abuh La Ha NedaL]

Hello, in a 1xn matrix sequence it works very well for me, but then a question arose if I have a matrix, say 1000x50, how would the logic and the conversion be done? in the code below it seems to me that only one column is being computed, how to compute the 50 columns? very grateful for your help.

r    = 3.99; 
x(1) = 0.225;
N    = 49999;
for ii = 1:N
   x(ii+1) = r*x(ii)*(1 - x(ii)); % logistic map
end

C=median(x);

for i=1:length(x)
    
if x(i)>=C
    x(i)=1;
else
    x(i)=0;
end 
    
end 

B = reshape(x,[1000,50]); 

L=6;
cp= length(B)-(L-1);           

for k=1:cp
    
     sb(k,:) = x(k:k+5);
     
end

[numRows,numCols] = size(sb);
v=2.^(5:-1:0);
sc=sb.*v;

[Star Strider]

That is not my code, and most of it does not appear to me to bear any relation to my code.

r    = 3.99; 
x(1) = 0.225;
N    = 49999;
for ii = 1:N
   x(ii+1) = r*x(ii)*(1 - x(ii)); % logistic map
end

C=median(x);

for i=1:length(x)
    
if x(i)>=C
    x(i)=1;
else
    x(i)=0;
end 
    
end 

B = reshape(x,[1000,50]); 

L=6;
cp= length(B)-(L-1);           

for k=1:cp
    
     sb(k,:) = x(k:k+5);
     
end

[numRows,numCols] = size(sb);
v=2.^(5:-1:0);
sc=sb.*v

sc = 995×6

0 16 8 0 2 0 32 16 0 4 0 0 32 0 8 0 0 0 0 16 0 0 0 1 32 0 0 0 2 1 0 0 0 4 2 1 0 0 8 4 2 0 0 16 8 4 0 1 32 16 8 0 2 0 32 16 0 4 0 1

This appears to me to be an appropriate result.

It would seem that summing the rows would approximate the result of my code, so:

ssc = sum(sc,2)

ssc = 995×1

26 52 40 17 35 7 14 29 58 53

.

[Abuh La Ha NedaL]

Yes you are right. I expressed myself badly. Your code works. I can't expand to an array with two or more columns. The problem is to generate a symbolic series from a time series, for this I need to convert the series into binary 0 and 1. My series (Logistic map) was built with a number of points N=50000. The second step is to slice the signal into 50 parts with each part containing 1000 points, that is, a length l=1000. With a length l=1000 using the logic of the figure, for an L=6 you would be able to create 995 groups with 6 bits, according to the formula l-(L-1), after this step you would convert the 995 groups of 6 bits to decimal, the which would generate me a sequence of 995 integers. The next step would be to count how many times the numbers of the series are repeated and plot a graph, this for a column. In total I will have 50 overlay graphics for the 50 computed columns. The problem may be in the last for loop, as it is not computing all columns. My english doesn't help, sorry.

[Star Strider]

‘The next step would be to count how many times the numbers of the series are repeated and plot a graph, this for a column.’

If I understand correctly what you want to do, the histcounts function would be best for this. The numbers would go from 0 to 63, so there would be 64 bins.

Not repeating all the code, that would go something like this:

ssc = randi([0 63], 995, 1);                        % Integers (0, 63)
[N,edges] = histcounts(ssc, 64)                     % Counts

N = 1×64

14 12 17 23 11 17 16 13 13 15 20 14 20 18 15 12 14 16 8 19 15 13 9 19 22 10 9 13 18 11

edges = 1×65

0 0.9900 1.9800 2.9700 3.9600 4.9500 5.9400 6.9300 7.9200 8.9100 9.9000 10.8900 11.8800 12.8700 13.8600 14.8500 15.8400 16.8300 17.8200 18.8100 19.8000 20.7900 21.7800 22.7700 23.7600 24.7500 25.7400 26.7300 27.7200 28.7100

See if that does what you want. Put this in a loop for each column, and store the ‘N’ results in a matrix.

.

[Abuh La Ha NedaL]

This looks promising, I'll check it out. The heart of the problem at the moment is how to compute each column of the 1000x50 matrix each generating 995 integers. In case you are interested, example 1 of the article in the attached link is what I intend to reproduce.

https://ur.booksc.me/book/64254012/24dd02

[Star Strider]

Thank you.

(I cannot access the the book or the article.)

[Abuh La Ha NedaL]

Star, seems to work, but still no solution on how to compute for the 50 columns.

Artigo : DOI:10.1109/iembs.1998.745176

clear
clc

r    = 3.99; 
x(1) = 0.225;
N    = 49999;
for ii = 1:N
   x(ii+1) = r*x(ii)*(1 - x(ii)); % logistic map
end

H=x;
C=median(x);
%BS = reshape(x,[1000,50]);

for i=1:length(x)
    
if x(i)>=C
    x(i)=1;
else
    x(i)=0;
end 
    
end 

B = reshape(x,[1000,50]); % reshape signal 

L=6;
cp= length(B)-(L-1);      %cp=l-(L-1)

for k=1:cp      
    
     sb(k,:) = B(k:k+5);    % 1 column 
     sb1(k,:) = B(k+1:k+6); % 2 column
     
     % how to make this part automatic, sweeping the 50 columns???
   
end


[numRows,numCols] = size(sb);
[numRows1,numCols1] = size(sb1);
v=2.^(5:-1:0);

sc=sb.*v;
sc1=sb1.*v;

bc = mat2cell(sb,ones(cp,1),L)';
bc1 = mat2cell(sb1,ones(cp,1),L)';
bcd = cellfun(@(x)bin2dec(sprintf('%d',x)),bc, 'Unif',0);
bcd1 = cellfun(@(x)bin2dec(sprintf('%d',x)),bc1, 'Unif',0);
fo = bcd';
fo1 = bcd1';
ty = reshape(fo,[199,5]);
ty1 = reshape(fo1,[199,5]);

T=cell2table(ty);
T1=cell2table(ty1);
A = table2array(T);
A1 = table2array(T1);
A=A';
A1=A1';

%M=mode(A)
[M,F] = mode(A);
[M1,F1] = mode(A1);

%edges = unique(A);
%edges1 = unique(A1);
%counts = histc(A(:), edges);
%counts1 = histc(A1(:), edges1);

[N, arestas] = histcounts(A, 64);
[N1, arestas1] = histcounts(A1, 64);

figure;
plot(N)
hold on
plot(N1)
hold off
xlabel ('N')
ylabel ('F')

[Star Strider]

To work with it, use a loop —

r    = 3.99; 
x(1) = 0.225;
N    = 49999;
for ii = 1:N
   x(ii+1) = r*x(ii)*(1 - x(ii)); % logistic map
end

H=x;
C=median(x);
%BS = reshape(x,[1000,50]);

for i=1:length(x)
    
if x(i)>=C
    x(i)=1;
else
    x(i)=0;
end 
    
end 

B = reshape(x,[1000,50]); % reshape signal 

L=6;
cp= length(B)-(L-1);      %cp=l-(L-1)

for k=1:cp      
    
     sb(k,:) = B(k:k+5);    % 1 column 
     sb1(k,:) = B(k+1:k+6); % 2 column
     
     % how to make this part automatic, sweeping the 50 columns???
   
end


[numRows,numCols] = size(sb);
[numRows1,numCols1] = size(sb1);
v=2.^(5:-1:0);

sc=sb.*v;
sc1=sb1.*v;

bc = mat2cell(sb,ones(cp,1),L)';
bc1 = mat2cell(sb1,ones(cp,1),L)';
bcd = cellfun(@(x)bin2dec(sprintf('%d',x)),bc, 'Unif',0);
bcd1 = cellfun(@(x)bin2dec(sprintf('%d',x)),bc1, 'Unif',0);
fo = bcd';
fo1 = bcd1';
ty = reshape(fo,[199,5]);
ty1 = reshape(fo1,[199,5]);

T=cell2table(ty);
T1=cell2table(ty1);
A = table2array(T);
A1 = table2array(T1);
A=A';
A1=A1';

%M=mode(A)
[M,F] = mode(A);
[M1,F1] = mode(A1);

%edges = unique(A);
%edges1 = unique(A1);
%counts = histc(A(:), edges);
%counts1 = histc(A1(:), edges1);

nc = size(A,2);                             % Controls Iteration Count
Nv = zeros(nc,64);                          % Preallocate
N1v = zeros(nc,64);                         % Preallocate
arestas = zeros(nc,65);                     % Preallocate
arestas1 = zeros(nc,65);                    % Preallocate
for k = 1:nc
    [Nv(k,:), arestas(k,:)] = histcounts(A(:,k), 64);
    [N1v(k,:), arestas1(k,:)] = histcounts(A1(:,k), 64);
end

nc = 5                                      % Show First 5 (Delete This Later To Show All)

nc = 5

figure
for k = 1:nc
    subplot(nc,1,k)
    plot(Nv(k,:))
    hold on
    plot(N1v(k,:))
    hold off
    grid
    xlabel ('N')
    ylabel ('F')
    title(sprintf('Column # %2d',k))
end

It is of course possible to plot all in a single figure, however this option offers the ability to see each individually.

.

[Abuh La Ha NedaL]

the end result should look like this graph. I will analyze the code, thanks.

[Star Strider]

O.K. Then forget the subplot approach and plot everything together.

I have no idea how this is supposed to work. I am just doing what you indicate would be best, so experiment with looping through the rows of ‘A’ and ‘A1’ instead of the columns, since even when plotted together, the combined plot does not look like the posted image. .

[Abuh La Ha NedaL]

more clearly, all the process done so far is being executed for only one column, I would like to execute it for all columns. It's a bottleneck that I'm not able to think of a possible way out. Maybe the second loop is the point, but I don't know how to make the task automatic. Thanks.

[Star Strider]

I am not certain that I understand the problem.

My code loops over all the columns in the ‘A’ and ‘A1’ matrices and calculates the histograms. That is the best I can do.

[Abuh La Ha NedaL]

yes, but the A and A1 was generated manually, A would be the first column A1 the second column, but doing A until A49 manually is unfeasible. Take a look at the third loop, you will understand. Did you get access to the article? try looking for the DOI number.

[Abuh La Ha NedaL]

I have attached the article to the message above.

[Walter Roberson]

Who imposed the requirement that each column must be stored in a separate numbered variable? If that requirement were not present then the task would not be difficult.

[Star Strider]

@Abuh La Ha NedaL — I agree with Walter.

The ‘A’ and ‘A1’ variables are matrices, not vectors, so I am now lost. Are they to be summed each over the rows (dimension 2) to create column vectors?

To create more of them, loop through the code that created each one of them here, do the appropriate summation to generate a column vector for each one, and then store it appropriately in a matrix. Then do the histogram on each column.

For the record, I do not understand what the paper does. It also seems to be fairly far removed from the subject of the initial post.

[Abuh La Ha NedaL]

1- Step - generate the logistic map;

2- Step -convert to a symbolic series;

3- Step - subdivide the symbolic series into 50 parts. Each part will have l=1000 points;

4- for 'each column', Use an L-6, (means group the values ​​in group of 6, following the logic of the image of the first post;

5- According to the article, using the formula l-(L-1) it is possible to find the number of possible combinations. For this case 995 combinations and consequently 995 integers, formed after conversion to decimal base;

6- after converting to decimal base, it is necessary to count the number of times each partition number is repeated. These values ​​must be stored in order to generate a graph;

7- steps 5 and 6 must be performed for all 50 columns at the end of the process I will have 50 overlaid graphs generating the symbol spectrum graph.

with the step by step, I hope I have helped to better understand the problem. Thank you for your help.

[Star Strider]

I will let you do that.

I have done all I can.

(I will delete my answer in the next day or so.)

[Abuh La Ha NedaL]

I really appreciate your help Star, it really is a more complex problem. The part of the initial logic you solved well.

[Star Strider]

My pleasure!

Answer 2

[Walter Roberson]

In the case where what you care about is the decimal output, and not how it is arrived at, then:

b = [0 1 0 1 1 0 0 0 0 1 1 1 0]

b = 1×13

0 1 0 1 1 0 0 0 0 1 1 1 0

results = conv(b, 2.^(0:5), 'valid')

results = 1×8

22 44 24 48 33 3 7 14

crosscheck = [0b010110, 0b101100, 0b011000]

crosscheck = 1×3

22 44 24