How can I align a sequence of three data points?

like I have a data sequence

data_A = [0 4 5 7 8 9], data_B = [4 5 7 8 1 5 6], data_C = [ 2 5 3 4 5 7 8]

I want result in this form..............Kindly help................

red square indicate matched sequence........

Answer 1

[Mathieu NOE]

hello

try this

data_A  = [0 4 5 7 8 9];
data_B = [4  5 7 8 1 5 6];
data_C = [ 2 5 3  4 5 7 8];

% check that all intersects are identical (make the code robust)
out1 = intersect(data_A,data_B);
out2 = intersect(data_B,data_C);
out3 = intersect(data_A,data_C);

tol = 1e-6;

if (sum(out1-out2) + sum(out1-out3) + sum(out2-out3)) < tol % they are all 3 identical
    
    % find max length of all sequences
    n = max([numel(data_A) numel(data_B) numel(data_C)]);
    % strfind(A,B) % will find the starting indices where B is embedded in A.
    s1 = strfind(data_A,out1);
    s2 = strfind(data_B,out1);
    s3 = strfind(data_C,out1);
    
    % how much leading zeros do we need ? 
    m = max([s1 s2 s3]); % 
    lza = m - s1; % is the number of leading zeros to padd to the data_A sequence
    lzb = m - s2; % is the number of leading zeros to padd to the data_A sequence
    lzc = m - s3; % is the number of leading zeros to padd to the data_A sequence
    data_A  = [zeros(1,lza) data_A];
    data_B  = [zeros(1,lzb) data_B];
    data_C  = [zeros(1,lzc) data_C];
    
    % how much trailing zeros do we need ? 
    % find max length of all sequences
    q = max([numel(data_A) numel(data_B) numel(data_C)]);
    tza = q - numel(data_A); % is the number of trailing zeros to padd to the data_A sequence
    tzb = q - numel(data_B); % is the number of trailing zeros to padd to the data_A sequence
    tzc = q - numel(data_C); % is the number of trailing zeros to padd to the data_A sequence
    data_A  = [data_A zeros(1,tza)];
    data_B  = [data_B zeros(1,tzb)];
    data_C  = [data_C zeros(1,tzc)];

    
end

% finally concatenate all 3 sequences 
out = [data_A;data_B;data_C]

out = 3×10

0 0 0 4 5 7 8 9 0 0 0 0 0 4 5 7 8 1 5 6 2 5 3 4 5 7 8 0 0 0

[CM Sahu]

Thank you for your answer.

but this method works for a particular type of data set.

I will br happy if any generic solution with some tolerance value is provided.

data type like..........

data_A = [0 4.1 5 7 8 9 10];

data_B = [4.2 5 7 8 1 5 6 10];

data_C = [2 5 3 4.3 5 7 8 10];

[Mathieu NOE]

hello

this is a new code , jope it helps

data_A  = [0 4.1 5 7 8 9 10];
data_B = [4.2 5 7 8 1 5 6 10];
data_C = [2 5 3 4.3 5 7 8 10];

% round the values to nearest integer 
data_Ar  = round(data_A);
data_Br = round(data_B);
data_Cr = round(data_C);

% check that all intersects are identical (make the code robust)
[out1,IA,IB] = intersect(data_Ar,data_Br);
ci = find(diff(IA) == 1);ci = [ci; ci(end)+1];IA = IA(ci); % we keep only IA values that are distant with delta = 1
out1 = data_Ar(IA);

[out2,IA,IB] = intersect(data_Br,data_Cr);
ci = find(diff(IA) == 1);ci = [ci; ci(end)+1];IA = IA(ci);% we keep only IA values that are distant with delta = 1
out2 = data_Br(IA);

[out3,IA,IB] = intersect(data_Ar,data_Cr);
ci = find(diff(IA) == 1);ci = [ci; ci(end)+1];IA = IA(ci);% we keep only IA values that are distant with delta = 1
out3 = data_Ar(IA);

tol = 1e-6;
if (sum(out1-out2) + sum(out1-out3) + sum(out2-out3)) < tol % they are all 3 identical
    
    % find max length of all sequences
    n = max([numel(data_A) numel(data_B) numel(data_C)]);
    % strfind(A,B) % will find the starting indices where B is embedded in A.
    s1 = strfind(data_Ar,out1);
    s2 = strfind(data_Br,out1);
    s3 = strfind(data_Cr,out1);
    
    % how much leading zeros do we need ? 
    m = max([s1 s2 s3]); % 
    lza = m - s1; % is the number of leading zeros to padd to the data_A sequence
    lzb = m - s2; % is the number of leading zeros to padd to the data_A sequence
    lzc = m - s3; % is the number of leading zeros to padd to the data_A sequence
    data_A  = [zeros(1,lza) data_A];
    data_B  = [zeros(1,lzb) data_B];
    data_C  = [zeros(1,lzc) data_C];
    
    % how much trailing zeros do we need ? 
    % find max length of all sequences
    q = max([numel(data_A) numel(data_B) numel(data_C)]);
    tza = q - numel(data_A); % is the number of trailing zeros to padd to the data_A sequence
    tzb = q - numel(data_B); % is the number of trailing zeros to padd to the data_A sequence
    tzc = q - numel(data_C); % is the number of trailing zeros to padd to the data_A sequence
    data_A  = [data_A zeros(1,tza)];
    data_B  = [data_B zeros(1,tzb)];
    data_C  = [data_C zeros(1,tzc)];
    
end

% finally concatenate all 3 sequences 
out = [data_A;data_B;data_C]

out = 3×11

0 0 0 4.1000 5.0000 7.0000 8.0000 9.0000 10.0000 0 0 0 0 0 4.2000 5.0000 7.0000 8.0000 1.0000 5.0000 6.0000 10.0000 2.0000 5.0000 3.0000 4.3000 5.0000 7.0000 8.0000 10.0000 0 0 0

[CM Sahu]

Thank you

[Mathieu NOE]

my pleasure !

Answer 2

[Dyuman Joshi]

data_A  = [0 4 5 7 8 9];
data_B = [4 5 7 8 1 5 6];
data_C = [2 5 3 4 5 7 8];

Note that this is not the best method to obtain the common sub-sequence, however, this works for this particular data set

y = intersect(intersect(data_A,data_B),data_C)

y = 1×4

4 5 7 8

data = {data_A;data_B;data_C};
%number of elements in each vector
n = cellfun('length', data);
%the starting index of the sub-sequence in each vector
k = cellfun(@(x) strfind(x,y), data);
m = max(k);
nd = numel(data);
%preallocation
out = zeros(nd,max(n+m-k));
for idx = 1:nd
    out(idx,m-k(idx)+(1:n(idx))) = data{idx};
end
out

out = 3×10

0 0 0 4 5 7 8 9 0 0 0 0 0 4 5 7 8 1 5 6 2 5 3 4 5 7 8 0 0 0

[CM Sahu]

Thank you for your answer

but yes this method work only for particular set of data

if data_A = [0 4 5 7 8 9 10];

data_B = [4 5 7 8 1 5 6 10];

data_C = [2 5 3 4 5 7 8 10]; changes

then it will not work because the intersection data 10 is not this [4 5 7 8] sequence.

[Dyuman Joshi]

Yes, the trickiest part of this question is obtaining the common sub-sequence.

I will update my answer, if I am able to find a solution to that.

Answer 3

[Shaik]

Hi Sahu,

Hope it helps.

% define the data sequences
data_A = [0 4 5 7 8 9];
data_B = [4 5 7 8 1 5 6];
data_C = [2 5 3 4 5 7 8];

% find the overlapping elements between data_A, data_B, and data_C
overlap_AB = intersect(data_A, data_B);
overlap_AC = intersect(data_A, data_C);
overlap_BC = intersect(data_B, data_C);
overlap_ABC = intersect(overlap_AB, overlap_AC);
overlap_ABC = intersect(overlap_ABC, overlap_BC);

% find the indices of the overlapping elements in data_A, data_B, and data_C
idx_overlap_A = ismember(data_A, overlap_ABC);
idx_overlap_B = ismember(data_B, overlap_ABC);
idx_overlap_C = ismember(data_C, overlap_ABC);

% plot the data sequences with red squares indicating the overlapping elements
plot(1:length(data_A), data_A, '-o', 'Color', 'b');
hold on;
plot(1:length(data_B), data_B, '-o', 'Color', 'g');
plot(1:length(data_C), data_C, '-o', 'Color', 'r');
scatter(find(idx_overlap_A), data_A(idx_overlap_A), 'MarkerFaceColor', 'r', 'MarkerEdgeColor', 'r');
scatter(find(idx_overlap_B), data_B(idx_overlap_B), 'MarkerFaceColor', 'r', 'MarkerEdgeColor', 'r');
scatter(find(idx_overlap_C), data_C(idx_overlap_C), 'MarkerFaceColor', 'r', 'MarkerEdgeColor', 'r');
hold off;

[CM Sahu]

Thank you for your try