Extract sequences from vector

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Peter
Peter 2013-2-27
回答： Saeid Abrari 2020-10-6
Dear Forum members,
I have a question I could not find any solution for. I have a vector, let's say
U=[2,2,5,7,3,4,4,4,2,2,2,3].
I also have a sequence, which is
s=[2,2,3].
I now want to find all the indexes in U, where my sequence s could be built from. In my case, that would be
(1,2,5), (9,10,12), (10,11,12), (9,11,12). So the problem is the sequences have to be build first from U and do not exist as in s. I find this quite tough and would be happy if someone had a solution.
I have Matlab 7.13.
Thanks a lot! Peter
  2 个评论
Azzi Abdelmalek
Azzi Abdelmalek 2013-2-27
编辑：Azzi Abdelmalek 2013-2-27
Are you looking for all the sequences?, for example (2,5,7,3)
Peter
Peter 2013-2-28
Hello Azzi!
I do not really understand what you mean with (2,5,7,3). Maybe my example was a bit confusing! I search for all values in U that could be used to build s, but only in chronological order, so I realized I forgot some in my example. U(1,1,12) would also correspond to s, or U(2,9,12), U(1,2,5) etc.
At the end of the day, I want to find the values (2,2,3) in U that are closest together, so in my example I would search for U(10,11,12). But: it can happen in my real data that the values of s are only in U with interruptions, otherwise it would be a simple task. My plan was to to first find all possible triples in U that could be used to build s and then find out via their indexes which triple is closest together.
Is this clearer now?
Kind regards and many thanks, Peter

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回答（4 个）

Babak
Babak 2013-2-27
http://www.mathworks.com/matlabcentral/answers/44977-how-can-i-generate-all-possible-combinations-from-multiple-sets-of-nchoosek
perms
nchoosek
might be helpful in finding combinations of the elements of different size vectors.
Peter
Peter 2013-2-28
编辑：Peter 2013-2-28
I think I could come up with a solution myself
x=[2,2,5,7,3,4,4,4,2,2,2,3]'; y=[2,2,3]';
p=0; s=0; t=0;
for i=1:length(x);
for t=1+s:length(x); % here the code goes down one line
if p<length(y);
if x(t,1)==y(p+1,1);
codes(p+1,s+1)=t;
p=p+1;
end;
end;
end;
s=s+1;
p=0;
end;
codes=codes';
codes=unique(codes,'rows');
k=find(sum(codes==0,2)>0);
codes(k,:)=[]; % incomplete triples from end of the code are removed
codes=codes';
The code goes through all records in U (here called x) and looks up s (here called y). The final matrix in codes contains all indices in x from which y can be built from.
  1 个评论
Babak
Babak 2013-2-28
Nice solution. I rather let "codes" be a cell object with option of having empty cells. Then you can support finding zero elements in y and your
k=find(sum(codes==0,2)>0);
line wouldn't fail for length(y)>3.

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Saeid Abrari
Saeid Abrari 2020-10-6
for dig=1:16
ndata = length(Frame);
ndata_1=length(digit(dig,:));
n_diff=abs(ndata-ndata_1);
Sound_1=[digit(dig,:)';zeros(n_diff,1)];
sx = sum(Frame.^2);
sy = sum(Sound_1.^2);
maxlag = ndata-1;
cnt = 0;
for lag = -maxlag:maxlag
cnt = cnt + 1;
sxy = 0;
for i=1:ndata
j = i + lag;
if j>0 && j<=ndata
sxy = sxy + Frame(i) * SoundVec_1(j);
end
end
cc(cnt) = sxy / sqrt(sx*sy); % correlation coefficient
ck(cnt) = sxy; % cross-correlation value
lags(cnt) = lag;
end
simil(dig)=max(cc);
[~,i] = max(cc);
Ts=1/Fs;
t1=Ts:Ts:FrameLen*Ts;
norm_ck=cc/max(cc);
lx = (length(norm_ck));
half = ceil(lx/2);
norm_ck_p=norm_ck(1:half);
td = i - ndata;
tau1=lags/Fs;
Saeid Abrari
Saeid Abrari 2020-10-6
for dig=1:12
ndata = length(Frame);
ndata_1=length(digit(dig,:));
n_diff=abs(ndata-ndata_1);
Sound_1=[digit(dig,:)';zeros(n_diff,1)];
sx = sum(Frame.^2);
sy = sum(Sound_1.^2);
maxlag = ndata-1;
cnt = 0;
for lag = -maxlag:maxlag
cnt = cnt + 1;
sxy = 0;
for i=1:ndata
j = i + lag;
if j>0 && j<=ndata
sxy = sxy + Frame(i) * SoundVec_1(j);
end
end
cc(cnt) = sxy / sqrt(sx*sy); % correlation coefficient
ck(cnt) = sxy; % cross-correlation value
lags(cnt) = lag;
end
simil(dig)=max(cc);
[~,i] = max(cc);
Ts=1/Fs;
t1=Ts:Ts:FrameLen*Ts;
norm_ck=cc/max(cc);
lx = (length(norm_ck));
half = ceil(lx/2);
norm_ck_p=norm_ck(1:half);
td = i - ndata;
tau1=lags/Fs;
%Find the maximum corellation index(digit)
if(max(simil)>0.5)
[seqm sequ(s_loc)]=max(simil);
s_loc=s_loc+1;
end

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