Embedding an array into another: vectorization method

I have two arrays. The first one is a consecutive sequential one, like:

      seq1 =

     1     0
     2     0
     3     0
     4     0
     5     0
     6     0
     7     0
     8     0
     9     0
    10     0 
    ...continues

The second one is like:

      seq2 =

     2   250
     3   260
     5   267
     6   270
     8   280
    10   290
    13   300
    18   310
    20   320
    21   330 
    ...continues

I need to embed `seq2` into `seq1` in such a way that I end up with the sequence:

      seq3 =

     1     0
     2   250
     3   260
     4   260
     5   267
     6   270
     7   270
     8   280
     9   280
    10   290
    11   290
    ... continues

I could do this with loops but the arrays are really big so I don't want to use two `for` loops, it is taking too long. How can I do this in a vectorised manner?

[dpb]

What are p, q, r, ... ? Constants, additional vectors, ... ?

[Tanmay]

Constant numbers. I've edited the question so its clearer.

Answer 1

[Andrei Bobrov]

seq1 = (1:10);
seq2 = {2,'p';5,'q';7,'r';10,'s';33,'t'};

n = [seq2{:,1}]';
s = seq1(:);
i0 = (min(min(s),min(n)):max(max(s),max(n)))';
[l0,ii] = ismember(i0,n);
idx = cumsum(l0);
t = idx ~= 0;
out = num2cell(i0);
out(t,2) = seq2(idx(t),2);
out(~t,2) = {0};

ADD

A = [
     1     0
     2     0
     3     0
     4     0
     5     0
     6     0
     7     0
     8     0
     9     0
    10     0 ];
B = [  
     2   250
     3   260
     5   267
     6   270
     8   280
    10   290
    13   300
    18   310
    20   320
    21   330 ];

Z = (min([A(:,1);B(:,1)]):max([A(:,1);B(:,1)]))';
l0 = ismember(Z,B(:,1));
ii = cumsum(l0);
t = ii ~= 0;
Z(t,2) = B(ii(t),2);

[Tanmay]

Very elegant, thank you so much!

Answer 2

[Stephen23]

One very simple and efficient way is to use interp1 with its previous option:

>> A = [
     1     0
     2     0
     3     0
     4     0
     5     0
     6     0
     7     0
     8     0
     9     0
    10     0 ];
>> B = [  
     2   250
     3   260
     5   267
     6   270
     8   280
    10   290
    13   300
    18   310
    20   320
    21   330 ];
>> Z = A;
>> Z(:,2) = interp1(B(:,1),B(:,2),A(:,1),'previous')
>> Z =
     1    NA
     2   250
     3   260
     4   260
     5   267
     6   270
     7   270
     8   280
     9   280
    10   290

[Andrei Bobrov]

+1

Z = (min([A(:,1);B(:,1)]):max([A(:,1);B(:,1)]))';
Z(:,2) = interp1(B(:,1),B(:,2),Z,'previous');

[Tanmay]

Wasn't aware of this function, thank you!