How can I add a fixed interval to a set of data?

I have a vector with N numbers, the vector has downward/upward/downward trend (comes from a set of data). :

A=[4 3 1 0 1 2 5....]

I want to add some extra data (missing ones) to my vector, with interval of 1,for example:

B=[4 3 2 1 0 2 3 4 5 ...]

How can I obtain vector B from vector A?

[Jan]

I assume, you mean:

B=[4 3 2 1 0 *1* 2 3 4 5 ...]

Answer 1

[arich82]

I'm assuming that there's a typo in your expected result, and that B should be

B=[4 3 2 1 0 1 2 3 4 5 ...]

A simple (but ugly) for loop can do this, e.g.

A = [4 3 1 0 1 2 5];
B = A(1);
for k = 2:numel(A)
    d = diff(A(k - 1:k));
    if abs(d) == 1
        B(end + 1) = A(k);
    else
        B = [B(1:end - 1), B(end):sign(d):A(k)];
    end
end % for k

Note that the above is terribly inefficient if numel(A) is large, since B is growing inside the loop at each iteration.

If your data sets are large, or if you'll be running this numerous times, there are potentially more efficient (though also more obtuse) answers. For example, a variation on run length encoding:

A = [4 3 1 0 1 2 5];
d = diff(A);
val = sign(d); 
len = abs(d);  
ind = [0, cumsum(len)] + 1; % add one for 1-based indexing; note: A == B(ind);
n = ind(end);  % note: numel(B) == sum(abs(diff(A))) + 1 == n;
mask = false(1, n-1); 
mask(ind(1:end-1)) = true; % ind(end) == numel(B), not start of new phase
diffB = val(cumsum(mask)); % cumsum(mask) gives the rle phase number, i.e. index into val
B = A(1) + cumsum([0, diffB]);

Note that this method is NOT robust against repeated values in A, i.e. it fails if any(diff(A) == 0); a workaround would be to sanitize A, e.g. A = A([true, abs(diff(A)) > 0]), but this would obviously throw away data if the repeats are significant.

Please accept this answer if it works for you, or let me know in the comments if I've missed something.

[arich82]

Note: the if statement in the first method can be eliminated, i.e.

A = [4 3 1 0 1 2 5];
B = A(1);
for k = 2:numel(A)
    d = diff(A(k - 1:k));
    B = [B(1:end - 1), B(end):sign(d):A(k)];
end % for k

would give identical results, but this method is still much less efficient, especially for large datasets.

As an alternative to the RLE approach, you could also interpolate:

d = diff(A);
val = sign(d); 
len = abs(d);  
ind = [0, cumsum(len)] + 1; % add one for 1-based indexing; note: A == B(ind);
B = interp1(ind, A, 1:ind(end));

Note that, to get identical results with the previous methods, you'd need to do a round(B) to eliminate the numerical errors due to interpolation.

[Amin Gan]

Actually, Ive faced another problem which I thought that solution would resolve it, but the size of vector becomes big. I have two set of different data(1000 each). when I plot x Vs y axis, for each point of x axes, there is one or more value for y axes. I want to sum the repeated values of A(x-axis) for each value of B(y-axis). I use following code for that:

 [uA a b] = unique(A);
 sB = arrayfun(@(x) (sum(B(b==x))), 1:numel(a));
 X = [uA sB'];

The problem is that, sB does not give me the best answer because my x axis has downward/upward/downward trend (comes from a set of data).for example at x=2 (attached file), I should have 4 different values for y and sum them up and get around y=11.2, but my data has only two values of 2 and the next close number is 3. However, by using above method at x=2, y=5.6 and x=3, y=5.6 which is not correct( from the curve). Do you know how to solve that problem (finding sB),is using average or median a good idea?

[arich82]

If you want to sum multiple y variables for a given x variable, it sounds like you should be able to use accumarray (or possibly histc). Whether summing or averaging is more appropriate will depend on your application, but either is possible with accumarray.

Unless I'm misunderstanding, this doesn't seem closely related to the current question. It might be better to post this as a new question with a small set of example data, and the expected result (especially since I get an error when I try to open the .fig file you uploaded).

[arich82]

Not sure why I couldn't open the .fig file you posted, but I see the .jpg from your other post here.

I now understand how the two questions are related. For posterity, I'll post two addenda here, and post a full answer in the other question, linked above.

To open the .fig file, I first renamed it to curve.mat, and loaded it. This allowed me to load the x and y data using the following:

data = load('curve.mat');
x = data.hgS_070000.children(1).children.properties.XData;
y = data.hgS_070000.children(1).children.properties.YData;

From the x data, I see that you do indeed need a method which tolerates repeated values in A, i.e. when any(diff(A) == 0) returns true. A simple modification of the run-length encoding scheme does this:

diffx = diff(x);
val = sign(diffx); 
len = abs(diffx) + 1; % add one to include diff==0 as new phase; need to subtract off cumsum in ind
ind = [0, cumsum(len) - cumsum(abs(val))] + 1; %  add one for 1-based indexing; note: x == X(ind);
n = ind(end); % note: numel(X) == sum(abs(diff(x))) + 1 == n;
mask = false(1, n-1); 
mask(ind(1:end-1)) = true; % ind(end) == numel(X), not start of new phase
diffX = val(cumsum(mask)); % cumsum(mask) gives the rle phase number, i.e. index into val
X = x(1) + cumsum([0, diffX]);
K = 1:numel(X);
Y = interp1(K(ind), y, K);

I haven't had a chance to thoroughly vet this, so there might be an error, but essentially, since X is now uniformly spaced (though still non-monotonic), we can interpolate against the index (i.e. treat X and Y as parameterized by K, which is monotonic). I have to run now, but I'll post a full solution in the other thread tomorrow.

I hope this helps.

Answer 2

[Amin Gan]

Thank you