creating new vector that is a combination of two interpolated vectors

Currently I have a vector called A [28 56 84 112 140 168 196 224 252 280 308 336 364 392 420 448 476 504 532 560 588 616 644 672 700 728 756 784 812 840 868 896 924] and another called B that is [82.5 127.5,182.5, 230, 277.5, 330, 380, 432.5, 480, 530, 582.5, 632.5, 682.5, 732.5, 830.5, 882.5 930.5, 980, 1032.5, 1080]

How do I go by interpolating the two vectors such that they will be in units of 1 i.e. [28,29,30…924] for A and [83,84, 85…1080] for B then combining them such that the third new vector will only have points that are common in vector A and vector B after interpolation? After which how would i go about adding zeroes to the end of the third new vector such that it will have the same length as another vector C that is 1xN long?

Answer 1

[Star Strider]

You don’t need to interpolate, simply redefine the A and B vectors as vectors with a step of 1, then use the intersect function to get the values in common to both:

A1 = min(round(A)):max(round(A));
B1 = min(round(B)):max(round(B));
C = intersect(A1,B1);

[Vaultec]

since Im using the new combined vector as the color scale for a scatter plot, ive been told that its better to interpolate the data to have more accurate results. Any idea how to do this?

[dpb]

You have no data to use to interpolate as described...and there's no need for interesect either, the range uniform between the two limits as shown below.

[Star Strider]

Your interpolated data would be the same as A1 and B1 here, since that is how you defined them. They proceed linearly from minimum to maximum (rounded) with an increment = 1. You can experiment with interp1, but you will see that the result is the same.

[dpb]

NB:

Star's solution--

>> A1 = min(round(A)):max(round(A));
   B1 = min(round(B)):max(round(B));
   C = intersect(A1,B1);

dpb's...

>> i1=ceil(max(A(1),B(1)));
   i2=floor(min(A(end),B(end)));
   C1=[i1:i2];

Compare the two...

>> all(C1==C)
ans =
   1
>>

Identical. QED--don't need intersect nor even to create the other series; simply find the upper/lower range limits and fill in between.

Answer 2

[dpb]

i1=ceil(max(A(1),B(1));       % greater of the two start values rounded up
i2=floor(min(A(end),B(end))); % end is minimum of last rounded down
C=[i1:i2];                    % the values between the bounds, inclusive