How to make a new array based on two arrays?

A=[1 4 3 1 5 9 1 11 14 15 16];

B=[1 4 3 1 5 6 7 8 9 1 11 12 13 14 15 16];

How to make a new array like this:

C=[1 4 3 1 5 nan nan nan 9 1 11 nan nan 14 15 16];

The new array are to be like A but in addition has nan where the numbers are missing so it will end up with the same length as B.

Answer 1

[Walter Roberson]

You need a matching algorithm to find the best match. You can use dynamic programming, or you can use Boyer-Moore or one of the improvements of it.

http://www.mathcs.emory.edu/~cheung/Courses/323/Syllabus/Text/Matching-Boyer-Moore1.html

http://www.mathcs.emory.edu/~cheung/Courses/323/Syllabus/Text/Matching-Boyer-Moore2.html

You cannot just take the first available match. Your text clearly permits repetition, and therefore has to permit repetition (and partial repetition) of patterns. If you are positioned at [4 5 6 7] now and the target has [15 4 5 18 4 5 6 7] then if you match the first available 4 5 after the 15, then you only get to match 2 positions there, whereas if you held off judgement to beyond the 18 then you get to match 4 positions there. Is the "right" output [nan 4 5 nan nan nan 6 7] or is the "right" output [nan nan nan nan 4 5 6 7] ?

If you have the Bioinformatics Toolbox, you might be able to use https://www.mathworks.com/help/bioinfo/sequence-alignment.html Sequence Alignment -- but if you do, be sure to examine the assumptions it has about what the "best" sequence is. Would it produce the [nan 4 5 nan nan nan 6 7] or would it produce the [nan nan nan nan 4 5 6 7], and which is better for your purpose ?

[Rikke]

Thank you for your respons!

I want it to produce [nan 4 5 nan nan nan 6 7]. I forgot to say that actually every number in the array are different, so for example 4 and 5 will never show up twice (even though the example I wrote didn't show that). And also the nan can be a number as well, for example zero, it does't matter: [0 4 5 0 0 0 6 7]. As long as the array gets extended at the right place.

[Walter Roberson]

In the special case where the elements are unique, then

C = B .* ismember(B, A);

[Rikke]

Thanks!