finding a numeric pattern in an array

Hi all,
I have a big numeric vector and I am trying to find a pattern in the vector.
Example:
my numeric vector = [11 5 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 11 2 31 5 1 2 11 2 31 5 1 2 11 2 31 5 1 2 11 2 31 5 1 2 11 2 31 5 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 11 2 31 5 1 2 11 2 31 5 1 2 11 2]
pattern : [1 2 3 4 5 *any numbers* 1 2 3 4 5 1 2 11 2 31]
I know strfind works well if I know the exact sequence of pattern. But for the above case I am unable to find the solution.

 Accepted Answer

V = [11,5,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,11,2,31,5,1,2,11,2,31,5,1,2,11,2,31,5,1,2,11,2,31,5,1,2,11,2,31,5,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,11,2,31,5,1,2,11,2,31,5,1,2,11,2]
V = 1×133
11 5 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5
F = @(v) regexptranslate('escape',char(v));
P = sprintf('%s.*?%s',F([1,2,3,4,5]),F([1,2,3,4,5,1,2,11,2,31]))
P = '□□□□□.*?□□□□□□□□□□'
[X,Y,M] = regexp(char(V),P, 'start','end','match') % start & end indices, matched text.
X = 1×2
6 64
Y = 1×2
35 122
M = 1×2 cell array
{'□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□'} {'□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□□'}
Checking the matched content:
double(M{1})
ans = 1×30
1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 11 2 31
double(M{2})
ans = 1×59
1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

3 Comments

@Stephen23 can't we just take char([1 2 3 4 5 '.*?' 1 2 3 4 5 1 2 11 2 31])?
"can't we just take char([1 2 3 4 5 '.*?' 1 2 3 4 5 1 2 11 2 31])?"
That should work the same as my original answer, but ... I realized that we also need to account for special characters that need escaping, so you might also need to use REGEXPTRANSLATE something like this:
F = @(v) regexptranslate('escape',char(v));
P = sprintf('%s.*?%s',F([1,2,3,4,5]),F([1,2,3,4,5,1,2,11,2,31]))
P = '□□□□□.*?□□□□□□□□□□'
I will update my answer with this.
@Stephen23 excellent! Thank you :) that worked for me perfectly

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on 27 Mar 2023

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