Filtering sequences in MATLAB - matlab

Question

Is it possible to do something like regular expressions with MATLAB to filter things out? Basically I'm looking for something that will let me take a vector like:
[1 2 1 1 1 2 1 3 3 3 3 1 1 4 4 4 1 1]
and will return:
[3 3 3 3 4 4 4]
These are the uninterrupted sequences (where there's no interspersion).
Is this possible?

Answer 1

Using regular expressions
Use MATLAB's built-in regexp function for regular expression matching. However, you have to convert the input array to a string first, and only then feed it to regexp:
C = regexp(sprintf('%d ', x), '(.+ )(\1)+', 'match')
Note that I separated the values with spaces so that regexp can match multiple digit numbers as well. Then convert the result back to a numerical array:
res = str2num([C{:}])
The dot (.) in the pattern string represents any character. To find sequences of certain digits only, specify them in brackets ([]). For instance, the pattern to find only sequences of 3 and 4 would be:
([34]+ )(\1)+
A simpler approach
You can filter out successively repeating values by checking the similarity between adjacent elements using diff:
res = x((diff([NaN; x(:)])' == 0) | (diff([x(:); NaN])' == 0))
Optionally, you can keep only certain values from the result, for example:
res(res == 3 | res == 4)

Answer 2

You can do it like this:
v=[1 2 1 1 1 2 1 3 3 3 3 1 1 4 4 4 1 1];
vals=unique(v); % find all unique values in the vector
mask=[]; % mask of values to retain
for i=1:length(vals)
indices=find(v==vals(i)); % find indices of each unique value
% if the maximum difference between indices containing
% a given value is 1, it is contiguous
% --> add this value to the mask
if max(indices(2:end)-indices(1:end-1))==1
mask=[mask vals(i)];
end
end
% filter out what's necessary
vproc=v(ismember(v,mask))
Result:
vproc =
3 3 3 3 4 4 4

Answer 3

This can be another approach, although a little bit too elaborated.
If you see the plot of your array, you want to retain its level sets (i.e. a == const) which are topologically connected (i.e. made by one piece).
Coherently, such level sets are exactly the ones corresponding to a==3 and a==4.
Here is a possible implementation
a = [1 2 1 1 1 2 1 3 3 3 3 1 1 4 4 4 1 1]
r = []; % result
b = a; % b will contain the union of the level sets not parsed yet
while ~isempty(b)
m = a == b(1); % m is the current level set
connected = sum(diff([0 m 0]).^2) == 2; % a condition for being a connected set:
% the derivative must have 1 positive and 1
% negative jump
if connected == true % if the level set is connected we add it to the result
r = [r a(m)];
end
b = b(b~=b(1));
end

Answer 4

If you try something like
a = [1 2 1 1 1 2 1 3 3 3 3 1 1 4 4 4 1 1] %initial vector
b = a>=3 %apply filter condition
a = a(b) %keep values that satisfy filter
a will output
a = [3 3 3 3 4 4 4]