Split an array according to a logic

Is there a way to split a vector ,

say;

[1 2 3 6 7 8 12 13 14 15 ] ....( the actual vector is much longer )

into preassigned groups or cells such that cosecutive numbers

{1 2 3 } { 6 7 8} {12 13 14 15} .... are grouped in different cells.

Thanks a ton.

Answer 1

[Jon]

If the numbers are ordered as you show them you can just use reshape() to make them into a m by 3 array. If you really need it to be a cell you can convert that using mat2cell

So for example

x = 1:30
X = reshape(x,10,3)
Xcell = mat2cell(X,ones(10,1))

[TV8]

Its not always a sequence of 3 and I need a more general solution, where say a 20000 point vector , once sorted into ascending values can be grouped into smaller chunks of consecutive numbers.

[Jon]

I'm looking a little more closely at your example. I thought at first that that all of your groups had 3 elements. In which case you can use the approach I show above. Now I see that your last group has 4 elements. This would need a somewhat different solution. I'm not quite clear what your input and output are here. You mention consceutive numbers but in you example your input vector has 12,13,14,15 but then you make the group {11,13,14,15}. Is this just a typo.Can you describe further if it isn't simple groups of 3 e.g. {1,2,3}, {4,5,6},{7,8,9},...

[TV8]

Oh yeah that was a typo, sorry my bad, It needs to be a seperate group for each set of consective numbers.

[TV8]

Further explanation : It needs to be sorted into groups of consecutive numbers and it maynot be the same size so say further in the vector there there is 27 28 29 30 31 32 it need to be sorted into a unique group of 6 {27 28 29 30 31 32}. Then we have the groups : {1 2 3 } { 6 7 8} {12 13 14 15} and {27 28 29 30 31 32}

[Jon]

I think this will work for your situation

% example vector containing groups of consectutive numbers
x = [1 2 3 6 7 8 12 13 14 15 18 19 20 31];

% mark the consecutive elements, difference between consecutive elements is
% 1
idl = diff(x)== 1;

% mark where beginning of each group of consecutive elements occurs, these
% are jumps from 0 to 1
iStart = diff([0 idl])==1;

% count up jumps to form group numbers
groupCount = cumsum(iStart);

% assign group number, but only to ones that belong to a group (the
% consecutive values)
% (for non-members groupCount.*idl is zero so these are assigned to group
% zero)
group = groupCount.*idl ;


% loop through to assign to cell array by group
numGroups = max(group);
X = cell(numGroups,1);
for k = 1:numGroups
    X{k} = x(k==group);
end

X =

4×1 cell array

{[ 1 2]}

{[ 6 7]}

{[12 13 14]}

{[ 18 19]}

[TV8]

Thanks a ton. The logic is sound, so thats perfect, the output seems to skip the last digit in the group. I'll investigate and post the correction here. But this was what i needed. Sent matlab into a frenzy with my loops. Thanks @Jon , hope you have a great day ahead. : )

[Jon]

OK glad that helped. I meant to add in my post that there might be some edge cases to consider at the very beginning and end of the vector, but hopefully you can work that out. If not I can have another look.

[Jon]

Oh, I see, I don't catch the last element in each group. I have to think about how to fix that.

[Jon]

I think this now catches the end of each consecutive group too. Most of the logic is the same, but I realized you have to look forward and backward to see if they are consecutive and not miss the ends of the groups

% example vector containing groups of consectutive numbers
x = [1 2 3 6 7 8 12 13 14 15 18 19 20 31 35 36 40 41];

% elements are consecutive if they the next element is one larger than they
% are
idlFwd = [diff(x)== 1 0];
% elements are also consecutive if the one before them is one less than they are
idlBwd = [0 flip(diff(flip(x))== -1)] 

% elements are consecutive if either criteria is met
idl = idlFwd|idlBwd;

% mark where beginning of each group of consecutive elements occurs, these
% are jumps from 0 to 1
iStart = diff([0 idlFwd])==1;

% count up jumps to form group numbers
groupCount = cumsum(iStart);

% assign group number, but only to ones that belong to a group (the
% consecutive values)
% (for non-members groupCount.*idl is zero so these are assigned to group
% zero)
group = groupCount.*idl ;


% loop through to assign to cell array by group
numGroups = max(group);
X = cell(numGroups,1);
for k = 1:numGroups
    X{k} = x(k==group);
end
disp(X)

1 2 3 6 7 8 12 13 14 15 18 19 20 31 35 36 40 41

{[ 1 2 3]}

{[ 6 7 8]}

{[12 13 14 15]}

{[ 18 19 20]}

{[ 35 36]}

{[ 40 41]}