Split an array according to a logic
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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.
采纳的回答
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))
9 个评论
TV8
2021-11-1
Jon
2021-11-1
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
2021-11-1
TV8
2021-11-1
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
2021-11-1
Jon
2021-11-1
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
2021-11-1
Oh, I see, I don't catch the last element in each group. I have to think about how to fix that.
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]}
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