How to split array into sub arrays?

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how to split principal vector "a" into sub vectors a1, a2, a3 and a4
exemple:
a=[1 3 5 8 11 12 15 17 18 19 20 21 24 29 31 32 33 34 35 36 38 39];
a1=[1 3 5 8];
a2=[11 12 15 17 18 19 20];
a3=[21 24 29];
a4=[31 32 33 34 35 36 38 39];
thnkx
  3 Comments
benghenia aek
benghenia aek on 30 Jan 2024
I want to split the big array "a" into:
a1 "the numbers between 0 and 10"
a2 "the numbers between 11 and 20"
a3 "the numbers between 21 and 30"
and make it into cell arrays

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Accepted Answer

Dyuman Joshi
Dyuman Joshi on 30 Jan 2024
Edited: Dyuman Joshi on 19 Feb 2024
a = [1 3 5 8 11 12 15 17 18 19 20 21 24 29 31 32 33 34 35 36 38 39 69].';
%bin the data into groups of 10
k = findgroups(floor((a-1)/10));
z = accumarray(k, a, [], @(x) {x.'})
z = 5×1 cell array
{[ 1 3 5 8]} {[ 11 12 15 17 18 19 20]} {[ 21 24 29]} {[31 32 33 34 35 36 38 39]} {[ 69]}

More Answers (1)

Matt J
Matt J on 30 Jan 2024
Edited: Matt J on 19 Feb 2024
a = [1 3 5 8 11 12 15 17 18 19 20 21 24 29 31 32 33 34 35 36 38 39];
G=findgroups(discretize(a,[0:10:max(a)+10],'Include','right'));
z=splitapply(@(x){x}, a, G)
z = 1x4 cell array
{[1 3 5 8]} {[11 12 15 17 18 19 20]} {[21 24 29]} {[31 32 33 34 35 36 38 39]}
  2 Comments
Dyuman Joshi
Dyuman Joshi on 19 Feb 2024
splitapply() won't work if there is a gap in the data -
a = [1 3 5 8 11 12 15 17 18 19 20 21 24 29 31 32 33 34 35 36 38 39 69];
G=discretize(a,[0:10:max(a)+10],'Include','right');
z=splitapply(@(x){x}, a, G)
Error using splitapply
For N groups, every integer between 1 and N must occur at least once in the vector of group numbers.
Matt J
Matt J on 19 Feb 2024
Small modification,
a = [1 3 5 8 11 12 15 17 18 19 20 21 24 29 31 32 33 34 35 36 38 39 69];
G=findgroups(discretize(a,[0:10:max(a)+10],'Include','right'));
z=splitapply(@(x){x}, a, G)
z = 1x5 cell array
{[1 3 5 8]} {[11 12 15 17 18 19 20]} {[21 24 29]} {[31 32 33 34 35 36 38 39]} {[69]}

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