I want to split an array in sub-arrays of consecutive elements. For example, for the array:
a=[1 2 3 6 7 9 10 15]
I want an output 1 2 3
, 6 7
, 9 10
, 15
.
I think that the natural choice is to use a struct
for this:
[v,x] = find(diff(a)>1) %find "jumps"
xx=[0 x length(a)]
for ii=1:length(xx)-1
cs{ii}=a(xx(ii)+1:xx(ii+1)); %output struct array
end
v =
1 1 1
x =
3 5 7
xx =
0 3 5 7 8
The code works correctly but I was wondering if there are smarter ways to do this.
-
\$\begingroup\$ You're using a cell array, not a struct. \$\endgroup\$Cris Luengo– Cris Luengo2017年12月10日 02:22:57 +00:00Commented Dec 10, 2017 at 2:22
2 Answers 2
You should preallocate the cs
cell array:
[v,x] = find(diff(a)>1); %find "jumps"
xx = [0 x length(a)];
cs = cell(length(a)+1,1);
for ii = 1:length(xx)-1
cs{ii} = a(xx(ii)+1:xx(ii+1));
end
Style comments:
Try to keep consistent formatting, either put spaces around all equal signs, or around none.
Terminate statements with a semicolon to prevent your function producing output to the command window.
You could take advantage of accessing array's elements via colon notation. For example, if you need to extract the first three elements of an array a=[7 9 6 4 5 8]
, you type b=a(1:3)
. These statements are equivalent a(1:1:3)=a(1:3)=a([1 2 3])
which return the first three elements. In your example, it is a matter of controlling the indexes for colon notation, therefore,
a=[1 2 3 6 7 9 10 15];
b=a(1:3) % --> [1 2 3]
c=a(4:5) % --> [6 7]
d=a(6:7) % --> [9 10]
e=a(end) % --> [15]