extract multiple slices of difference sizes from a vector

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LH
LH am 10 Jul. 2024
Kommentiert: Aquatris am 10 Jul. 2024
Hi all,
I have a vector A of the shape:
A = [1 4 6 5 4 8 7 8 6 4];
and I want to split this vector into slices. The number of slices is 3 and each slice contains a certain number of elements as follows:
num_groups = 3;
num_points_in_each_group = [2 3 5];
I want the result to be: . I have tried with the code below, but it's not working:
for i = 1:num_groups
end_count = num_points_in_each_group(i);
if i == 1
start_count = 1;
else
start_count = num_points_in_each_group(i-1) + 1;
end
A_slice = A(start_count:end_count);
end
Any help would be appreicted.
Thanks
  1 Kommentar
Aquatris
Aquatris am 10 Jul. 2024
Here is a simple modification to your code to work as intended to increase your understanding hopefully.
Since you are new, for loops might be tempting since it is easy to understand. However, keep in mind they are not efficient and you should work towards vectorization whereever possible.
A = [1 4 6 5 4 8 7 8 6 4];
num_groups = 3;
num_points_in_each_group = [2 3 5];
for i = 1:num_groups
% your end count is sum of previous group sizes and current group size
end_count = sum(num_points_in_each_group(1:i));
% your start count is 1+numberOfPointsOfPreviousGroup
start_count = 1+sum(num_points_in_each_group(1:(i-1)));
% you need to store it as cell since groups can have different number of elements
A_slice{i} = A(start_count:end_count);
end
A_slice
A_slice = 1x3 cell array
{[1 4]} {[6 5 4]} {[8 7 8 6 4]}

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Antworten (2)

Stephen23
Stephen23 am 10 Jul. 2024
Bearbeitet: Stephen23 am 10 Jul. 2024
The MATLAB approach:
A = [1 4 6 5 4 8 7 8 6 4];
N = [2 3 5];
C = mat2cell(A,1,N);
Checking:
C{:}
ans = 1x2
1 4
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ans = 1x3
6 5 4
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<mw-icon class=""></mw-icon>
ans = 1x5
8 7 8 6 4
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LH
LH am 10 Jul. 2024
I think I have solved it:
A = [1 4 6 5 4 8 7 8 6 4]';
groups = 3;
end_count = cumsum(num(:));
for i = 1:groups
if i == 1
start_count = 1;
else
start_count = end_count(i-1) + 1;
end
A_slice = A(start_count:end_count(i));
end
Thanks for your help.

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