padding an array with zeros in between elements

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Gillean
Gillean 2024-1-29
评论： Gillean 2024-1-29
Basically, I would like to pad an array of location peak values with zeros inbetween each, the catch is the number of zeros between each should be the difference of the two ascending elements. For example, in an array A = [7 15 21 29], with a loop I would like to pad this so it turns into,
A = [ 7 0 0 0 0 0 0 0 15 0 0 0 0 0 21 0 0 0 0 0 0 0 29 ]
I'm aware I'll have to minus 1 to the equation somewhere too. As you can see, I'm basically making an ascending integer array with the elements that were not included now included as zeros... I've been trying with for loops and if statements but to no avail unfortuantely
Thanks in advance!

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采纳的回答

Stephen23
Stephen23 2024-1-29
编辑：Stephen23 2024-1-29
Ran in:
Assuming that A is sorted, here is a simple approach:
A = [7,15,21,29];
B = A(1):A(end);
B(~ismember(B,A)) = 0
B = 1×23
7 0 0 0 0 0 0 0 15 0 0 0 0 0 21 0 0 0 0 0 0 0 29
If A is not sorted, then replace A(1):A(end) with min(A):max(A)

更多回答（1 个）

Rishi
Rishi 2024-1-29
Ran in:
Hi Gillean,
I understand from your query that you want to modify your array such that the missing numbers between two elements are replaced by '0'.
You can preallocate an array of 0s the required size. Then, using a loop, you can calculate the number of 0s required between two elements, and shift the second element by that difference.
Here is the code to implement the same:
A = [7, 15, 21, 29];
total_length = A(end) - A(1) + 1;
% Preallocate the padded array with zeros
padded_A = zeros(1, total_length);
padded_A(1) = A(1);
insert_index = 1;
for i = 1:length(A)-1
% Calculate the number of positions to move forward
insert_index = insert_index + (A(i+1) - A(i));
padded_A(insert_index) = A(i+1);
end
disp(padded_A);
7 0 0 0 0 0 0 0 15 0 0 0 0 0 21 0 0 0 0 0 0 0 29
Hope this helps!
  1 个评论
Gillean
Gillean 2024-1-29
Hi, thanks so much this works wonderfully! Appreciate the code as it goes through the process nicely, cheers!

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