How to add zeros to the end of an array

I have two arrays like A=[ 1 5 3 8 9 4 7 6 5 2 ] & B=[ 5 2 3 9 7 4 5 ]
In 'A', I have 10 elements(say m) and in 'B', I have 7(say n) elements. I need to add 10-7=3 (m-n) zeros to the end of B.
Please help.

2 commentaires

Becky CNS
Becky CNS le 15 Mar 2018
Déplacé(e) : Dyuman Joshi le 29 Nov 2023
I am trying to do this but with A and B as matrices. The error message I get is 'Dimensions of matrices being concatenated are not consistent'. All I want to do is add zeros to another vector the length of A-B. How would I change the above code?
Jan
Jan le 15 Mar 2018
Déplacé(e) : Dyuman Joshi le 29 Nov 2023
If you post your code, we could fix it directly. This would be more convenient and useful than to create an artificial example.
A = rand(3,4);
B = rand(2,2);
B(size(A,1), size(A,2)) = 0;
A - B

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 Réponse acceptée

Jan
Jan le 29 Jan 2013
Modifié(e) : Jan le 29 Jan 2013
Matlab fills missing elements with 0 automatically:
A = [1 5 3 8 9 4 7 6 5 2];
B = [5 2 3 9 7 4 5];
B(numel(A)) = 0;
Less efficient, but working also:
B = [B, zeros(1, length(A) - length(B))];

12 commentaires

Azzi Abdelmalek
Azzi Abdelmalek le 29 Jan 2013
What if A and B have the same size? (for the first code)
Jan
Jan le 29 Jan 2013
Modifié(e) : Jan le 29 Jan 2013
In the OP's problem, the sizes of A and B are 10 and 7. For more general problems:
nA = numel(A);
nB = numel(B);
if nA > nB
B(nA) = 0;
elseif nB > nA
A(nB) = 0;
end
Mori
Mori le 1 Juil 2015
This adds zero at the end, what if we need first zeros and then elements of B?
B = [zeros(1, length(A) - length(B)), B];
Mori
Mori le 1 Juil 2015
this command works perfectly for raw array but not for column, any idea?
B=[zeros(length(A) - length(B),1);B]
John Lutz
John Lutz le 13 Oct 2017
Thanks Andrei Bobrov,for
B = [zeros(1, length(A) - length(B)), B];
this worked perfectly for me. Maybe someone will see this and provide an explanation! I couldnt make sense of it from the matlab zero doc
Jan
Jan le 13 Oct 2017
@John Lutz: zeros(1,n) creates a row vector of n zeros. [z,B] concatenates the arrays z and B horizontally. length(A)-length(B) determines the difference of the lengths of the two arrays.
Black4Ghost
Black4Ghost le 18 Mar 2020
I am sorry I am commenting so late on this Thread but I have the same Issue with the only difference that I don't want zeros to fill out my cell array. I want it filled out with chars of the value '0'. Also sorry for not opening a new question but its almost the same problem I am afraid.
kay Dee
kay Dee le 2 Avr 2020
thx worked for me
Md Rezaul Karim
Md Rezaul Karim le 26 Jan 2021
Thank you Jan
Jan
Jan le 26 Jan 2021
@Black4Ghost: Filling with characters works by:
[YourData, rempmat('0', 1, wantedWidth - numel(YourData))]

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Plus de réponses (3)

Azzi Abdelmalek
Azzi Abdelmalek le 29 Jan 2013
Modifié(e) : Azzi Abdelmalek le 29 Jan 2013
A=[ 1 5 3 8 9 4 7 6 5 2 ];
B=[ 5 2 3 9 7 4 5 ];
B(end:numel(A))=0

2 commentaires

Jan
Jan le 29 Jan 2013
B(end) overwrites the last element of B.
B(end+1:numel(A))=0

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If you're using release R2023b or later, you could use paddata or resize to do this.
A=[ 1 5 3 8 9 4 7 6 5 2 ];
B=[ 5 2 3 9 7 4 5 ];
n = max(length(A), length(B)); % using length is okay since A and B are both vectors
If we always want to make the vector longer or keep it the same size, use paddata.
A2 = paddata(A, n)
A2 = 1×10
1 5 3 8 9 4 7 6 5 2
B2 = paddata(B, n)
B2 = 1×10
5 2 3 9 7 4 5 0 0 0
If you want to append to the vector or remove elements depending on whether it's shorter or longer than your desired size, use resize.
A3 = resize(A, 8) % drop last 2 elements
A3 = 1×8
1 5 3 8 9 4 7 6
B3 = resize(B, 8) % add one 0 at the end
B3 = 1×8
5 2 3 9 7 4 5 0
There's also trimdata to always make it the desired length or shorter.

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