Shift array to left or right, keep length and feel zero empty area

605 Ansichten (letzte 30 Tage)
Hello,
I have an array:
A = [1 2 3 4 5 6 7 8 9];
I want create B array from A
B = [0 0 1 2 3 4 5 6 7];
or
B = [3 4 5 6 7 8 9 0 0];
How it is possible?
circshift dont feel zeros

Akzeptierte Antwort

Cris LaPierre
Cris LaPierre am 6 Dez. 2020
I'm not aware of a function that will do exactly what you describe. However, the final result is still possible if you are willing to break the process up into steps.
A = [1 2 3 4 5 6 7 8 9];
B = zeros(size(A));
B(3:end) = A(1:7)
B = 1×9
0 0 1 2 3 4 5 6 7
% or
n = -2;
B = circshift(A,n);
if n>0
B(1:n) = 0
else
B(end+n+1:end) = 0
end
B = 1×9
3 4 5 6 7 8 9 0 0
  2 Kommentare
Stephen23
Stephen23 am 12 Nov. 2023
Note that the IF..ELSE..END can be replaced with one line:
A = 1:9;
n = -2;
B = circshift(A,n);
B([1:n,end+n+1:end]) = 0
B = 1×9
3 4 5 6 7 8 9 0 0
n = +2;
B = circshift(A,n);
B([1:n,end+n+1:end]) = 0
B = 1×9
0 0 1 2 3 4 5 6 7

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (2)

Aamod
Aamod am 13 Okt. 2023
Same code for a 2D matrix - shift by shift values and zero pad the extra area.
function Eout = circshiftzeropad(Ein,shifty,shiftx)
%function Eout = circshiftzeropad(Ein,shifty,shiftx)
%shifts and zeropads an array
Eout = circshift(Ein,[shifty shiftx]);
if shiftx <0
Eout(:,(end+shiftx):end)= 0;
else
Eout(:,(1:shiftx))= 0;
end
if shifty <0
Eout((end+shifty):end,:)= 0;
else
Eout((1:shifty),:)= 0;
end

Steven Lord
Steven Lord am 13 Nov. 2023
If you're using release R2023b or later you could use paddata.
A = [1 2 3 4 5 6 7 8 9];
B = paddata(A(1:end-2), numel(A), Side="leading")
B = 1×9
0 0 1 2 3 4 5 6 7
C = paddata(A(3:end), numel(A)) % Default is Side="trailing"
C = 1×9
3 4 5 6 7 8 9 0 0
If you object to the fact that I hard-coded "1:end-2" and "3:end":
n = 2;
B = paddata(A(1:(end-n)), numel(A), Side="leading")
B = 1×9
0 0 1 2 3 4 5 6 7
C = paddata(A((1+n):end), numel(A))
C = 1×9
3 4 5 6 7 8 9 0 0

Kategorien

Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by