Filter löschen
Filter löschen

flip a matrix using loops

3 Ansichten (letzte 30 Tage)
Ariela Glikman
Ariela Glikman am 9 Dez. 2018
Kommentiert: Stephen23 am 9 Dez. 2018
hi,
im tring to flip a matrix usind withot the flip function
for ex:
[1 3 -2; -4 4 8; 7 -1 3] will cange to [7 -1 3; -4 4 8; 1 3 -2]
my script is
outFlip(i,:)= inMatrix(size(inMatrix,1)+1-i,:);

Antworten (2)

Jan
Jan am 9 Dez. 2018
Bearbeitet: Jan am 9 Dez. 2018
Your line is correct already. Only the loop is missing:
n = size(matrix, 1);
for k = 1:n
outFlip(k,:) = inMatrix(n - k + 1, :);
end
Without the loop:
n = size(matrix, 1);
outFlip = inMatrix(n:-1:1, :);
  1 Kommentar
Stephen23
Stephen23 am 9 Dez. 2018
+1 simple without the loop.

Melden Sie sich an, um zu kommentieren.


madhan ravi
madhan ravi am 9 Dez. 2018
matrix=randi([0,4],2);
outFlip=zeros(size(matrix,1),size(matrix,2));
for i= 1:size(matrix,1)
for j=1:size(matrix,2)
outFlip(j,i)=matrix(i,j); % just swap the indices
end
end
  2 Kommentare
Ariela Glikman
Ariela Glikman am 9 Dez. 2018
no, i mean:
[1 3 -2; -4 4 8; 7 -1 3] will cange to [7 -1 3; -4 4 8; 1 3 -2]
madhan ravi
madhan ravi am 9 Dez. 2018
ok then:
matrix=[1 3 -2; -4 4 8; 7 -1 3];
ctr=1;
outFlip=zeros(size(matrix,1),size(matrix,2));
for i=size(matrix,1):-1:1
outFlip(i,:)=matrix(ctr,:); % just swap the indices
ctr=ctr+1;
end
outFlip

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange

Tags

Produkte

Community Treasure Hunt

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

Start Hunting!

Translated by