필터 지우기
필터 지우기

how to create a series of changes over the loop variable and can assign a value to it

조회 수: 1 (최근 30일)
how to create a series of changes over the loop variable and can assign a value to it Example
for r=1:3
A|r| = r
B|r| = A|r+1| + A|r|
end
the result
A1=1
A2=2
A3=3
B1=3
B2=5
...

채택된 답변

Youssef  Khmou
Youssef Khmou 2013년 5월 19일
try :
n=1:30;
A=n;
for r=1:length(n)-1
B(r)=A(r+1)+A(r);
end
plot(A), hold on, plot(B)
  댓글 수: 3
Youssef  Khmou
Youssef Khmou 2013년 5월 19일
hi, In this case, the dimension grows inside the loop so you need to use the cell :
for i=1:4
A{i}=rand(3,4+i);
end
Pham Ngoc Thanh
Pham Ngoc Thanh 2013년 5월 19일
Hi I have tried
for i=1:4
A{i}=rand(3,4+i);
end but have an error: "Cell contents assignment to a non-cell array object" What can I do?

댓글을 달려면 로그인하십시오.

추가 답변 (2개)

Image Analyst
Image Analyst 2013년 5월 19일
You don't have a good condition for when r = the last value, for example, what is A(4) when r only goes to 3. But ignoring that last element, this will work for you:
n=10
A=1:n
B = A + [A(2:end) 0]
In command window:
n =
10
A =
1 2 3 4 5 6 7 8 9 10
B =
3 5 7 9 11 13 15 17 19 10
If you want to add the last element of A then turn the 0 into A(n).

Image Analyst
Image Analyst 2013년 5월 19일
This works just fine (if that's what you want to do):
clc;
for i=1:4
A{i} = rand(3,4+i)
end
whos A
A =
[3x5 double]
A =
[3x5 double] [3x6 double]
A =
[3x5 double] [3x6 double] [3x7 double]
A =
[3x5 double] [3x6 double] [3x7 double] [3x8 double]
Name Size Bytes Class Attributes
A 1x4 1072 cell
See how it grows with each iteration and the size of the last-added cell is bigger each time? Just what you wanted, right?
  댓글 수: 3
Image Analyst
Image Analyst 2013년 5월 22일
You can't multiply a 3x5 by a 6x3 - it doesn't work. Anyway, I see you've already accepted another answer so I guess you got it solved that way.
Pham Ngoc Thanh
Pham Ngoc Thanh 2013년 5월 22일
I'm sorry. B{i} = A{i}'*{A{i+1} [5x3]*[3x6}
Now i can understand some things. Thanks you a lot.

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Creating and Concatenating Matrices에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by