How to create a vector in which numbers increase at first, then remain constant, then reduce back to 1 ?

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Shounak Shastri
Shounak Shastri 2014-4-25
评论： Shounak Shastri 2014-4-26
I need to create vectors of length 255 which have the following format:
{1,2,3,4,5,5,5,5,......,5,5,4,3,2,1}
{1,2,3,4,5,6,6,6,....,6,6,5,4,3,2,1}
what i did was, create a for loop and then use a series of while loops as shown below :
for l=1:255
count=1;
%insert the increasing numbers, numbers go upto l
while count<=l
ul(count)=count;
count=count+1;
end
%After reaching l, they should be constant
while count<(255-l)
ul(count)=l;
count=count+1;
end
%Last part where numbers should decrease to 1
x=0;
while count>(254-l)&&count<256
ul(count)=l-x;
x=x+1;
count=count+1;
end
U(l,:)=ul;
end
The problem : 1- in the output, instead of getting 1 as the 255th element, i am getting a 0
2- after l=128(half of 255), the series should go upto 128 and reduce back to 1. Instead it goes above 128 and then starts reducing.
  1 个评论
Navid
Navid 2014-4-25
lets say that you want to create a vector length "m" and want to go up to "n" and then be constant and then decrease:
function y=increasedecreasefun(m,n)
A=n*ones(1,m);
for i=1:n-1
A(i)=i;
A(m-i+1)=i;
end
y=A
end

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采纳的回答

Image Analyst
Image Analyst 2014-4-25
编辑：Image Analyst 2014-4-25
How about:
% Do for n=5,6,7,8,9.
% Can extend to other numbers
% Method #1: put into a cell array.
for n = 5:9
U{n-4} = [1:n, 5*ones(1,255-2*n), n:-1:1]
end
% Method #2: put into a 2D array.
U2D = zeros(5, 255);
for n = 5:9
U2D(n-4,:) = [1:n, 5*ones(1,255-2*n), n:-1:1]
end
[EDITED TO SHOW ALTERNAT, 2-D ARRAY METHOD]
  1 个评论
Shounak Shastri
Shounak Shastri 2014-4-26
Thanks.... Got a clue from your first answer itself. But your Method 2 also works. This is what i used in my program
for l = 1:255
if l<(255/2)
u(l,:) = [1:l, l*ones(1,255-2*l), l:-1:1];
else
u(l,:) = [1:128, 127:-1:1];
end
end
The if-else block is to accommodate the problem #2. The matrix u stores the series till l=128. After l=128, it goes up till 128 and reduces back to 1.

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更多回答（2 个）

Youssef Khmou
Youssef Khmou 2014-4-25
To accomplish this task without loop, you need two variables, the maximum value m and the number of redundancy n :
First example :
m=5;
n=245;
A=[1:m m*ones(1,n) m:-1:1];
Second example :
m=6;
n=255-12;
B=[1:m m*ones(1,n) m:-1:1];
  3 个评论
显示 1更早的评论
Shounak Shastri
Shounak Shastri 2014-4-26
Won't I have to implement a loop to calculate n for 255 iterations ? Otherwise I can create an array for all values of n
n=255:-2:1
and then do the rest using each element of n.

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Jos (10584)
Jos (10584) 2014-4-25
If N is your fixed length and M is the maximum number (-> (1 2 … M-1 M M M M-1 … 2 1"), here is a simply code:
N = 20 , M = 6
A = min(N/2-abs((0:N)-N/2)+1, M)
To get all the arrays for M is 1 up till (N+1)/2, try this:
N = 11
AA = bsxfun(@min, N/2-abs((0:N)-N/2)+1, (1:(N/2)+1).') % AA is a ceil(N+1)/2-by-N matrix

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