how to write this for cycle

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salvatore liberto
salvatore liberto 2016-11-10
评论： Thorsten 2016-11-10
Hi everyone, i need to create this for cycle:
x(64) = 0:1/64:1;
x(128) = 0:1/128:1;
.
.
.
x(4096) = 0:1/4096:1;

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回答（6 个）

Star Strider
Star Strider 2016-11-10
I’m not certain what you want. It is straightforward to define an anonymous function to do that:
x = @(n) 0 : 1/n : 1;
and to call it, for example:
xv64 = x(64);
and so for the others.
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Star Strider
Star Strider 2016-11-10
You can have a constant step, or a constant length, but not both.
The best you could do is to store them in a cell array, or create a matrix of NaN values equal to the length of the longest vector, and then fill the matrix rows with shorter vectors.
For example:
x = @(n) 0 : 1/n : 1;
M = nan(7, length(x(2^12)));
for k1 = 1:size(M,1)
v = x(2^(k1+5));
lv = length(v);
M(k1,1:lv) = v;
end
This is likely the most efficient way to create a matrix.
What do you want to do?
Guillaume
Guillaume 2016-11-10
As I pointed out
"every matrix contains n (=2^i) values which vary from 0 to 1, with a step of 1/(2^i)"
is not possible. You either have n+1 values with a step of 1/n, or you have n values with a step of 1/(n-1), or the values go from 0 to 1-1/n to get n values with 1/n step.

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Guillaume
Guillaume 2016-11-10
arrayfun(@(n) linspace(0, 1, n), 2.^(6:12), 'UniformOutput', false)
will create 7 (not 6!) matrices with the exponent varying from 6 to 12, stored in a cell array
Note that each matrix will have n elements and therefore the step will be 1/(n-1) not 1/n (as it's not possible to generate n elements from 0 to 1 included with a step of 1/n)
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Guillaume
Guillaume 2016-11-10
If you're really insistent you want a step of 1/n, then:
x = arrayfun(@(n) linspace(0, 1, n+1), 2.^(6:12), 'UniformOutput', false); %n+1 elements will create a step of 1/n
If you're really insistent that you want the 7 matrices at indices 64,128, 256, ..., therefore wasting plenty of memory:
x(2.^(6:12)) = arrayfun(@(n) linspace(0, 1, n), 2.^(6:12), 'UniformOutput', false); %or n+1 in linspace

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salvatore liberto
salvatore liberto 2016-11-10
In this way it is certainly more clear.
I have a n = 64.
and
x=0:h:1; {h = 1/n)
i want to create a for loop that, instead of writing the same for loop (changing the number of n, and the name of the variable), generates 6 matrices [n*1]; more specifically, this six:
x(64) = 0 : h : 1; {h = 1/64}
x(128) = 0 : h : 1; {h = 1/128}
x(256) = 0 : h : 1; {h = 1/256}
x(512) = 0 : h : 1; {h = 1/512}
x(1024) = 0 : h : 1; {h = 1/1024}
x(2048) = 0 : h : 1; {h = 1/2048}
x(4096) = 0 : h : 1; {h = 1/4096}
  1 个评论
Guillaume
Guillaume 2016-11-10
编辑：Guillaume 2016-11-10
The arrayfun function I used in my answer is exactly equivalent to a for loop, it just requires less typing and is in my opinion easier to read.
If you're really insistent you want a step of 1/n per matrix (and therefore n+1 numbers), then replace the n in the linspace call by n+1.
As also pointed out your 6 matrices are actually seven matrices.

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Thorsten
Thorsten 2016-11-10
编辑：Thorsten 2016-11-10
You have to use a cell array because your indices have different sizes
e = 6:12;
for i = 1:numel(e), x{2^e(i)} = 0:1/2^e(i):1; end
if you do not need to address x{64}, x{128}, ... but if you can also use x{1}, x{2}, ... you can save some memory by using
for i = 1:numel(e), x{i} = 0:1/2^e(i):1; end
salvatore liberto
salvatore liberto 2016-11-10
Cell contents assignment to a non-cell array object. This is that appears with the code of thorsten
salvatore liberto
salvatore liberto 2016-11-10
it's not necessary to write x(64) but also x64
  1 个评论
Guillaume
Guillaume 2016-11-10
编辑：Guillaume 2016-11-10
Please use Comment on this Answer to actually comment on an answer, not the Answer this question box. It's impossible to follow the thread of your comments

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