how to write this for cycle

Hi everyone, i need to create this for cycle:

x(64) = 0:1/64:1;

x(128) = 0:1/128:1;

.

.

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x(4096) = 0:1/4096:1;

Answer 1

[Star Strider]

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.

[salvatore liberto]

I need to create 6 matrices [n*1]

where n change from 2^i, where i = 6, to 2^12.

every matrix contains n values which vary from 0 to 1, with a step of 1/(2^i).

[Guillaume]

Yes, it's really not clear what the question is.

The following may be a clearer way of generating the vector:

x = @(n) linspace(0, 1, n+1);

At least, it makes it clearer that xv64 = x(64) has 65 elements (which is the number of elements that the colon expression will generate).

Of course, maybe the desired expression was actually:

 x = @(n) linspace(0, 1, n);

[Star Strider]

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]

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.

Answer 2

[Guillaume]

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)

[Guillaume]

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

Answer 3

[salvatore liberto]

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}

[Guillaume]

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.

Answer 4

[Thorsten]

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]

the address is for ease of reading.

Answer 5

[salvatore liberto]

Cell contents assignment to a non-cell array object. This is that appears with the code of thorsten

[Thorsten]

Use

clear x

before you run the code.

Answer 6

[salvatore liberto]

it's not necessary to write x(64) but also x64

[Guillaume]

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