How to convert logical into decimal number?

I have matrix C size 1x1013 cell. Each cell has different size.

 <1x16 logical> <1x53 logical> <1x41 logical> <1x37 logical> <1x50 logical> <1x48 logical> and so on.

I want to convert each cell on matrix C into decimal number. I have tried use bin2dec but it doesn't worked.

For anyhelp, thank you.

[Jan]

What is the longest logical vector in your cell? Logical vectors with more than 53 elements cannot be represented as DOUBLE number exactly:

2^53 - (2^53 + 1)  %   replies 0 !

[Anisa]

the longest is 103.

[Jan]

@Anisa: Whenever you write "it doesn't work" in the forum, be sure to add the reuiqred details: Do the results differ from your expectations (if so, how), or do you get an error message (if so, post a complete copy)?

[Anisa]

I have matrix C size 1x1013 cell. Each cell consists of binary bit (logical) with different size (the longest is 103 bits). I want to convert each cell on matrix C{i} into decimal number. For example:

C{i} = {[1 0 0 0 1 1 0 1 0 1], [0 1 1], [1 0 0 1 1], [1 0 0 1 1 1 0 0 0 1 1 1 0 0 0 0 1 0 1 0 0 0 1 1], [1 0 1 0 1 0 1 0]}

I want the results on decimal. Like this:

C{} = {[565],[3],[19],[10252451],[170]}

I have tried used bin2dec but it works only 52 bits or less, since my longest bit is 103. So i try this manual converting.

z = [1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 0 0 0 0 0 1 1 1 0 0 0 0 1 1 0 1];
z=logical(z);
x=length(z);
for i=1:x;
  f(i)=(z(i)*(2^(x-i)));
end
dec=sum(f)

And here is the results:

dec = 2.9171e+010

What does the 'e+010' mean? How to apply the function into each cell? If use 'cellfun(fun, A)', how to do it? Since i have no idea how to set up the 'fun' correctly.

[Jan]

The notation "2.9171e+010" means 2.9171 * 10^10.

As explained already, the result is not exact. It is not only DEC2BIN whcih stops working at 52 bits, even the precision of numbers store in double variables is limited to this range. If the number has more bits, only the 52 most significant bits are considered. Try this:

2^54 + 1 == 2^54

I have posted a solution already, which works until 52 bits and handles a cell string also. It can be easily expanded to more bits, but again with the same limitation in the accuracy or the output.

There are still open questions for clarifications. When you want help, it would be a good idea not to ignore them, because they are essential for a solution. Imagine the effects, when you do not care about our questions.

[Anisa]

I'm so sorry Jan. I didn't mean to ignore them. I'm a newbie here and still learn how to use matlab. I'm sorry.

And thank you for the solution. I already try it. Thank you.

[Carlos Lara]

I think .. It could be solved by using the 'bitset' function

Answer 1

[Jan]

Assuming that the highest significant bit comes on the left:

C = {logical(randi(2,1,10) - 1), logical(randi(2,1,40) - 1)};

P2 = transpose(2 .^ (52:-1:0));
D  = zeros(size(C));
for iD = 1:numel(D)
  aC    = C{iD};
  D(iD) = aC * P2(54 - length(aC):53);
end

This is much more efficient than converting the logical vector to a string, the string to a double vector (inside bin2dec) and wrap this by cellfun and a slow anoymous function. I am really a fan of cellfun, but only, if it is efficient.

[Andrei Bobrov]

+1

[Anisa]

how do I do this since the longest logical value is 103? is there another way?

[Azzi Abdelmalek]

Uing Jan's idea

C = {logical(randi(2,1,103) - 1), logical(randi(2,1,100) - 1)};
idx=2.^(0:102)';
out=zeros(size(C))
for k=1:numel(C)
out{k}=C{k}*flipud(idx(1:numel(C{k})))
end

[Jan]

@Azzi: My first name is Jan. It is fine to use it in the forum. Instead of flipping the data, it would be more efficient to flip the vector of the powers of two once only. Using the dot product is faster than performing sum(a.*b). If the data vae 103 bits, it is enough to get the powers of two from 0 to 102, instead of 103.

@Anisa: You cannot store a value until 2^102 in a DOUBLE without loosing information. This works until 2^52 only. If this does not matter, you can consider the first 53 bits only. So please explain, what you want to achieve exactly.

[José-Luis]

Well, not to be pedantic or anything :), but you can store values larger than 2^53 - 1 without losing information, provided that they don't have more than 53 bits (15 digits) of significant values. This doesn't help Anisa, though.

[Jan]

@Jose-Luis: Correct. The binary vector [ones(1,52), zeros(1,102)] can be represented exactly (or perhaps ones(1,53)?).

[Anisa]

Actually i want to convert it into different base-N number.

[Walter Roberson]

Are you looking for strings as output, or numeric values? If you are looking for numeric values as output, you are not going to be able to produce that for more than 53 bits, fewer if you want to encode your numeric base in decimal (e.g., 333 decimal to mean 4^3 - 1 in base 4).

Longer numbers can be constructed as symbolic numbers if you have the symbolic toolbox, or there is John D'Errico's vpi (Variable Precision Integer) contribution.

[Jan]

@Anisa: It would be helpful, if you answer the questions in the comments. Please explain, what you exactly want as output and do not let us guess the details. Thanks.

Answer 2

[José-Luis]

a = randi(2,1,10)-1;
a = logical(a);
a = [{a};{a}]; %some cell array with logical values inside
%Turning into a string and then into a decimal number
your_vals = cellfun(@(x) bin2dec(sprintf('%i',x)),a);