问题
I'm trying to convert a base-10 integer k into a base-q integer, but not in the standard way. Firstly, I'd like my result to be a vectors (or a string 'a,b,c,...' so that it can be converted to a vector, but not 'abc...'). Most importantly, I'd like each 'digit' to be in base-10. As an example, suppose I have the number 23 (in base-10) and I want to convert it to base-12. This would be 1B in the standard 1,...,9,A,B notation; however, I want it to come out as [1, 11]. I'm only interested in numbers k with 0 \le k \le n^q - 1, where n is fixed in advance.
Put another way, I wish to find coefficients a(r) such that
k = \sum_{r=0}^{n-1} a(r) q^r
where each a(r) is in base-10. (Note that 0 \le a(r) \le q-1.)
I know I could do this with a for-loop -- struggling to get the exact formula at the moment! -- but I want to do it vectorised, or with a fast internal function.
However, I want to be able to take n to be large, so would prefer a faster way than this. (Of course, I could change this to a parfor-loop or do it on the GPU; these aren't practical for my current situation, so I'd prefer a more direct version.)
I've looked at stuff like dec2base, num2str, str2num, base2dec and so on, but with no luck. Any suggestion would be most appreciated.
Regarding speed and space, any preallocation for integers in the range [0, q-1] or similar would also be good.
To be clear, I am looking for an algorithm that works for any q and n, converting any number in the range [0,q^n - 1].
回答1:
You can use dec2base and replace the characters by numbers:
x = 23;
b = 12;
[~, result] = ismember(dec2base(x,b), ['0':'9' 'A':'Z']);
result = result -1;
gives
>> result
result =
1 11
This works for base up to 36 only, due to dec2base limitations.
For any base (possibly above 36) you need to do the conversion manually. I once wrote a base2base function to do that (it's essentially long division). The number should be input as a vector of digits in the origin base, so you need dec2base(...,10) first. For example:
x = 125;
b = 6;
result = base2base(dec2base(x,10), '0':'9', b); % origin nunber, origin base, target base
gives
result =
3 2 5
Or if you need to specify the number of digits:
x = 125;
b = 6;
d = 5;
result = base2base(dec2base(x,10), '0':'9', b, d)
result =
0 0 3 2 5
EDIT (August 15, 2017): Corrected two bugs: handling of input consisting of all "zeros" (thanks to @Sanchises for noticing), and properly left-padding the output with "zeros" if needed.
function Z = base2base(varargin)
% Three inputs: origin array, origin base, target base
% If a base is specified by a number, say b, the digits are [0,1,...,d-1].
% The base can also be directly an array with the digits
% Fourth input, optional: how many digits the output should have as a
% minimum (padding with leading zeros, i.e with the first digit)
% Non-valid digits in origin array are discarded.
% It works with cell arrays. In this case it gives a matrix in which each
% row is padded with leading zeros if needed
% If the base is specified as a number, digits are numbers, not
% characters as in `dec2base` and `base2dec`
if ~iscell(varargin{1}), varargin{1} = varargin(1); end
if numel(varargin{2})>1, ax = varargin{2}; bx=numel(ax); else bx = varargin{2}; ax = 0:bx-1; end
if numel(varargin{3})>1, az = varargin{3}; bz=numel(az); else bz = varargin{3}; az = 0:bz-1; end
Z = cell(size(varargin{1}));
for c = 1:numel(varargin{1})
x = varargin{1}{c}; [valid, x] = ismember(x,ax); x = x(valid)-1;
if ~isempty(x) && ~any(x) % Non-empty input, all zeros
z = 0;
elseif ~isempty(x) % Non-empty input, at least a nonzero
z = NaN(1,ceil(numel(x)*log2(bx)/log2(bz))); done_outer = false;
n = 0;
while ~done_outer
n = n + 1;
x = [0 x(find(x,1):end)];
y = NaN(size(x)); done_inner = false;
m = 0;
while ~done_inner
m = m + 1;
t = x(1)*bx+x(2);
r = mod(t, bz); q = (t-r)/bz;
y(m) = q; x = [r x(3:end)];
done_inner = numel(x) < 2;
end
y = y(1:m);
z(n) = r; x = y; done_outer = ~any(x);
end
z = z(n:-1:1);
else % Empty input
z = []; % output will be empty (unless user has required left-padding) with the
% appropriate class
end
if numel(varargin)>=4 && numel(z)<varargin{4}, z = [zeros(1,varargin{4}-numel(z)) z]; end
% left-pad if required by user
Z{c} = z;
end
L = max(cellfun(@numel, Z));
Z = cellfun(@(x) [zeros(1, L-numel(x)) x], Z, 'uniformoutput', false); % left-pad so that
% result will be a matrix
Z = vertcat(Z{:});
Z = az(Z+1);
来源：https://stackoverflow.com/questions/40683265/convert-integer-to-generic-base-matlab