我正在寻找如何在1x10数组中随机选择10个正非零元素，其总和为1
示例：
A=[0.0973 0.1071 0.0983 0.0933 0.1110 0.0942 0.1062 0.0970 0.0981 0.0974]
注意：如果我们对上面矩阵中的元素求和它将是1.我需要matlab来生成这样的矩阵随机
答案 0 :(得分：2)
尝试使用Roger的fex提交：http://www.mathworks.com/matlabcentral/fileexchange/9700-random-vectors-with-fixed-sum
这是文件内容的副本（如果链接死亡）。
所有功劳显然都归功于原始海报Roger Stafford ：
function [x,v] = randfixedsum(n,m,s,a,b)
% [x,v] = randfixedsum(n,m,s,a,b)
%
% This generates an n by m array x, each of whose m columns
% contains n random values lying in the interval [a,b], but
% subject to the condition that their sum be equal to s. The
% scalar value s must accordingly satisfy n*a <= s <= n*b. The
% distribution of values is uniform in the sense that it has the
% conditional probability distribution of a uniform distribution
% over the whole n-cube, given that the sum of the x's is s.
%
% The scalar v, if requested, returns with the total
% n-1 dimensional volume (content) of the subset satisfying
% this condition. Consequently if v, considered as a function
% of s and divided by sqrt(n), is integrated with respect to s
% from s = a to s = b, the result would necessarily be the
% n-dimensional volume of the whole cube, namely (b-a)^n.
%
% This algorithm does no "rejecting" on the sets of x's it
% obtains. It is designed to generate only those that satisfy all
% the above conditions and to do so with a uniform distribution.
% It accomplishes this by decomposing the space of all possible x
% sets (columns) into n-1 dimensional simplexes. (Line segments,
% triangles, and tetrahedra, are one-, two-, and three-dimensional
% examples of simplexes, respectively.) It makes use of three
% different sets of 'rand' variables, one to locate values
% uniformly within each type of simplex, another to randomly
% select representatives of each different type of simplex in
% proportion to their volume, and a third to perform random
% permutations to provide an even distribution of simplex choices
% among like types. For example, with n equal to 3 and s set at,
% say, 40% of the way from a towards b, there will be 2 different
% types of simplex, in this case triangles, each with its own
% area, and 6 different versions of each from permutations, for
% a total of 12 triangles, and these all fit together to form a
% particular planar non-regular hexagon in 3 dimensions, with v
% returned set equal to the hexagon's area.
%
% Roger Stafford - Jan. 19, 2006
% Check the arguments.
if (m~=round(m))|(n~=round(n))|(m<0)|(n<1)
error('n must be a whole number and m a non-negative integer.')
elseif (s<n*a)|(s>n*b)|(a>=b)
error('Inequalities n*a <= s <= n*b and a < b must hold.')
end
% Rescale to a unit cube: 0 <= x(i) <= 1
s = (s-n*a)/(b-a);
% Construct the transition probability table, t.
% t(i,j) will be utilized only in the region where j <= i + 1.
k = max(min(floor(s),n-1),0); % Must have 0 <= k <= n-1
s = max(min(s,k+1),k); % Must have k <= s <= k+1
s1 = s - [k:-1:k-n+1]; % s1 & s2 will never be negative
s2 = [k+n:-1:k+1] - s;
w = zeros(n,n+1); w(1,2) = realmax; % Scale for full 'double' range
t = zeros(n-1,n);
tiny = 2^(-1074); % The smallest positive matlab 'double' no.
for i = 2:n
tmp1 = w(i-1,2:i+1).*s1(1:i)/i;
tmp2 = w(i-1,1:i).*s2(n-i+1:n)/i;
w(i,2:i+1) = tmp1 + tmp2;
tmp3 = w(i,2:i+1) + tiny; % In case tmp1 & tmp2 are both 0,
tmp4 = (s2(n-i+1:n) > s1(1:i)); % then t is 0 on left & 1 on right
t(i-1,1:i) = (tmp2./tmp3).*tmp4 + (1-tmp1./tmp3).*(~tmp4);
end
% Derive the polytope volume v from the appropriate
% element in the bottom row of w.
v = n^(3/2)*(w(n,k+2)/realmax)*(b-a)^(n-1);
% Now compute the matrix x.
x = zeros(n,m);
if m == 0, return, end % If m is zero, quit with x = []
rt = rand(n-1,m); % For random selection of simplex type
rs = rand(n-1,m); % For random location within a simplex
s = repmat(s,1,m);
j = repmat(k+1,1,m); % For indexing in the t table
sm = zeros(1,m); pr = ones(1,m); % Start with sum zero & product 1
for i = n-1:-1:1 % Work backwards in the t table
e = (rt(n-i,:)<=t(i,j)); % Use rt to choose a transition
sx = rs(n-i,:).^(1/i); % Use rs to compute next simplex coord.
sm = sm + (1-sx).*pr.*s/(i+1); % Update sum
pr = sx.*pr; % Update product
x(n-i,:) = sm + pr.*e; % Calculate x using simplex coords.
s = s - e; j = j - e; % Transition adjustment
end
x(n,:) = sm + pr.*s; % Compute the last x
% Randomly permute the order in the columns of x and rescale.
rp = rand(n,m); % Use rp to carry out a matrix 'randperm'
[ig,p] = sort(rp); % The values placed in ig are ignored
x = (b-a)*x(p+repmat([0:n:n*(m-1)],n,1))+a; % Permute & rescale x
return