I have the matrix a=[5 5 10 10 2 4 5],I want the result to be d=[5 1 2 7;10 3 4 0;2 5 0 0;4 6 0 0],

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I have the matrix
a=[5 5 10 10 2 4 5],
I want the result to be
d=[ 5 1 2 7;
10 3 4 0;
2 5 0 0;
4 6 0 0],
  3 个评论

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回答(4 个)

Image Analyst
Image Analyst 2013-11-25
How about this:
a=[5 5 10 10 2 4 5]
d = transforma(a); % Call transforma() in the main program.
function d = transforma(a)
d=[ 5 1 2 7;
10 3 4 0;
2 5 0 0;
4 6 0 0];
It meets all the criteria you've given.
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Sean de Wolski
Sean de Wolski 2013-11-25
a=[5 5 10 10 2 4 5],
clear d;
[u,~,iy] = unique(a,'stable');
for ii = numel(u):-1:1
lhs = [u(ii) find(iy==ii).'];
d(ii,1:numel(lhs)) = lhs;
end
  6 个评论
Sean de Wolski
Sean de Wolski 2013-11-26
编辑:Sean de Wolski 2013-11-26
@Walter, I'd be curious to time bsxfun(...) against find. I agree that any opportunity for a bsxfun, it should be used just 'cuz it's awesome.
@IA, you know how silent we are on future features :)
@Yousef, I do see the duplicated 20 in column two above. But I don't know how you got it. Please give me full reproduction steps.
When I run something like the following, it passes as I expect:
a = randi(700,[1 10000]);
% a=[5 5 10 10 2 4 5],
clear d;
[u,~,iy] = unique(a,'stable');
for ii = numel(u):-1:1
lhs = [u(ii) find(iy==ii).'];
d(ii,1:numel(lhs)) = lhs;
end
T = matlab.unittest.TestCase;
import matlab.unittest.constraints.IsEqualTo
T.verifyThat(ismember(1:numel(a),d(:,2:end)),IsEqualTo(true(1,numel(a))));
T.verifyThat(nnz(d(:,2:end)),IsEqualTo(numel(a)));
Interactive verification passed.
Interactive verification passed.

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Andrei Bobrov
Andrei Bobrov 2013-11-26
编辑:Andrei Bobrov 2013-11-26
[a1,b,c] = unique(a(:),'first')
[~,ii] = sort(b);
a1 = a1(ii);
c = ii(c);
n = max(histc(c,(1:max(c))'));
f1 = @(x){[x(:)',zeros(1,n-numel(x))]};
p = accumarray(c,(1:numel(c))',[],f1)
out = [a1, cat(1,p{:})];
or with Nan
[a1,b,c] = unique(a(:),'first')
[~,ii] = sort(b);
a1 = a1(ii);
c = ii(c);
idx = bsxfun(@times,1:numel(a),bsxfun(@eq,a1,a));
idx(idx==0) = nan;
ii = sort(idx,2);
out = [a1,ii(:,~all(isnan(ii)))];

Roger Stafford
Roger Stafford 2013-11-27
编辑:Roger Stafford 2013-11-27
Here's another version. Let a be your array and d be the desired result.
[u,t,q] = unique(a(:),'first','stable');
[q,p] = sort(q);
r = p;
n = size(p,1);
r(p) = (1:n)';
t = r(t);
r = cumsum(1-accumarray(t(2:end),diff(t),[n,1]));
d = [i.accumarray([q,r],p)];

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