# I don't know how to avoid eval

조회 수: 7 (최근 30일)
Wan Ji 2021년 9월 3일
댓글: Chunru 2021년 9월 4일
Hi friends from community,
I am very confused on how to manage 'eval' function existing in my code, i need it to be removed from my code and run the function as fast as possible.
The subIndex function runs as to solve such a problem.
Given an integer N, I want the summation of these M non-negative integers be less or equal than N.
List all cases of , such that .
For example
N = 2, M = 2;
Then p becomes
p = [0,0; 1,0; 0,1; 2,0; 0,2; 1,1];
Following is the function with 'eval':
function p = subIndex(M,N)
s = (N+1)*ones(1,M);
r = (1:prod(s))';
ch = '[';
for i = 1:1:M
ch = [ch,'f',num2str(i),','];
end
eval([ch(1:end-1),']=ind2sub(s,r);']);
p = eval([ch(1:end-1),']-1']);
q = sum(p,2);
[~,idx] = sort(q);
p = p(idx,:);
q = sum(p,2)<=N&sum(p,2)>=1;
p = p(q,:);
end
##### 댓글 수: 7이전 댓글 5개 표시이전 댓글 5개 숨기기
Stephen23 2021년 9월 4일
편집: Stephen23 2021년 9월 4일
@Chunru: this is easy to implement, if you really need this feature.
If you take a copy of the function you can see that the output uses VARARGOUT.
Chunru 2021년 9월 4일
@Stephen Sure it is not difficult to implement it. But we have to maintain the code ourselves that is very much similar to a built-in code.

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### 채택된 답변

Robert U 2021년 9월 3일
Hi Wan Ji,
you can use cell-arrays. You don't need to assign variables with different names to multiple outputs.
function p = subIndex(M,N)
% Given an integer N, I want the summation of these M non-negative integers be less or equal than N.
% List all cases of p1,p2,...pM, such that sum(p1...pM) <= N.
s = (N+1)*ones(1,M);
r = (1:prod(s))';
f = cell(1,M);
[f{:}] = ind2sub(s,r);
f = [f{:}];
p = f-1;
q = sum(p,2);
[~,idx] = sort(q);
p = p(idx,:);
q = sum(p,2)<=N&sum(p,2)>=1;
p = p(q,:);
end
Kind regards,
Robert
##### 댓글 수: 1이전 댓글 -1개 표시이전 댓글 -1개 숨기기
Stephen23 2021년 9월 4일

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### 추가 답변 (2개)

Walter Roberson 2021년 9월 3일
[f{1:M-1}] = ind2sub(s, r);
p = cell2mat(f)-1;
However, it looks to me as if you are doing "integer partitions" of M. People have posted functions for that, including https://www.mathworks.com/matlabcentral/answers/226437-obtain-all-integer-partitions-for-a-given-integer#answer_497158
##### 댓글 수: 2없음 표시없음 숨기기
Robert U 2021년 9월 3일
Hi Walter,
I am not convinced that "M-1" is correct. For the case M = 2 you would get only one cell with the x values of ind2sub. You would miss the second column of p if I am not mistaken.
Kind regards,
Robert
Walter Roberson 2021년 9월 3일
Ah yes I was confused about what the -1 was doing in the original code. I see now that it was skipping a final comma. eval() is such a mess to use!
Also... strjoin() would have avoided having to remove the comma. And the loop could have been avoided with
strjoin("f"+(1:M), ',')

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Chunru 2021년 9월 3일
N = 2, M = 2;
%p = [0,0; 1,0; 0,1; 2,0; 0,2; 1,1];
p = subIndex(M,N)
p = subIndex1(M,N)
function p = subIndex(M,N)
s = (N+1)*ones(1,M);
r = (1:prod(s))';
ch = '[';
for i = 1:1:M
ch = [ch,'f',num2str(i),','];
end
eval([ch(1:end-1),']=ind2sub(s,r);']);
p = eval([ch(1:end-1),']-1']);
q = sum(p,2);
[~,idx] = sort(q);
p = p(idx,:);
q = sum(p,2)<=N&sum(p,2)>=1;
p = p(q,:);
end
function p = subIndex1(M,N)
% Consider the M numbers are index pf M-D matrix, each dim has N+1 elements
s = (N+1)*ones(1,M);
k = cumprod(s);
ii = (0:k(end)-1)'; % linear index (0 based)
idx = zeros(length(ii), M); % sub []xM
% linear index i--> sub idx of size Mx1 (0 based)
for j=M:-1:2
ir = rem(ii, k(j-1));
idx(:, j) = (ii - ir) /k(j-1);
ii = ir;
end
idx(:, 1) = ii;
p = idx(sum(idx,2)<=N, :);
end

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