I don't know how to avoid eval
이전 댓글 표시
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
.
, 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: 7
Wan Ji
2021년 9월 3일
Stephen23
2021년 9월 4일
@Wan Ji: the simple and efficient MATLAB approach is to use comma-separated lists (just as Robert U showed):
When you start to mess around with evaluating text then you are doing something wrong.
Chunru
2021년 9월 4일
Could it be a feature request to Mathworks such that ind2sub will return an array of sub when there is only one output argument? The solution of "[f{:}]=..." is good. But returning an array is better.
A = sub2ind([5 1], [2 4])
So a single output is possible and meaningful... but probably not at all common.
So instead of returning a cell when only one output is requested, it would probably make more sense to add an option, such as 'OutputFormat', 'cell'
In the above example, the code is for sub->ind. The syntax should be IND = sub2ind(SIZ,I,J).
A = sub2ind([5 7], [2 4], [3 2])
What we are looking for should be ind->sub:
A = ind2sub([2,4], 1:6)
If the number of output argument is 1, it just return the linear index. It should be more meaningful to return an arrray of subscripts. However, there might be side effects when the second input argumet (linear index) is an array.
[A, B] = ind2sub([2,4], 1:6)
채택된 답변
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: 2
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), ',')
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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