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

7 件のコメント

Wan Ji
Wan Ji 2021 年 9 月 3 日
@Stephen @Jan I need your help.
Stephen23
Stephen23 2021 年 9 月 4 日
@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
Chunru 2021 年 9 月 4 日
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.
Walter Roberson
Walter Roberson 2021 年 9 月 4 日
A = sub2ind([5 1], [2 4])
A = 1×2
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'
Chunru
Chunru 2021 年 9 月 4 日
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])
A = 1×2
12 9
What we are looking for should be ind->sub:
A = ind2sub([2,4], 1:6)
A = 1×6
1 2 3 4 5 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)
A = 1×6
1 2 1 2 1 2
B = 1×6
1 1 2 2 3 3
Stephen23
Stephen23 2021 年 9 月 4 日
編集済み: Stephen23 2021 年 9 月 4 日
@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
Chunru 2021 年 9 月 4 日
@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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 採用された回答

Robert U
Robert U 2021 年 9 月 3 日

2 投票

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

その他の回答 (2 件)

Walter Roberson
Walter Roberson 2021 年 9 月 3 日

1 投票

[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

2 件のコメント

Robert U
Robert U 2021 年 9 月 3 日
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
Walter Roberson 2021 年 9 月 3 日
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
Chunru 2021 年 9 月 3 日

1 投票

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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