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Output shape inconsistent when indexing with empty vector

Nathan Jessurun さんによって質問されました 2019 年 7 月 17 日
最新アクティビティ Nathan Jessurun さんによって コメントされました 2019 年 8 月 7 日
I am working with code that maintains three lists -- two exist in the same matrix and one is separate:
twoLists = 1:9;
twoLists = [twoLists' 2*twoLists'];
singleList = (1:9)';
I then index into these lists as follows:
entries = [1;2];
first = twoLists(entries, 1);
second = twoLists(entries, 2);
xponding = singleList(entries);
This is all well and good until 'entries' is empty and both sets of lists only contain one entry:
twoLists = [1 1];
singleList = [2];
entries = ones(1, 0);
first = twoLists(entries, 1);
second = twoLists(entries, 2);
xponding = singleList(entries);
In this case, the shape of 'xponding' doesn't match 'first' and 'second'. I could fix the problem by indexing into singleList like:
singleList(entries, 1);
but I was wondering if there was a better way to solve the problem / elminate it entirely, since it would be a pain to index like this for the whole function.

  1 件のコメント

madhan ravi
2019 年 7 月 17 日
show how your desired result should look like

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1 件の回答

回答者: Nikhil Sonavane 2019 年 8 月 2 日

You are encountering the undesired output because of the following line in the code-
entries = ones(1, 0);
Indexing matrices in MATLAB should always be done using positive integers. I suggest you to always start indexing all the matrices from 1 and not zero as you tried it in your first example. In your second example you are creating a matrix with dimensions 1x0 and later assessing another matrix using the previous matrix which has undefined dimensions.

  1 件のコメント

Nathan Jessurun 2019 年 8 月 7 日
I tried to simulate a situation that appears in my code. 'entries' is formed from a logical index of potential values. So if
values = 1:3
solutions = logical([0 0 0]);
entries = values(solutions);
I get the situation I described.
It doesn't change with how I initialize 'entries'. Does that make sense? Sorry, the code base is large and I attempted to minimize the complexity of the question.

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