Searching and Manipulating Cell Array - Part 2

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Sam Da 2012-10-19

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/51352-searching-and-manipulating-cell-array-part-2

采纳的回答： Matt J

This is a new problem but the problem data is similar.

I have a cell array A:

[3,2]

[2,4]

[4,5]

[2,5]

[6]

and cell array B:

[2,3,4]

[2,6]

[4,5,6]

In the array B I can see that [2,3,4] is made of elements from [3,2] and [2,4] from A. So, I will assign these two elements from A to B and remove them from A.

A now has:

[4,5]

[2,5]

[6]

[2,5] in B is made from elements of [4,5] and [2,5] from A. So, we assign [4,5] and [2,5] to second element of B and so on. Please observe in second example that 4 was an extra element from A which was not present in B at all but still I want to assign [4,5] and [2,5] to second element of B. We delete them from A.

Now A has only:

[6]

So, how do I achieve this?

Thanks guys

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Matt J 2012-10-20

编辑：Matt J 2012-10-20

Quote: "[2,5] in B is made from elements of [4,5] and [2,5] from A."

B as you've listed it does not show a [2,5]. It does show a [2,6]

Also, even if B did include [2,5], it wouldn't require elements from [4,5] in A since the [2,5] in A completely covers the [2,5] in B.

Matt J 2012-10-20

Quote: "Please observe in second example that 4 was an extra element from A"

If extra elements don't matter, why not use all of A to cover B{1}. Also, what if there are multiple different ways to cover a B{i} with members of A? Does it matter which one we pick?

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Answer 1

Matt J 2012-10-20

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/51352-searching-and-manipulating-cell-array-part-2#answer_62632

编辑：Matt J 2012-10-20

This shouldn't require very much change to the solution I gave you in Part 1

In the outer loop over B, you can delete things from A. Better yet, instead of deleting, overwrite the appropriate A{i} with nans to avoid a lot of memory reallocation.