# could anyone help me how to convert double to cell in an array.

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jaah navi 2021 年 6 月 24 日
コメント済み: jaah navi 2021 年 7 月 16 日
I am having a cell array A in the folllowing manner
where A=3x1 cell
1x3 double - [1,1,1]
1x3 double - [1,2,2]
1x3 double - [1,1,2]
now, I want to convert A into B as given below
B=3x1 cell
1x1cell - [1,2,3]
1x2 cell -[1] [2,3]
1x2 cell - [1,2] [3]
i.e.,
A=3x1 cell to B=3x1 cell
1x3 double - [1,1,1] to 1x1cell - [1,2,3]
1x3 double - [1,2,2] to 1x2 cell -[1] [2,3]
1x3 double - [1,1,2] to 1x2 cell - [1,2] [3]
Could anyone please help me on this to do it on a general manner as my cell array size is larger.
##### 2 件のコメント表示非表示 1 件の古いコメント
jaah navi 2021 年 6 月 25 日
the pattern for the algorthm is as follows:
A has all rows in double that needs to be converted as cell as shown in each rows of B.
In specific each rows of A has three values for example i am considering the first row of A 1x3 double - [1,1,1] .
Here as all the three values are 1 (in double), it need to be converted to cell as [1,2,3]
For 1x3 double - [1,2,2] the first value is 1 and the second and third value is 2. So to convert, the first value that corresponds to 1 should be [1] and the remaining two values corresponds to 2 should be [2,3].
For 1x3 double -[1,1,2] the first two values are 1 and the third value is 2. So to convert, the value corresponds to first two values should be [1,2] and the third value corresponds to 2 should be [3].

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### 採用された回答

Matt J 2021 年 6 月 24 日
B=cell(size(A));
for i=1:numel(A)
B{i}=splitapply(@(x){x}, 1:numel(A{i}) ,A{i});
end
##### 3 件のコメント表示非表示 2 件の古いコメント
jaah navi 2021 年 7 月 16 日
B=cell(size(A));
for i=1:numel(A)
B{i}=splitapply(@(x){x}, 1:numel(A{i}) ,A{i});
disp("A{" +i+ "}:"), A{i}
for j=1:numel(B{i})
disp("B{" +i+ "}{"+j+"}:"), B{i}{j}
end
disp ' '
end
when A is [4x3 double; 4x3 double; 4x3 double] instead of [1x3 double; 1x3 double; 1x3 double]

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