Accessing the nth dimension in a variable sized multidimensional array
显示 更早的评论
I have a large array of values, and before the execution I don't know how many dimension I need.
How can I extract a single column from this? Or even a specific value?
I tried things like writing a vector of the indices, but that didn't return a single value.
Example: I have a matrix A, how do I get A(1,1,...,1,:) when I don't know how many dimensions A has?
And what if I want to set a single dimension to a specific value like A(1,:,:,...,:)
Alternatively: is there another good way of storing high-dimensional scalar fields? I tried a list of index vectors like in a sparse matrix, but finding a specific value becomes very inefficient that way right?
采纳的回答
"...how do I get A(1,1,...,1,:) when I don't know how many dimensions A has?"
You use a comma-separated list, which you can easily generate from a cell array:
C = {1,1,1,...1,,':'};
A(C{:})
You can create the cell array using repmat or num2cell or similar. The function ndims will also be useful here.
"...what if I want to set a single dimension to a specific value like A(1,:,:,...,:)"
Use a comma-separated list. Lets try a more practical example, where we define the size of C automatically:
C = repmat({':'},1,ndims(A));
C{1} = 1; % C{dim} = index
val = A(C{:})
4 个评论
Jan Bode
2020-7-14
"...is there a way to use things like 'n:end' somehow?"
I guess you mean for some arbitrary dimension. As far as I know, there is no easy way to use end with this comma-separated list syntax. You would have to get the size of that dimension yourself and create a numeric vector within that size, and then use that vector in another comma-separated list to obtain the subset that you want.
It is worth noting that using the end method in indexing already does something similar, accessing the array first to get the size and then with the indexing to get the required subarray.
Walter Roberson
2020-7-14
subsref twice? That does not sound right. end calls the end method; https://www.mathworks.com/help/matlab/matlab_oop/object-end-indexing.html
Jan Bode
2020-7-15
更多回答（1 个）
Walter Roberson
2020-7-13
nd = ndims(YourArray);
dim_wanted = randi(nd);
proto = repmat({1}, 1, nd);
proto{dim_wanted} = ':';
just_that_dim = YourArray(proto{:});
dim_one = randi(nd);
proto = repmat({':'}, 1, nd);
proto{dim_one} = 1;
all_except_that_dim = YourArray(proto{:});
proto = arrayfun(@randi, size(YourArray), 'uniform', 0);
dim_wanted = randi(nd);
proto{dim_wanted} = ':';
random_vector_in_N_space = YourArray(proto{:});
1 个评论
Walter Roberson
2020-7-15
My Answer showed how to index by any arbitrary dimension: repmat() the most common index, and replace the specific choices, then use cell expansion as the index expression. You could make a helper function, such as
IndexDim = @(A,n,index) A(struct('t',[repmat({':'},1,n-1),index,repmat({':'},1,ndims(A)-2+(n==1))]).t)
Note: doing this on one line requires a fairly new version of MATLAB; I do not recall at the moment if it needs R2020a or R2019b, but R2019a would be expected to fail for this.
类别
在 帮助中心 和 File Exchange 中查找有关 Creating and Concatenating Matrices 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
