arr=[ [],[],[] ];
for i=1:3
arr(i)=[i,i+1];
end
Expected output: arr=[[1,2],[2,3],[3,4]];
I am trying to put multiple arrays into a single array and I have tried the above code but it does not work
How can I nest arrays?
What you are trying is very Pythonic, nesting lists in lists. This doesn't work in MATLAB. You have basically two easy options: a multi-dimensional array, or a cell array with arrays.
arr = zeros(3,2); % initialise empty array
arr_cell = cell(3,1); % initialise cell
for ii = 1:3
arr(ii,:) = [ii, ii+1]; % store output as columns
arr_cell{ii} = [ii, ii+1]; % store cells
end
arr =
1 2
2 3
3 4
celldisp(arr_cell)
arr_cell{1} =
1 2
arr_cell{2} =
2 3
arr_cell{3} =
3 4
Cells can take various sized (or even types of) arguments, whereas a multi-dimensional matrix has to have the same number of columns on each row. This makes a cell array more flexible, but the numerical array is a lot faster and has more numerical functions available to it.
A small side-note, I suggest to not use i as a variable name in MATLAB, as it's a built-in function.
Related
If I have a square matrix of arrays such as:
[1,2], [2,3]
[5,9], [1,4]
And I want to get the mean of the first values in the arrays of each row such:
1.5
3
Is this possible in Matlab?
I've used the mean(matrix, 2) command to do this with a matrix of single values, but I'm not sure how to extend this to deal with the arrays.
Get the first elements in all arrays of matrix, then call mean function
mean(matrix(:,:,1))
maybe you need to reshape before call mean
a = matrix(:,:,1);
mean(a(:))
You can apply mean function inside mean function to get the total mean value of the 2D array at index 1. You can do similary with array at index 2. Consider the following snapshot.
After staring at your problem for a long time, it looks like your input is a 3D matrix where each row of your formatting corresponds to a 2D matrix slice. Therefore, in proper MATLAB syntax, your matrix is actually:
M = cat(3, [1,2; 2,3], [5,9; 1,4]);
We thus get:
>> M = cat(3, [1,2; 2,3], [5,9; 1,4])
M(:,:,1) =
1 2
2 3
M(:,:,2) =
5 9
1 4
The first slice is the matrix [1,2; 2,3] and the second slice is [5,9; 1,4]. From what it looks like, you would like the mean of only the first column of every slice and return this as a single vector of values. Therefore, use the mean function and index into the first column for all rows and slices. This will unfortunately become a singleton 3D array so you'll need to squeeze out the singleton dimensions.
Without further ado:
O = squeeze(mean(M(:,1,:)))
We thus get:
>> O = squeeze(mean(M(:,1,:)))
O =
1.5000
3.0000
I have a two arrays within a <1x2 cell>. I want to permute those arrays. Of course, I could use a loop to permute each one, but is there any way to do that task at once, without using loops?
Example:
>> whos('M')
Name Size Bytes Class Attributes
M 1x2 9624 cell
>> permute(M,p_matrix)
This does not permute the contents of the two arrays within M.
I could use something like:
>> for k=1:size(M,2), M{k} = permute(M{k},p_matrix); end
but I'd prefer not to use loops.
Thanks.
This seems to work -
num_cells = numel(M) %// Number of cells in input cell array
size_cell = size(M{1}) %// Get sizes
%// Get size of the numeric array that will hold all of the data from the
%// input cell array with the second dimension representing the index of
%// each cell from the input cell array
size_num_arr = [size_cell(1) num_cells size_cell(2:end)]
%// Dimensions array for permuting with the numeric array holding all data
perm_dim = [1 3:numel(size_cell)+1 2]
%// Store data from input M into a vertically concatenated numeric array
num_array = vertcat(M{:})
%// Reshape and permute the numeric array such that the index to be used
%// for indexing data from different cells ends up as the final dimension
num_array = permute(reshape(num_array,size_num_arr),perm_dim)
num_array = permute(num_array,[p_matrix numel(size_cell)+1])
%// Save the numeric array as a cell array with each block from
%// thus obtained numeric array from its first to the second last dimension
%// forming each cell
size_num_arr2 = size(num_array)
size_num_arr2c = num2cell(size_num_arr2(1:end-1))
M = squeeze(mat2cell(num_array,size_num_arr2c{:},ones(1,num_cells)))
Some quick tests show that mat2cell would prove to be the bottleneck, so if you don't mind indexing into the intermediate numeric array variable num_array and use it's last dimension for an equivalent indexing into M, then this approach could be useful.
Now, another approach if you would like to preserve the cell format would be with arrayfun, assuming each cell of M to be a 4D numeric array -
M = arrayfun(#(x) num_array(:,:,:,:,x),1:N,'Uniform',0)
This seems to perform much better than with mat2cell in terms of performance.
Please note that arrayfun isn't a vectorized solution as most certainly it uses loops behind-the-scenes and seems like mat2cell is using for loops inside its source code, so please do keep all these issues in mind.
I'v 3 cell arrays of same number of cells, However I want to reduce in only one cell array having multiple dimensions so that they can be access by column wise like array's column. I'm working in Matlab's environment, However I'v tried to do so, but unfortunately I'v found no cell access as column as in matrices. Any Suggestions for handle such case?
My Code:
P = cell(1,10);
Pd = cell(1,10);
Pdd = cell(1,10);
for ii=1:10
P{ii}= [repmat([0 0 0],2,1)];
Pd{ii} = [repmat([1 1 1],2,1)];
Pdd{ii} = [repmat([2 2 2],2,1)];
end
Concatenate the three cell arrays vertically:
Pall = [P; Pd; Pdd]
This gives a 3x10 cell array Pall, such that Pall(1,:) is P, etc.
If you want to avoid creating multiple cell arrays from the beginning: do something similar to this:
for ii=1:10
Pall{1,ii}= [repmat([0 0 0],2,1)];
Pall{2,ii} = [repmat([1 1 1],2,1)];
Pall{3,ii} = [repmat([2 2 2],2,1)];
end
To avoid loops, with the values in your example:
Pall = mat2cell(repmat([repmat([0 0 0],2,1); repmat([1 1 1],2,1); repmat([2 2 2],2,1)],1,10),[2 2 2],3*ones(1,10))
I perhaps am going about this wrong, but I have data{1}, data{2}...data{i}. Within each, I have .type1, .type2.... .typeN. The arrays are different lengths, so horizontal concatenation does not work.
For simplicity sake
>> data{1}.type1
ans =
1
2
3
>> data{2}.type1
ans =
2
4
5
6
Results should be [1;2;3;2;4;5;6]
I've been trying to loop it but not sure how? I will have a variable number of files (a,b..). How do I go about looping and concatenating? Ultimately I need a 1xN array of all of this..
My working code, thanks..figured it out..
for i = 1:Types
currentType = nTypes{i}
allData.(currentType)=[];
for j = 1:nData
allData.(currentType) = [allData.(currentType); data{j}.(currentType)(:,3)]; %3rd column
end
end
Look at cat, the first argument is the dimension. In your simple example it would be:
result = cat(1,a,b);
Which is equivalent to:
result = [a;b];
Or you can concatenate them as row vectors and transpose back to a column vector:
result = [a',b']';
For the case of a structure inside a cell array I don't think there will be any way around looping. Let's say you have a cell array with M elements and N "types" as the structure fields for each element. You could do:
M=length(data);
newData=struct;
for i=1:M
for j=1:N
field=sprintf('type%d',j); % //field name
if (M==1), newData.(field)=[]; end % //if this is a new field, create it
newData.(field)=[newData.(field);data{i}.(field)];
end
end
I have three 1-d arrays where elements are some values and I want to compare every element in one array to all elements in other two.
For example:
a=[2,4,6,8,12]
b=[1,3,5,9,10]
c=[3,5,8,11,15]
I want to know if there are same values in different arrays (in this case there are 3,5,8)
The answer given by AB is correct, but it is specific for the case when you have 3 arrays that you are comparing. There is another alternative that will easily scale to any number of arrays of arbitrary size. The only assumption is that each individual array contains unique (i.e. non-repeated) values:
>> allValues = sort([a(:); b(:); c(:)]); %# Collect all of the arrays
>> repeatedValues = allValues(diff(allValues) == 0) %# Find repeated values
repeatedValues =
3
5
8
If the arrays contains repeated values, you will need to call UNIQUE on each of them before using the above solution.
Leo is almost right, should be
unique([intersect(a,[b,c]), intersect(b,c)])
c(ismember(c,a)|ismember(c,b)),
ans =
3 5 8
I think this works for all matrices.
Define what you mean by compare. If the arrays are of the same length, and you are comparing equality then you can just do foo == bar -- it's vectorized. If you need to compare in the less than/greater than sense, you can do sign(foo-bar). If the arrays are not the same length and/or you aren't comparing element-wise -- please clarify what you'd like the output of the comparison to be. For instance,
foo = 1:3;
bar = [1,2,4];
baz = 1:2;
sign(repmat(foo',1,length([bar,baz])) - repmat([bar, baz],length(foo),1))
# or, more concisely:
bsxfun(#(x,y)sign(x-y),foo',[bar,baz])
does what you ask for, but there is probably a better way depending on what you want as an output.
EDIT (OP clarified question):
To find common elements in the 3 arrays, you can simply do:
>> [intersect(a,[b,c]), intersect(b,c)]
ans =
8 3 5