I have two cell arrays C & D , they contains numerical data (but some cells are empty) . the data that is inside each cell may be a 2D array, I want to find the intersection of each cell in C with each cell in D
How can I do such thing?
for example : if the size of C & D is 10-by10
C= [ {1 2 } ,{ 3 4},.... etc]
D = [ { 1 34 7} , {2 5},... etc]
Out = c intersect D
out= [ { 1} , {},.... etc]
>> C = {1 [2 3 4; 5 6 7] [] [] 5};
>> D = {1:2 3:5 6 7:9 []};
>> R = cellfun(#(c, d) intersect(c(:), d(:)), C, D, 'uniformoutput', 0);
>> R{:}
ans =
1
ans =
3
4
5
ans =
Empty matrix: 1-by-0
ans =
Empty matrix: 0-by-1
ans =
Empty matrix: 1-by-0
If your data is only numeric (each cell contains a numeric value and empties) I suggest you to change that to a numeric array and use the intersect function. It is easy to represent missing values as NaN.
To convert to double:
tmp = {1, 2, 3, 4, 5, []};
% // Getting rid of the empties
index_empties = cellfun(#isempty, tmp);
tmp(index_empties) = {NaN};
% // converting to double
tmp_double = cellfun(#double, tmp);
Cell values are there to easily create a vector with non-homogeneous data types (Strings and numbers, for example). It is quite common see people using cells to store numbers. While this might be valid in some cases, using cells to store homogeneous data will waste memory and will complicate some operations. For example, you cannot easily sum two cell-vectors with numeric data while summing two double-vectors is trivial.
Related
Suppose I have a cell of arrays of the same size, for example
arr = {[1 NaN 2 ], ...
[NaN 4 7 ], ...
[3 4 NaN] };
and I also have a vector, for example
vec = [1 2 2];
How do I find the corresponding cell entry that matches the vector vec. Matching means the entries in the same location are the same, except for NaNs?
For this particular vector vec I would like to have 1 returned, since it matches the first row.
Another vector [5 4 7] would return 2.
Vectors that don't match like [7 7 7] and vectors that match more than one entry like [3 4 7] should throw an error.
Note that the vector [3 7 4] does not match the second entry, because the order is important.
For each cell element, just check if
all(isnan(cellElement) | cellElement == vec)
is true, which means, you found a match. If you convert your cell to a matrix checkMatrix with multiple rows and each row corresponding to one cellElement, you can even do it without implementing a loop by repeating vec vertically and comparing the whole matrix in a single step. You will have to tell all() to check along dimension 2 rather than dimension 1 and have find() detect all the matches, like so:
find( all( ...
isnan(checkMatrix) | checkMatrix == repmat(vec,size(checkMatrix, 1),1) ...
, 2)); % all() along dimension 2
So I thought about it and came up with this:
matching_ind = #(x, arr) find(...
cellfun(#(y) max(abs(not(x-y==0).*not(isnan(x-y)))),...
arr) == 0);
inds = matching_ind(vec, arr);
if length(inds) ~= 1
error('42');
end
See if this bsxfun based approach works for you -
A = vertcat(arr{:});
matching_ind = find(all(bsxfun(#eq,A,vec(:).') | isnan(A),2)) %//'
if numel(matching_ind)~=1
error('Error ID : 42.')
else
out = matching_ind(1);
end
Suppose, we have a cell array consisting of ids and one attribute, e.g.
A{1,1}=[1 2;2 4]
A{1,2}=[2 3 5;8 5 6]
Now, I'd like to have a final output consisting of unique ids of two cells (first row values) and corresponding columns have attribute value of each cell separately.
i.e.
C =
[1] [ 2]
[2] [1x2 double] % 4 in first cell and 8 in second cell
[3] [ 5]
[5] [ 6]
it seems that it's not possible to use something like C=[unique(A{1,:}(1,:)')]. Any help is greatly appreciated.
Assuming that each cell has two rows and a variable amount of columns where the first row is the ID and the second row is an attribute, I'd consolidate all of the cells into a single 2D matrix and use accumarray. accumarray is very suitable here because you want to group values that belong to the same ID together and apply a function to it. In our case, our function will simply place the values in a cell array and we'll make sure that the values are sorted because the values that are grouped by accumarray per ID come into the function in random order.
Use cell2mat to convert the cells into a 2D matrix, transpose it so that it's compatible for accumarray, and use it. One thing I'll need to note is that should any IDs be missing, accumarray will make this slot empty. What I meant by missing is that in your example, the ID 4 is missing as there is a gap between 3 and 5 and also the ID 6 between 5 and 7 (I added the example in your comment to me). Because the largest ID in your data is 7, accumarray works by assigning outputs from ID 1 up to ID 7 in increments of 1. The last thing we would need to tackle is to eliminate any empty cells from the output of accumarray to complete the grouping.
BTW, I'm going to assume that your cell array consists of a single row of cells like your example.... so:
%// Setup
A{1,1}=[1 2;2 4];
A{1,2}=[2 3 5;8 5 6];
A{1,3}=[7;8];
%// Convert row of cell arrays to a single 2D matrix, then transpose for accumarray
B = cell2mat(A).';
%// Group IDs together and ensure they're sorted
out = accumarray(B(:,1), B(:,2), [], #(x) {sort(x)});
%// Add a column of IDs and concatenate with the previous output
IDs = num2cell((1:numel(out)).');
out = [IDs out];
%// Any cells from the grouping that are empty, eliminate
ind = cellfun(#isempty, out(:,2));
out(ind,:) = [];
We get:
out =
[1] [ 2]
[2] [2x1 double]
[3] [ 5]
[5] [ 6]
[7] [ 8]
>> celldisp(out(2,:))
ans{1} =
2
ans{2} =
4
8
If you'd like this done on a 2D cell array, where each row of this cell array represents a separate instance of the same problem, one suggestion I have is to perhaps loop over each row. Something like this, given your example in the comments:
%// Setup
A{1,1}=[1 2;2 4];
A{1,2}=[2 3 5;8 5 6];
A{1,3}=[7;8];
A{2,1}=[1 2;2 4];
A{2,2}=[1;7];
%// Make a cell array that will contain the output per row
out = cell(size(A,1),1);
for idx = 1 : size(A,1)
%// Convert row of cell arrays to a single 2D matrix, then transpose for accumarray
B = cell2mat(A(idx,:)).';
%// Group IDs together and ensure they're sorted
out{idx} = accumarray(B(:,1), B(:,2), [], #(x) {sort(x)});
%// Add a column of IDs and concatenate with the previous output
IDs = num2cell((1:numel(out{idx})).');
out{idx} = [IDs out{idx}];
%// Any cells from the grouping that are empty, eliminate
ind = cellfun(#isempty, out{idx}(:,2));
out{idx}(ind,:) = [];
end
We get:
>> out{1}
ans =
[1] [ 2]
[2] [2x1 double]
[3] [ 5]
[5] [ 6]
[7] [ 8]
>> out{2}
ans =
[1] [2x1 double]
[2] [ 4]
>> celldisp(out{1}(2,:))
ans{1} =
2
ans{2} =
4
8
>> celldisp(out{2}(1,:))
ans{1} =
1
ans{2} =
2
7
I have a (row)vector of some size, containing the values 1,2 and 3. They are in there in no 'specific' order, so a sample of the array would be [1,1,1,1,2,2,2,1,1,2,2,3,3]. What I want to do is find the consecutive number of identical elements, but with some restrictions.
What I want is to make new arrays of which the elements denote:
The number of consecutive 1's before it changes into a 2
The number of consecutive 2's before it changes into a 1
The number of consecutive 2's before it changes into a 3
The number of consecutive 3's before it changes into a 2
So for the example I have given, the arrays would be
[4,2]
[3]
[2]
[]
I'm not sure how to tackle this. I can use the diff function to find where it changes sign, but then it'll be a little tough to figure out exactly what change has occured, right?
The method does not have to be super fast, as I only have to do this a few times for around 10^5 datapoints.
This approach will group things the way you specified in the question:
a=[1,1,1,1,2,2,2,1,1,2,2,3,3]
b = diff(a)
c = find(b)
d = diff([0,c]);
type1 = d(b(c) == 1 & a(c) == 1);
type2 = d(b(c) == -1 & a(c) == 2);
type3 = d(b(c) == 1 & a(c) == 2);
type4 = d(b(c) == -1 & a(c) == 3);
type1 =
4 2
type2 =
3
type3 =
2
type4 =
Empty matrix: 1-by-0
Use the standard procedure with diff to detect changes and run lengths, and then apply accumarray to group run lengths according to each pair of values before and after the change:
x = [1,1,1,1,2,2,2,1,1,2,2,3,3];
x = x.'; %'// easier to work with a column vector
ind = find(diff(x))+1; %// index of positions where x changes
%// To detect what change has ocurred, we'll use x(ind-1) and x(ind)
len = diff([1; ind]); %// length of each run
result = accumarray([x(ind-1) x(ind)], len, [], #(v) {v}); %// group lengths
Note the order within each result vector may be altered, as per accumarray.
In your example, this gives
>> result
result =
[] [2x1 double] []
[3] [] [2]
>> result{1,2}
ans =
2
4
>> result{2,1}
ans =
3
>> result{2,3}
ans =
2
I believe this will do the trick (although it's not very pretty)
a=[1,1,1,1,1,2,2,2,2,1,1,1,2,2,3,3,3];
d=diff(a);
deltas=(d~=0);
d12=[];d23=[];d32=[];d21=[];
last=0;
for i=1:length(a)-1
if deltas(i)
if a(i)==1&&a(i+1)==2
d12=[d12,i-last];
last=i;
elseif a(i)==2&&a(i+1)==3
d23=[d23,i-last];
last=i;
elseif a(i)==3&&a(i+1)==2
d32=[d32,i-last];
last=i;
elseif a(i)==2&&a(i+1)==1
d21=[d21,i-last];
last=i;
end
end
end
I have a 3000x1 cell array of vectors of different lengths and am looking for a way to search them all for a number and return the cell indices for the first and last occurrence of that number.
So my data looks like this:
[1]
[1 2]
[1 2]
[3]
[6 7 8 9]
etc
And I want to my results to look like this when I search for the number 1:
ans = 1 3
All the indices (e.g. [1 2 3] for 1) would also work, though the above would be better. So far I'm unable to solve either problem.
I've tried
cellfun(#(x) x==1, positions, 'UniformOutput', 0)
This returns a logical array, effectively putting me back at square 1. I've tried using find(cellfun...) but this gives the error undefined function 'find' for input arguments of type 'cell'. Most of the help I can find is for searching for strings within a cell array. Do I need to convert all my vectors to strings for this to work?
C = {[1]
[1 2]
[1 2]
[3]
[6 7 8 9]}; %// example data
N = 1; %// sought number
ind = cellfun(#(v) any(v==N), C); %// gives 1 for cells which contain N
first = find(ind,1);
last = find(ind,1,'last');
result = [ first last ];
Suppose A = [1 2 3;4 5 6;7 8 9]
I want to convert it to B = [{[1,2,3]};{[4,5,6]};{[7,8,9]}]
How can I do that in an easy way?
You can use mat2cell function.
From the documentation:
C = mat2cell(A,dim1Dist,...,dimNDist) divides array A into smaller
arrays within cell array C. Vectors dim1Dist,...dimNDist specify how
to divide the rows, columns, and (when applicable) higher dimensions
of A.
You can do it like this:
A = [1 2 3; 4 5 6; 7 8 9];
B = mat2cell(A, [1 1 1], 3);
will give you:
B={[1 2 3];[4 5 6];[7 8 9]}
Documentation also says:
C = mat2cell(A,rowDist) divides array A into an n-by-1 cell array C,
where n == numel(rowDist).
So, if you are always going to split your matrix to rows, but not to columns, you can do it without the second parameter.
B = mat2cell(A, [1 1 1]);
A better, generalized way would be:
mat2cell(A, ones(1, size(A, 1)), size(A, 2));
You can't have a "matrix of cells" like your notation for B implied.
A cell array allows you to store "any data type" in the individual cells. You can't store a cell as a data type in an array.
So let's assume you meant to say you wanted B = {[1,2,3], [4,5,6], [7,8,9]};
If that is the case, then
B = cell(1,3);
for ii=1:3
B(ii) = {A(ii, :)};
end
should do the trick.
Note - edited based on Hadi's comment.