In matrix A, every column represents an output variable and every row represents a reading (6 rows in total). Every output has a certain subgroup size (groups of 3 rows). I need A's elements to be sorted in the vertical direction within every subgroup.
A = [ 1 7 4; 4 9 3; 8 5 7; 2 9 1; 7 4 4; 8 1 3];
% consecutive 3 rows is one subgroup, within which sorting is required.
B = [1 5 3; 4 7 4; 8 9 7; 2 1 1; 7 4 3; 8 9 4]; % the expected result.
I was considering something along the lines of B = splitapply(#sort,A,2), but splitapply cannot be called like that. How can I get the desired result?
Please note, the actual matrix contains 8 columns and 300 rows. An example is demonstrated above.
The easiest solution would be to reshape your data, sort, then permute it:
rps = 3; % rows per subgroup
B = permute(sort(reshape(A.',rps,size(A,2),[]),2),[2 1 3]);
The above results in a 3x3x2 arrays, which in my opinion are easier to work with, but if you want the output as in the example, you can do the following:
B = reshape(permute(sort(reshape(A.',rps,size(A,2),[]),2),[2 3 1]),size(A));
Alternatively, you are correct to think that splitapply can be useful here, but it requires a bit more work.
This command works on the sample data and should also work on your full dataset:
b = cell2mat( splitapply( #(x){sort(x,2).'}, A.', repelem( 1:size(A,1)/rps, rps ) ).' );
I'll explain what this does:
repelem( 1:size(A,1)/rps, rps ) returns a row vector of groups. The amount of groups is the total amount of rows divided by the group size. (For good measure, there should be an assertion that this is divisible with no remainder).
splitapply( #(x){sort(x,2).'}, ... since splitapply has to return a scalar object per group, it needs to be told that the output is a cell so that it can return a matrix. (This might not be the best explanation, but if you attempt to run it w/o a cell output, you will get the following error:
The function 'sort' returned a non-scalar value when applied to the 1st group of data.
To compute nonscalar values for each group, create an anonymous function to return each value in a scalar cell:
#(x1){sort(x1)}
I perform several transpose operations since this is what splitapply expects.
I used cell2mat to convert the output cells back to a numeric array.
Related
So, I have large matrix (let's say dimensions are 160x6 and the name of the matrix is datamatrix). Next, let's say I have another matrix called datamatrix2 which has dimensions 80x2. Here's what I want to do:
find every row of datamatrix where the value in column 2 is 2 and the value in column 5 is 1,
and then take the value for column 3 and the value from column 6 of each of those rows and place them in column 1 and column 2, respectively, of datamatrix2.
So, for example:
Let's say that row 3 of datamatrix is the first row in datamatrix with a 2 in column 2 and a 1 in column 5. Let's say there is a 3.096 in column 3 of that row and a 10 in column 6 of that row. So, 3.096 would be placed in position 1,1 of data matrix2 and 10 would be placed in position 1,2 of datamatrix2.
Next, let's say that row 25 of datamatrix is the next row in datamatrix with a 2 in column 2 and a 1 in column 5. Let's say there is a 16.432 in column 3 of that row and a 15 in column 6 of that row. So, 16.432 would be placed in position 2,1 of data matrix2 and 15 would be placed in position 2,2 of datamatrix2.
This process would continue until all of the rows of datamatrix with a 2 in column 2 and a 1 in column 5 have been found.
Please let me know if anyone has any suggestions.
Mucho thanks!
G
When I follow your explanation rightly, what you want is:
index = (datamatrix(:,2) == 2) & (datamatrix(:,5) == 1);
datamatrix2 = datamatrix(index,[4,6]);
This solution uses logical indexing.index stores 1's and 0's depending on wheter the condition is fullfilled or not.
To start of with logical indexing it is easiest to start with a vector. Lets take x = [2 5 3 4 6]; and you want the first, second and fifth entry. Then x([1 2 5]) = [2 5 6]; This can also be expressed logically with x(logical([1 1 0 0 1])) = [ 2 5 6]; Notice how we have a true at every position we want.
The same can be applied when we want to access a matrix.
Lets take A = [1 2 3; 4 5 6; 7 8 9] as a sample matrix. The obvious way to access part of this matrix is something like A([1,2],[1,3]) = [1 3; 4 6]. We can now replace the index with a logical notation:
A([1,2],[1,3]) = A(logical([1 1 0]),logical([1 0 1])) = [1 3; 4 6].
Logical and vector notation can also be combined. In my code index is a logical representation of the columns we want, [4 6] is the vector notation of the rows we want. Therefore the right columns and rows will be selected.
I don't really know what exactly you are trying, but lets take a shot
First: find any value (here 6) in any column (here 2) and then take another value of the found rows from any column (here 5) (here with an example magic-matrix)
datamatrix = [magic(5) magic(5)];
myvalue = 2;
values = datamatrix(find(datamatrix(:,2)==myvalue), 5);
If you have 2 conditions for a row, then use this here
values = datamatrix(find((datamatrix(:,column1)==value1).*(datamatrix(:,column2)==value2)),column3);
if you need to get more than one value, say from two different columns, then you just need to replace column3with something like this
column3 --> [col1 col2 col3]
Summarizing, try this code here
values = datamatrix(find((datamatrix(:,2)==2).*(datamatrix(:,5)==1)),[3 6]);
datamatrix2(:,[1 2]) = values
The second statement may fail because of dimension errors. If so, you should maybe provide the datamatrix entries.
I've two matrix a and b and I'd like to combine the rows in a way that in the first row I got no duplicate value and in the second value, columns in a and b which have the same row value get the maximum value in new matrix. i.e.
a = 1 2 3
8 2 5
b = 1 2 5 7
2 4 6 1
Desired output
c = 1 2 3 5 7
8 4 5 6 1
Any help is welcomed,please.( the case for accumulation is asked here)
Accumarray accepts functions both anonymous as well as built-in functions. It uses sum function as default. But you could change this to any in-built or anonymous functions like this:
In this case you could use max function.
in = horzcat(a,b).';
[uVal,~,idx] = unique(in(:,1));
out = [uVal,accumarray(idx,in(:,2),[],#max)].'
Based upon your previous question and looking at the help file for accumarray, which has this exact example.
[ii, ~, kk] = unique([a(1,:) b(1,:)]);
result = [ ii; accumarray(kk(:), [a(2,:) b(2,:)], [], #max).'];
The only difference is the anonymous function.
I am new to matlab and I was wondering what it meant to use logical indexing/masking to extract data from a matrix.
I am trying to write a function that accepts a matrix and a user-inputted value to compute and display the total number of values in column 2 of the matrix that match with the user input.
The function itself should have no return value and will be called on later in another loop.
But besides all that hubbub, someone suggested that I use logical indexing/masking in this situation but never told me exactly what it was or how I could use it in my particular situation.
EDIT: since you updated the question, I am updating this answer a little.
Logical indexing is explained really well in this and this. In general, I doubt, if I can do a better job, given available time. However, I would try to connect your problem and logical indexing.
Lets declare an array A which has 2 columns. First column is index (as 1,2,3,...) and second column is its corresponding value, a random number.
A(:,1)=1:10;
A(:,2)=randi(5,[10 1]); //declares a 10x1 array and puts it into second column of A
userInputtedValue=3; //self-explanatory
You want to check what values in second column of A are equal to 3. Imagine as if you are making a query and MATLAB is giving you binary response, YES (1) or NO (0).
q=A(:,2)==3 //the query, what values in second column of A equal 3?
Now, for the indices where answer is YES, you want to extract the numbers in the first column of A. Then do some processing.
values=A(q,2); //only those elements will be extracted: 1. which lie in the
//second column of A AND where q takes value 1.
Now, if you want to count total number of values, just do:
numValues=length(values);
I hope now logical indexing is clear to you. However, do read the Mathworks posts which I have mentioned earlier.
I over simplified the code, and wrote more code than required in order to explain things. It can be achieved in a single-liner:
sum(mat(:,2)==userInputtedValue)
I'll give you an example that may illustrate what logical indexing is about:
array = [1 2 3 0 4 2];
array > 2
ans: [0 0 1 0 1 0]
using logical indexing you could filter elements that fullfil a certain condition
array(array>2) will give: [3 4]
you could also perform alterations to only those elements:
array(array>2) = 100;
array(array<=2) = 0;
will result in "array" equal to
[0 0 100 0 100 0]
Logical indexing means to have a logical / Boolean matrix that is the same size as the matrix that you are considering. You would use this as input into the matrix you're considering, and any locations that are true would be part of the output. Any locations that are false are not part of the output. To perform logical indexing, you would need to use logical / Boolean operators or conditions to facilitate the selection of elements in your matrix.
Let's concentrate on vectors as it's the easiest to deal with. Let's say we had the following vector:
>> A = 1:9
A =
1 2 3 4 5 6 7 8 9
Let's say I wanted to retrieve all values that are 5 or more. The logical condition for this would be A >= 5. We want to retrieve all values in A that are greater than or equal to 5. Therefore, if we did A >= 5, we get a logical vector which tells us which values in A satisfy the above condition:
>> A >= 5
ans =
0 0 0 0 1 1 1 1 1
This certainly tells us where in A the condition is satisfied. The last step would be to use this as input into A:
>> B = A(A >= 5)
B =
5 6 7 8 9
Cool! As you can see, there isn't a need for a for loop to help us select out elements that satisfy a condition. Let's go a step further. What if I want to find all even values of A? This would mean that if we divide by 2, the remainder would be zero, or mod(A,2) == 0. Let's extract out those elements:
>> C = A(mod(A,2) == 0)
C =
2 4 6 8
Nice! So let's go back to your question. Given your matrix A, let's extract out column 2.
>> col = A(:,2)
Now, we want to check to see if any of column #2 is equal to a certain value. Well we can generate a logical indexing array for that. Let's try with the value of 3:
>> ind = col == 3;
Now you'll have a logical vector that tells you which locations are equal to 3. If you want to determine how many are equal to 3, you just have to sum up the values:
>> s = sum(ind);
That's it! s contains how many values were equal to 3. Now, if you wanted to write a function that only displayed how many values were equal to some user defined input and displayed this event, you can do something like this:
function checkVal(A, val)
disp(sum(A(:,2) == val));
end
Quite simply, we extract the second column of A and see how many values are equal to val. This produces a logical array, and we simply sum up how many 1s there are. This would give you the total number of elements that are equal to val.
Troy Haskin pointed you to a very nice link that talks about logical indexing in more detail: http://www.mathworks.com/help/matlab/math/matrix-indexing.html?refresh=true#bq7eg38. Read that for more details on how to master logical indexing.
Good luck!
%% M is your Matrix
M = randi(10,4)
%% Val is the value that you are seeking to find
Val = 6
%% Col is the value of the matrix column that you wish to find it in
Col = 2
%% r is a vector that has zeros in all positions except when the Matrix value equals the user input it equals 1
r = M(:,Col)==Val
%% We can now sum all the non-zero values in r to get the number of matches
n = sum(r)
M =
4 2 2 5
3 6 7 1
4 4 1 6
5 8 7 8
Val =
6
Col =
2
r =
0
1
0
0
n =
1
As an example, I have a matrix [1,2,3,4,5]'. This matrix contains one column and 5 rows, and I have to generate a pair of points like (1,2),(1,3)(1,4)(1,5),(2,3)(2,4)(2,5),(3,4)(3,5)(4,5).
I have to store these values in 2 columns in a matrix. I have the following code, but it isn't quite giving me the right answer.
for s = 1:5;
for tb = (s+1):5;
if tb>s
in = sub2ind(size(pairpoints),(tb-1),1);
pairpoints(in) = s;
in = sub2ind(size(pairpoints),(tb-1),2);
pairpoints(in) = tb;
end
end
end
With this code, I got (1,2),(2,3),(3,4),(4,5). What should I do, and what is the general formula for the number of pairs?
One way, though is limited depending upon how many different elements there are to choose from, is to use nchoosek as follows
pairpoints = nchoosek([1:5],2)
pairpoints =
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5
See the limitations of this function in the provided link.
An alternative is to just iterate over each element and combine it with the remaining elements in the list (assumes that all are distinct)
pairpoints = [];
data = [1:5]';
len = length(data);
for k=1:len
pairpoints = [pairpoints ; [repmat(data(k),len-k,1) data(k+1:end)]];
end
This method just concatenates each element in data with the remaining elements in the list to get the desired pairs.
Try either of the above and see what happens!
Another suggestion I can add to the mix if you don't want to rely on nchoosek is to generate an upper triangular matrix full of ones, disregarding the diagonal, and use find to generate the rows and columns of where the matrix is equal to 1. You can then concatenate both of these into a single matrix. By generating an upper triangular matrix this way, the locations of the matrix where they're equal to 1 exactly correspond to the row and column pairs that you are seeking. As such:
%// Highest value in your data
N = 5;
[rows,cols] = find(triu(ones(N),1));
pairpoints = [rows,cols]
pairPoints =
1 2
1 3
2 3
1 4
2 4
3 4
1 5
2 5
3 5
4 5
Bear in mind that this will be unsorted (i.e. not in the order that you specified in your question). If order matters to you, then use the sortrows command in MATLAB so that we can get this into the proper order that you're expecting:
pairPoints = sortrows(pairPoints)
pairPoints =
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5
Take note that I specified an additional parameter to triu which denotes how much of an offset you want away from the diagonal. The default offset is 0, which includes the diagonal when you extract the upper triangular matrix. I specified 1 as the second parameter because I want to move away from the diagonal towards the right by 1 unit so I don't want to include the diagonal as part of the upper triangular decomposition.
for loop approach
If you truly desire the for loop approach, going with your model, you'll need two for loops and you need to keep track of the previous row we are at so that we can just skip over to the next column until the end using this. You can also use #GeoffHayes approach in using just a single for loop to generate your indices, but when you're new to a language, one key advice I will always give is to code for readability and not for efficiency. Once you get it working, if you have some way of measuring performance, you can then try and make the code faster and more efficient. This kind of programming is also endorsed by Jon Skeet, the resident StackOverflow ninja, and I got that from this post here.
As such, you can try this:
pairPoints = []; %// Initialize
N = 5; %// Highest value in your data
for row = 1 : N
for col = row + 1 : N
pairPoints = [pairPoints; [row col]]; %// Add row-column pair to matrix
end
end
We get the equivalent output:
pairPoints =
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5
Small caveat
This method will only work if your data is enumerated from 1 to N.
Edit - August 20th, 2014
You wish to generalize this to any array of values. You also want to stick with the for loop approach. You can still keep the original for loop code there. You would simply have to add a couple more lines to index your new array. As such, supposing your data array was:
dat = [12, 45, 56, 44, 62];
You would use the pairPoints matrix and use each column to subset the data array to access your values. Also, you need to make sure your data is a column vector, or this won't work. If we didn't, we would be creating a 1D array and concatenating rows and that's not obviously what we're looking for. In other words:
dat = [12, 45, 56, 44, 62];
dat = dat(:); %// Make column vector - Important!
N = numel(dat); %// Total number of elements in your data array
pairPoints = []; %// Initialize
%// Skip if the array is empty
if (N ~= 0)
for row = 1 : N
for col = row + 1 : N
pairPoints = [pairPoints; [row col]]; %// Add row-column pair to matrix
end
end
vals = [dat(pairPoints(:,1)) dat(pairPoints(:,2))];
else
vals = [];
Take note that I have made a provision where if the array is empty, don't even bother doing any calculations. Just output an empty matrix.
We thus get:
vals =
12 45
12 56
12 44
12 62
45 56
45 44
45 62
56 44
56 62
44 62
My question has two parts:
Split a given matrix into its columns
These columns should be stored into an array
eg,
A = [1 3 5
3 5 7
4 5 7
6 8 9]
Now, I know the solution to the first part:
the columns are obtained via
tempCol = A(:,iter), where iter = 1:end
Regarding the second part of the problem, I would like to have (something like this, maybe a different indexing into arraySplit array), but one full column of A should be stored at a single index in splitArray:
arraySplit(1) = A(:,1)
arraySplit(2) = A(:,2)
and so on...
for the example matrix A,
arraySplit(1) should give me [ 1 3 4 6 ]'
arraySplit(2) should give me [ 3 5 5 8 ]'
I am getting the following error, when i try to assign the column vector to my array.
In an assignment A(I) = B, the number of elements in B and I must be the same.
I am doing the allocation and access of arraySplit wrongly, please help me out ...
Really it sounds like A is alread what you want--I can't imagine a scenario where you gain anything by splitting them up. But if you do, then your best bet is likely a cell array, ie.
C = cell(1,3);
for i=1:3
C{i} = A(:,i);
end
Edit: See #EitanT's comment below for a more elegant way to do this. Also accessing the vector uses the same syntax as setting it, e.g. v = C{2}; will put the second column of A into v.
In a Matlab array, each element must have the same type. In most cases, that is a float type. An your example A(:, 1) is a 4 by 1 array. If you assign it to, say, B(:, 2) then B(:, 1) must also be a 4 by 1 array.
One common error that may be biting you is that a 4 by 1 array and a 1 by 4 array are not the same thing. One is a column vector and one is a row vector. Try transposing A(:, 1) to get a 1 by 4 row array.
You could try something like the following:
A = [1 3 5;
3 5 7;
4 5 7;
6 8 9]
arraySplit = zeros(4,1,3);
for i =1:3
arraySplit(:,:,i) = A(:,i);
end
and then call arraySplit(:,:,1) to get the first vector, but that seems to be an unnecessary step, since you can readily do that by accessing the exact same values as A(:,1).