I have two matrices in Matlab.
A =
and
B =
I want to assign the elements having the same cell-value according to it's corresponding column number in A matrix and move the elements there. I want to map the elements of B with A so that B elements also moves in that position.
I want this
A =
And therefore,
B =
Is there a way to do this?!
Thanks.
Easiest way I can think of is to create row/column pairs where the rows correspond row locations of the matrix and column locations are the actual elements of the matrix themselves. The values seen at these row/column pairs are again just the matrix values themselves.
You can very easily do this with sparse. Recreating the matrix above and storing this in A:
A = [1 2 5 8; 1 2 4 7];
... I would do it this way:
r = repmat((1:size(A,1)).', 1, size(A,2)); %'
S = full(sparse(r(:),A(:),A(:)));
The first line of code generates row locations for each value in the matrix A, then using sparse to specify row/column pairs and the associated values and we use full to convert to a proper numeric matrix.
We get:
S =
1 2 0 0 5 0 0 8
1 2 0 4 0 0 7 0
You can also do the same for the matrix B. You'd use sparse and specify the third parameter to be B instead:
B = [0.5 0.2 0.6 0.8; 0.4 0.6 0.8 0.9];
S2 = full(sparse(r(:),A(:),B(:)));
We get:
>> S2
S2 =
0.5000 0.2000 0 0 0.6000 0 0 0.8000
0.4000 0.6000 0 0.8000 0 0 0.9000 0
Related
i'm new to matlab, too used to python and having difficulty finding a way to filter a struct similar to how i can filter a pandas dataframe in python based on condition.
Matlab
a = arrayfun(#(x) x.value ==10, Data);
Data_10 = Data(a);
Error using arrayfun Non-scalar in Uniform output, at index 1, output
1. Set 'UniformOutput' to false.
How i would do so in python:
Data_10 = Data[Data.value == 10]
Try this:
Data_10 = zeros(size(Data.value));
Data_10(Data.value==10) == 10;
This should write into your array Data_10 the value 10 into each position, that has a 10 in Data and leave the rest as 0.
I am not sure if I fully understood your question. Here is my underestanding:
You want to filter certain values of an matrix.
Lets imagine we have a Matrix A filled with values. You want to filter values smaller than lowthresh = 0 and greater than upthresh = 5.
A = [3 6 -2.4 1; 0 34 4.76 0.5; 84 3 2.32 4; 1 -1 2 3.99];
lowthresh = 0;
upthresh = 5;
A(A<lowthresh | A>upthresh) = NaN; % Nan is a good flag
Output:
A =
3.0000 NaN NaN 1.0000
0 NaN 4.7600 0.5000
NaN 3.0000 2.3200 4.0000
1.0000 NaN 2.0000 3.9900
Having substituted your values you can do some basic functions ignoring NaNs:
For instance average:
mean(A,'omitnan')
ans =
1.3333 3.0000 3.0267 2.3725
I hope this adresses your question. Notice, that you can do this for any statement, that returns a boolean (isnan(), ... ) even if the boolean does not have anything to do with the matrix at all.
Lets say we have 2 matrizes that have the same size but different numbers:
A =
1 1 0
1 1 0
0 0 0
B =
0 0 0
0 0 0
0 0 0
We can easily say:
B(A==1) = 2
B =
2 2 0
2 2 0
0 0 0
I hope it helped a bit,
cheers Pablo
Given an sparse matrix A in MATLAB and the mean for the nonzero elements in its columns m, is there anyway to subtract the nonzero elements in each column from the mean of each column and avoid looping over columns?
I am looking for efficient solutions. Using 'bsxfun' could be one solution if it is possible to use.
Thanks
You can use the second output of find to get the column indices; use those to index into m to do the subtraction; and put the results back into A using logical indexing:
A = sparse([0 0 0 0; 1 0 3 2; 2 1 0 5]); %// example data
m = [1.5 1 3 3.5]; %// vector of mean of nonzero elements of each column
m = m(:);
[~, jj, vv] = find(A);
A(logical(A)) = vv - m(jj);
Original A:
>> full(A)
ans =
0 0 0 0
1 0 3 2
2 1 0 5
Final A:
>> full(A)
ans =
0 0 0 0
-0.5000 0 0 -1.5000
0.5000 0 0 1.5000
Let's say I have a mxn matrix of different features of a time series signal (column 1 represents linear regression of the last n samples, column 2 represents the average of the last n samples, column 3 represents the local max values of a different time series but correlated signal, etc). How should I normalize these inputs? All the inputs fall into different categories, so they have a different range. One ranges from 0,1, the other ranges from -5 to 50, etc etc.
Should I normalize the WHOLE matrix? Or should I normalize each set of inputs one by one individually?
Note: I usually use mapminmax function from MATLAB for the normalization.
You should normalise each vector/column of your matrix individually, they represent different data types and shouldn't be mixed up together.
You could for example transpose your matrix to have your 3 different data types in the rows instead of in the columns of your matrix and still use mapminmax:
A = [0 0.1 -5; 0.2 0.3 50; 0.8 0.8 10; 0.7 0.9 20];
A =
0 0.1000 -5.0000
0.2000 0.3000 50.0000
0.8000 0.8000 10.0000
0.7000 0.9000 20.0000
B = mapminmax(A')
B =
-1.0000 -0.5000 1.0000 0.7500
-1.0000 -0.5000 0.7500 1.0000
-1.0000 1.0000 -0.4545 -0.0909
You should normalize each feature independently.
column 1 represents linear regression of the last n samples, column 2 represents the average of the last n samples, column 3 represents the local max values of a different time series but correlated signal, etc
I can't say for sure about your particular problem, but generally, you should normalize each feature independently. So normalize column 1, then column 2 etc.
Should I normalize the WHOLE matrix? Or should I normalize each set of inputs one by one individually?
I'm not sure what you mean here. What is an input? If by that you mean an instance (a row of your matrix), then no, you should not normalize rows individually, but columns.
I don't know how you would do this in Matlab, but I took your question more as a theoretical one than an implementation one.
If you want to have a range of [0,1] for all the columns that normalized within each column, you can use mapminmax like so (assuming A as the 2D input array) -
out = mapminmax(A.',0,1).'
You can also use bsxfun for the same output, like so -
Aoffsetted = bsxfun(#minus,A,min(A,[],1))
out = bsxfun(#rdivide,Aoffsetted,max(Aoffsetted,[],1))
Sample run -
>> A
A =
3 7 4 2 7
1 3 4 5 7
1 9 7 5 3
8 1 8 6 7
>> mapminmax(A.',0,1).'
ans =
0.28571 0.75 0 0 1
0 0.25 0 0.75 1
0 1 0.75 0.75 0
1 0 1 1 1
>> Aoffsetted = bsxfun(#minus,A,min(A,[],1));
>> bsxfun(#rdivide,Aoffsetted,max(Aoffsetted,[],1))
ans =
0.28571 0.75 0 0 1
0 0.25 0 0.75 1
0 1 0.75 0.75 0
1 0 1 1 1
if i have the following code in matlab
function adj=edgeL2adj(el)
nodes=sort(unique([el(:,1) el(:,2)])); % get all nodes, sorted
adj=zeros(numel(nodes)); % initialize adjacency matrix
% across all edges
for i=1:size(el,1);adj(find(nodes==el(i,1)),find(nodes==el(i,2)))=el(i,3);
end
that convert edge list m x 3 to adjacency list n x n but i have a matrix of edge list m x 2 so what is the required change in previous code that give me true result.
example:
if edge list =[1 2;2 3;2 4] then adjacency matrix=[0 1 0 0;0 0 1 1;0 0 0 0; 0 0 0 0]
I would personally do away with your loop code and take advantage of sparse then convert back to a full matrix if desired. The first column consists of the source node and the second column consists of the target node. You simply set all of these entries in the sparse matrix to 1. However, judging from your code, the third column of your edge list is also the corresponding weight, and so I'll write code that will assume both cases. Also, make sure you filter out duplicate rows using unique:
Example with weights just being 1
edgelist = [1 2;2 3;2 4];
edgelist = unique(edgelist, 'rows');
sz = max(edgelist(:));
A = sparse(edgelist(:,1), edgelist(:,2), 1, sz, sz);
The first line of code denotes your edge list where each row pair consists of two nodes incident with each other (i.e. they are connected by an edge). The second line removes any duplicate rows from the edge list. The third line determines how big the adjacency matrix should be. We need to figure out what the largest node ID is so that we can allocate a N x N sparse matrix where N is the largest node ID. The last line of code simply uses the first column and second column of the edge list to populate the entries in the sparse matrix, we set them to one and ensure that the size of the matrix is N x N.
We get this:
>> A
A =
(1,2) 1
(2,3) 1
(2,4) 1
You can optionally convert the matrix to full by using the full function:
>> full(A)
ans =
0 1 0 0
0 0 1 1
0 0 0 0
0 0 0 0
As you can see, this corresponds with your desired result.
Example with weights in the third column
edgelist = [1 2 0.1;2 3 0.2;2 4 0.3];
edgelist = unique(edgelist, 'rows');
sz = max(max(edgelist(:, 1:2)));
A = sparse(edgelist(:,1), edgelist(:,2), edgelist(:,3), sz, sz);
Same code as before, but you're changing the third parameter to sparse with the third column of edgelist.
This is what we get:
>> A
A =
(1,2) 0.1000
(2,3) 0.2000
(2,4) 0.3000
>> full(A)
ans =
0 0.1000 0 0
0 0 0.2000 0.3000
0 0 0 0
0 0 0 0
An adjacency matrix should only contain boolean values to indicate an edge is present between vertices. I think this function assumed the third column of el is all ones. In the comments it is clarified that perhaps they are in fact weights. The function can be simplified too. Here's the modified code:
function adj=edgeL2adj(el)
nodes=unique(el(:, 1:2)); % get all nodes, sorted
adj=zeros(numel(nodes)); % initialize adjacency matrix
% across all edges
for i=1:size(el,1)
adj(nodes==el(i,1),(nodes==el(i,2)))=1;
% if third column is weights,
% adj(nodes==el(i,1),(nodes==el(i,2)))=el(i, 3);
end
i want to find same values of number in different column,
for example i have a matrix array:
A = [1 11 0.17
2 1 78
3 4 90
45 5 14
10 10 1]
so as you can see no. 1 in column 1 have the same values in column 2 and column 3, so i want to pick that number and put into another cell or matrix cell
B= [1]
and perform another operation C/B, letting C is equal to:
C= [1
3
5
7
9]
and you will have:
D= [1 11 0.17 1
2 1 78 3
3 4 90 5
45 5 14 7
10 10 1 9]
then after that, values in column 4 have equivalent numbers that we can define, but we will choose only those number that have number 1, or B in theirs row
define:
1-->23
3 -->56
9 --> 78
then we have, see image below:
so how can i do that? is it possible? thanks
Let's tackle your problem into steps.
Step #1 - Determine if there is a value shared by all columns
We can do this intelligently by bsxfun, unique, permute and any and all.
We first need to use unique so that we can generate all possible unique values in the matrix A. Once we do this, we can look at each value of the unique values and see if all columns in A contain this value. If this is the case, then this is the number we need to focus on.
As such, do something like this first:
Aun = unique(A);
eqs_mat = bsxfun(#eq, A, permute(Aun, [3 2 1]));
eqs_mat would generate a 3D matrix where each slice figures out where a particular value in the unique array appeared. As such, for each slice, each column will have a bunch of false values but at least one true value where this true value tells you the position in the column that matched a unique value. The next thing you'll want to do is go through each slice of this result and determine whether there is at least one non-zero value for each column.
For a value to be shared along all columns, a slice should have a non-zero value per column.
We can eloquently determine which value we need to extract by:
ind = squeeze(all(any(eqs_mat,1),2));
Given your example data, we have this for our unique values:
>> B
B =
0.1700
1.0000
2.0000
3.0000
4.0000
5.0000
10.0000
11.0000
14.0000
45.0000
78.0000
90.0000
Also, the last statement I executed above gives us:
>> ind
ind =
0
1
0
0
0
0
0
0
0
0
0
0
The above means that the second location of the unique array is the value we want, and this corresponds to 1. Therefore, we can extract the particular value we want by:
val = Aun(ind);
val contains the value that is shared along all columns.
Step #2 - Given the value B, take a vector C and divide by B.
That's pretty straight forward. Make sure that C is the same size as the total number of rows as A, so:
C = [1 3 5 7 9].';
B = val;
col = C / B;
Step #3 - For each location in A that shares the common value, we want to generate a new fifth column that gives a new value for each corresponding row.
You can do that by declaring a vector of... say... zeroes, then find the right rows that share the common value and replace the values in this fifth column with the values you want:
zer = zeros(size(A,1), 1);
D = [23; 56; 78];
ind2 = any(A == val, 2);
zer(ind2) = D;
%// Create final matrix
fin = [A col zer];
We finally get:
>> fin
fin =
1.0000 11.0000 0.1700 1.0000 23.0000
2.0000 1.0000 78.0000 3.0000 56.0000
3.0000 4.0000 90.0000 5.0000 0
45.0000 5.0000 14.0000 7.0000 0
10.0000 10.0000 1.0000 9.0000 78.0000
Take note that you need to make sure that what you're assigning to the fifth column is the same size as the total number of columns in A.