I would like to ask a question about matlab transpose symbol. For example in this case:
input=input';
It makes transpose of input but i want to learn why we should use transpose via usin Artificial Neural Network in matlab?
Second Question is:
I am trying to create a classification using ANN in matlab. I showed results like that:
a=sim(neuralnetworkname,test)
test is represens my test data in Neural network.
and the results is like that:
a =
Columns 1 through 12
2.0374 3.9589 3.2162 2.0771 2.0931 3.9947 3.1718 3.9813 2.1528 3.9995 3.8968 3.9808
Columns 13 through 20
3.9996 3.7478 2.1088 3.9932 2.0966 2.0644 2.0377 2.0653
If the result of a is about 2, it would benign, if the result of a is about 4,it is malignant.
So, I want to calculate that :for example,there are 100 benign in 500 data.(100/500) How can i write screen this 100/500
I tried to be clear, but if i didn't clear enough, I can try to explain more.Thanks.
First Question
You don't need to transpose input values everytime. Matlab nntool normally gets input values column by column by default. So you have two choice: 1. Change dataset order 2. Transpose input
Second Question
Suppose you have matrix like this:
a=[1 2 3 4 5 6 7 8 9 0 0 0];
To count how many elements below 8, write this:
sum(a<8) %[1 2 3 4 5 6 7 0 0 0]
Output will be:
10
Related
I want to reduce a two dimensional matrix to row vector.
But using reshape with large matrices is really slow. The other alternative is to use colon, but i want matrix's transpose to be colon and not the matrix itself.
e.g.
A=magic(3)
A =
8 1 6
3 5 7
4 9 2
A(:) will stack up all the columns one by one. but i am looking for something like this:
AA=A(2:3,:)';
and then reshape or colon AA instead of A.
The issue is i dont want to define additional variable like AA.
Is there anyway to reduce dimension of two dimensional matrix without reshape?
You can avoid the additional variable by linear indexing. For your example:
A([2 5 8 3 6 9])
which gives
3 5 7 4 9 2
What's happening here is that you treat A as if it was already transformed into a vector, and the elements of this one-dimensional array are accessed through indices 1 through 9. Using the colon is a special case of linear indexing, A(:) is the same as A(1 : end).
Figuring out the right linear indices can be tricky, but sub2ind can help with that.
It is possible that this slightly speeds up the code, mainly because (as #Shai wrote) you avoid writing data to an intermediate variable. I wouldn't expect too much, though.
Try looking at subsref. For your example, you could use it as follows:
subsref(A',struct('type','()','subs',{{2:3,':'}}))
Update: I misunderstood the original question; I thought the OP wanted to select rows from 2:3 from the transposed A matrix keeping the columns as is. I will keep the previous answer in case it could be useful to others.
I think he/she could use the following to slice and flatten a matrix:
subsref(A(2:3,:)', struct('type','()','subs',{{':'}}))
This would give as output:
[3 5 7 4 9 2]'
I have used SOM tool box in MATLAB or iris data set. following example and using the plotsomhits i can see that how many data values are in each neuron of my neuron grid of the SOM . However I wish to know actual data values which are grouped in every neuron of the given SOM configuration .Is there any way to do it. this is the example I used.
net = selforgmap([8 8]);
view(net)
[net,tr] = train(net,x);
nntraintool
plotsomhits(net,x)
not that hard. plotsomhits just plots "fancily" the results of the simulation of the net.
so if you simulate it and add the "hits" you have the result!
basicaly:
hits=sum(sim(net,x)');
In your net case this was my results, that coincide with the numbers in the plotsomehits
hits= 6 5 0 7 15 7 4 0 8 20 3 3 9 3 0 8 6 3 11 4 5 5 7 10 1
PD: you can learn a lot in this amazing SO answer:
MATLAB: help needed with Self-Organizing Map (SOM) clustering
You need to convert vector to indices first and then you can see what input values a neuron correspond to.
>> input_neuron_mapping = vec2ind(net(x))';
Now, look into the neuron's inputs.
For example, you want to see neuron input values for neuron 2.
>> neuron_2_input_indices = find(input_neuron_mapping == 2)
>> neuron_2_input_values = x(neuron_2_input_indices)
It will display all the input values from your data.
Read here for more details:
https://bioinformaticsreview.com/20220603/how-to-get-input-values-from-som-sample-hits-plot-in-matlab/
I have data that is output from a computational chemistry program (Gaussian09) which contains sets of Force Constant data. The data is arranged with indexes as the first 2-4 columns (quadratic, cubic and quartic FC's are calculated). As an example the cubic FC's look something like this, and MatLab has read them in successfully so I have the correct matrix:
cube=[
1 1 1 5 5 5
1 1 2 6 6 6
.
.
4 1 1 8 8 8
4 2 1 9 9 9
4 3 1 7 7 7 ]
I need a way to access the last 3 columns when feeding in the indices of the first 3 columns. Something along the lines of
>>index=find([cube(:,1)==4 && cube(:,2)==3 && cube(:,3)==1]);
Which would give me the row number of the data that is index [ 4 3 1 ] and allow me to read out the values [7 7 7] which I need within loops to calculate anharmonic frequencies.
Is there a way to do this without a bunch of loops?
Thanks in advance,
Ben
You have already found one way to solve this, by using & in your expression (allowing you to make non-scalar comparisons).
Another way is to use ismember:
index = find(ismember(cube(:,1:3),[4 3 1]));
Note that in many cases, you may not even need the call to find: the binary vector returned by the comparisons or ismember can directly be used to index into another array.
I need to randomly scramble the values of an nx1 matrix in matlab. I'm not sure how to do this efficiently, I need to do it many times for n > 40,000.
Example
Matrix before:
1 2 2 2 3 4 5 5 4 3 2 1
Scrambled:
3 5 2 1 2 2 3 4 1 4 5 2
thank you
If your data is stored in matrix data, then you can generate "scrambled" data using randperm like so:
scrambled = data(randperm(numel(data)));
This is sampling without replacement, so every value in data will appear once in scrambled.
For sampling with replacement (values in data may appear in scrambled multiple times and some may not appear at all), you could use randi like this:
scrambled = data(randi(numel(data),1,numel(data)));
This is a question about Matlab/Octave.
I am seeing some results of the medfilt1(1D Median filter command in matlab) computation by which I am confused.
EDIT:Sorry forgot to mention:I am using Octave for Windows 3.2.4. This is where i see this behavior.
Please see the questions below, and point if I am missing something.
1] I have a 1D data array b=[ 3 5 -8 6 0];
out=medfilt1(b,3);
I expected the output to be [3 3 5 0 0] but it is showing the output as [4 3 5 0 3]
How come? What is wrong here?
FYI-Help says it pads the data at boundaries by 0(zero).
2] How does medfilt2(2D median filter command in matlab) work.
Help says "Each output pixel contains the median value in the m-by-n neighborhood around the corresponding pixel in the input image".
For m=3,n=3, So does it calculate a 3x3 matrix MAT for each of input pixels placed at its center and do median(median(MAT)) to compute its median value in the m-by-n neighbourhood?
Any pointers will help.
thank you. -AD
I was not able to replicate your error with Matlab 7.11.0, but from the information in your question it seems like your version of medfilt1 does not differentiate between an odd or even n.
When finding the median in a vector of even length, one usually take the mean of the two median values,
median([1 3 4 5]) = (3+4)/2 = 3.5
This seems to be what happens in your case. Instead of treating n as odd, and setting the value to be 3, n is treated as even and your first out value is calculated to be
median([0 3 5]) = (3+5)/2 = 4
and so on.. EDIT: This only seems to happen in the endpoints, which suggest that the padding with zeros is not properly working in your Octave code.
For your second question, you are almost right, it calculates a 3x3 matrix in each center, but it does not do median(median(MAT)), but median(MAT(:)). There is a difference!
A = [1 2 3
14 5 33
11 7 13];
median(median(A)) = 11
median(A(:)) = 7