I'm trying to build a neural network and am following a tutorial.
What do those two lines mean?
syn0 = 2*np.random.random((3,4)) -1
syn1 = 2*np.random.random((4,1)) -1
Specifically, Those values (3,4 | 4,1)
I just don't get it...
I think I know what the first synapse's values mean, but not the 2nd one...
np.random.random creates an array of random values between 0 and 1 - the parameter it takes is the desired shape of the array, which is what (3,4) and (4,1) are in your example.
Simple random weight initialisation is sufficient to train your neural network, but initialising them with a mean of 0 speeds up training, which is what 2*np.random.random((3,4)) -1 does:
np.random.random((3,4)) // array with values in range [0, 1) and mean of 0.5
2 * np.random.random((3,4)) // array with values in range [0, 2) and mean of 1
2 * np.random.random((3,4)) - 1 // array with values in range [-1, 1) and mean of 0
Related
From a binary matrix, I want to calculate a kind of adjacency/joint probability density matrix (not quite sure how to label it as so please feel free to rename).
For example, I start with this matrix:
A = [1 1 0 1 1
1 0 0 1 1
0 0 0 1 0]
I want to produce this output:
Output = [1 4/5 1/5
4/5 1 1/5
1/5 1/5 1]
Basically, for each row, I want to calculate the proportion of times where they agreed (1 and 1 or 0 and 0). A will always agree with itself and thus have it as 1 along the diagonal. No matter how many different js are added it will still result in a 3x3, but an extra i variable will result in a 4x4.
I like to think of the inputs along i in the A matrix as the person and Js as the question and so the final output is a 3x3 (number of persons) matrix.
I am having some trouble with this on matlab. If you could please help point me in the right direction that would be fabulous.
So, you can do this in two parts.
bothOnes = A*A';
gives you a matrix showing how many 1s each pair of rows share, and
bothZeros = (1-A)*(1-A)';
gives you a matrix showing how many 0s each pair of rows share.
If you just add them up, you get how many elements they share of either type:
bothSame = A*A' + (1-A)*(1-A)';
Then just divide by the row length to get the desired fractional representation:
output = (A*A' + (1-A)*(1-A)') / size(A, 2);
That should get you there.
Note that this only works if A contains only 1's and 0's, but it can be adapted for other cases.
Here are some alternatives, assuming A can only contain 0 and 1:
If you have the Statistics Toolbox:
result = 1-squareform(pdist(A, 'hamming'));
Manual approach with implicit expansion:
result = mean(permute(A, [1 3 2])==permute(A, [3 1 2]), 3);
Using bitwise operations. This is a more esoteric approach, and is only valid if A has at most 53 columns, due to floating-point limitations:
t = bin2dec(char(A+'0')); % convert each row from binary to decimal
u = bitxor(t, t.'); % bitwise xor
v = mean(dec2bin(u)-'0', 2); % compute desired values
result = 1 - reshape(v, size(A,1), []); % reshape to obtain result
I have two matrices.
mcaps which is a double 1698 x 2
index_g which is a logical 1698 x 2
When using the line of code below I get the error message that Index exceeds matrix dimensions. I don't see how this is the case though?
tsp = nansum(mcaps(index_g==1, :));
Update
Sorry I should have mentioned that I need the sum of each column in the mcaps vector
** Example of data **
mcaps index_g
5 6 0 0
4 3 0 0
6 5 1 1
4 6 0 1
8 7 0 0
There are two problems here. I missed one. Original answer is below.
What I missed is that when you use the logical index in this way, you are picking out elements of the matrix that may have different numbers of elements in each column, so MATLAB can't return a well formed matrix back to nansum, and so returns a vector. To get around this, use the fact that 0 + anything = 0
% create a mask of values you don't want to sum. Note that since
% index_g is already logical, you don't have to test equal to 1.
mask = ~index_g & isnan(mcaps)
% create a temporary variable
mcaps_to_sum = mcaps;
% change all of the values that you don't want to sum to zero
mcaps_to_sum(mask) = 0;
% do the sum
sum(mcaps_to_sum,1);
This is basically all that the nansum function does internally, is to set all of the NaN values to zero and then call the sum function.
index_g == 1 returns a 1698 x 2 logical matrix, but then you add in an extra dimension with the colon. To sum the columns, use the optional dim input. You want:
tsp = nansum(mcaps(index_g == 1),1);
i have a question regarding histc:
I choose the max and min of a sorted signal as my range.
ma = ssigPE(end);
mi = ssigPE(1);
range = mi:ma;
[bincountsO,indO2] = histc(ssigPE, range);
so the range i get back is:
range = [-1.097184703736132 -0.097184703736132 0.902815296263868]
my problem is that just 2 bins get develop, so bincountsO has 2 bins
and indO2 has values as 0, 1 and 2
What am I doing wrong? I guess I m using the range wrong. I read the text here:
http://de.mathworks.com/help/matlab/ref/histc.html#inputarg_binranges
but I don't get it.
The bin ranges tell you where do bins start and stop. So a value of [0 1 2 7]for example, will give 3 bins: [0 1] , [1 2] , [2 7]
In matlab if you do mi:ma it will create an array from the value mi to ma with a step of 1. With your values, that gives just 3 values, hence 2 bins. There are 2 ways of creating a given step size length vectors.
Step size if 100 as an example
range=mi:(ma-mi)/100:ma;
alternatively, and way clearer
range=linspace(mi,ma,100)
I have approximately 5000 integer vectors (=SIZE) that look like:
[1 0 4 2 0 1 3 ...]
They have the same length N=32 and their values ranges from 0 to 4 but let's say [0 MAX].
I created a NN that takes vectors as inputs and outputs a binary array corresponding to one of the desired output(number of possible outputs = M):
for instance [0 1 0 0 ...0] => 2nd output. array_length = M
I used a Multi Layer Perceptron in Neuroph with those integer values but it did not converge.
So I am guessing the problem is using integer values or using a MLP with 3 layers: input, hidden and output.
Can you advise me on the network structure? which type of NN is suitable? Should I remodel the input and output to simplify the learning process? I have been thinking about Gray encoding for the integers input.
Ive got a vector like this
a=[0 5 3 0 1]
and a corresponding vector, containing the same amount of numbers as there are zeros in the first vector
b=[2 4]
what I want to get is
x=[2 5 3 4 1]
I tried fiddling around with, and somewhat got the feeling that the find / full methods might help me here, but didn't get it to work
c=(a==0)
>[1 0 0 1 0]
Thank you!
It is as easy as this:
x=a;
As x==0 gives the vector of all the locations an element = 0, ie [0 1 0 0 1], x(x==0) is indexing x to get the actual elements of x that are equal to 0, which you can then assign values as if it were any other vector/matrix (where the values we are not interested in do not exist, and are not indexed), using the following:
x(x==0)=b;