Assume I have a data like this:
x = np.random.randn(4, 100000)
and I fit a histogram
hist = np.histogramdd(x, density=True)
What I want is to get the probability of number g, e.g. g=0.1. Assume some hypothetical function foo then.
g = 0.1
prob = foo(hist, g)
print(prob)
>> 0.2223124214
How could I do something like this, where I get probability back for a single or a vector of numbers for a fitted histogram? Especially histogram that is N-dimensional.
histogramdd takes O(r^D) memory, and unless you have a very large dataset or very small dimension you will have a poor estimate. Consider your example data, 100k points in 4-D space, the default histogram will be 10 x 10 x 10 x 10, so it will have 10k bins.
x = np.random.randn(4, 100000)
hist = np.histogramdd(x.transpose(), density=True)
np.mean(hist[0] == 0)
gives something arround 0.77 meaning that 77% of the bins in the histogram have no points.
You probably want to smooth the distribution. Unless you have a good reason to not do, I would suggest you to use Gaussian kernel-density Estimate
x = np.random.randn(4, 100000) # d x n array
f = scipy.stats.gaussian_kde(x) # d-dimensional PDF
f([1,2,3,4]) # evaluate the PDF in a given point
I have N Gaussian distributions (multivariate) with N different means (covariance is the same for all of them) in D dimensions.
I also have N evaluation points, where I want to evaluate each of these (log) PDFs.
This means I need to get a NxN matrix, call it "kernels". That is, the (i,j)-th entry is the j-th Gaussian evaluated at the i-th point. A naive approach is:
from torch.distributions.multivariate_normal import MultivariateNormal
import numpy as np
# means contains all N means as rows and is thus N x D
# same for eval_points
# cov is not a problem , just a DxD matrix that is equal for all N Gaussians
kernels = np.empty((N,N))
for i in range(N):
for j in range(N):
kernels[i][j] = MultivariateNormal(means[j], cov).log_prob(eval_points[i])
Now one for loop we can get rid of easily, since for example if we wanted all the evaluations of the first Gaussian , we simply do:
MultivariateNormal(means[0], cov).log_prob(eval_points).squeeze()
and this gives us a N x 1 list of values, that is the first Gaussian evaluated at all N points.
My problem is that , in order to get the full N x N matrix , this doesn't work:
kernels = MultivariateNormal(means, cov).log_prob(eval_points).squeeze()
It doesn't figure out that it should evaluate each mean with all evaluation points in eval_points, and it doesn't return a NxN matrix with these which would be what I want. Therefore, I am not able to get rid of the second for loop, over all N Gaussians.
You are passing wrong shaped tensors to MultivariateNormal's constructor. You should pass a collection of mean vectors of shape (N, D) and a collection of precision matrix cov of shape (N, D, D) for N D-dimensional gaussian.
You are passing mu of shape (N, D) but your precision matrix is not well-shaped. You will need to repeat the precision matrix N number of times before passing it to the MultivariateNormal constructor. Here's one way to do it.
N = 10
D = 3
# means contains all N means as rows and is thus N x D
# same for eval_points
# cov is not a problem , just a DxD matrix that is equal for all N Gaussians
mu = torch.from_numpy(np.random.randn(N, D))
cov = torch.from_numpy(make_spd_matrix(D, D))
cov_n = cov[None, ...].repeat_interleave(N, 0)
assert cov_n.shape == (N, D, D)
kernels = MultivariateNormal(mu, cov_n)
I have two 2-D matrices which have a shared axis.
I want to get a 3-D array that holds the results of every pairwise multiplication made between all the combinations of vectors from each matrix along that shared axis.
What is the best way to achieve this? (assuming that the matrices are big)
As an illustration, let's say I have 100 technicians and 1000 customers.
For each of these individuals I have a 1-D array with ones and zeros representing their availability on a each day of the week.
That's a 7x100 matrix for the technicians, a 7x1000 matrix for the customers.
import numpy as np
technicians = np.random.randint(low=0,high=2,size=(7,100))
customers = np.random.randint(low=0,high=2,size=(7,1000))
result = solution(technicians, customers)
result.shape # (7,100,1000)
I want to find for each technician-customer couple the days they are both available.
If I perform a pairwise multiplication between each combination of technician availability and customer availability I get a 1-D arrays that shows for each couple whether they are both available on these days. Together they create the 3-D array I'm aiming for, shaped something like 7x100x1000.
Thanks!
Try
ans = technicians.reshape((7, 1, 100)) * customers.reshape((7, 1000, 1))
We make use of numpy.broadcasting.
General Broadcasting Rules: When operating on two arrays, NumPy
compares their shapes element-wise. It starts with the trailing
dimensions, and works its way forward. Two dimensions are compatible
when
(1) they are equal, or (2) one of them is 1
Now, we are matching the shape of technicians and customers as
technician : 7 x 1 x 100
customers : 7 x 1000 x 1
Result (3d array): 7 x 1000 x 100
using reshape. Then, we can apply elementwise multiplication with *.
I have a function, say peaksdetect(), that will generate a 2-D array of unknown number of rows; I will call it a few times, let's say 3 and I would like to make of these 3 arrays, one 3-D array. Here is my start but it is very complicated with a lot of if statements, so I want to make things simpler if possible:
import numpy as np
dim3 = 3 # the number of times peaksdetect() will be called
# it is named dim3 because this number will determine
# the size of the third dimension of the result 3-D array
for num in range(dim3):
data = peaksdetect(dataset[num]) # generates a 2-D array of unknown number of rows
if num == 0:
3Darray = np.zeros([dim3, data.shape]) # in fact the new dimension is in position 0
# so dimensions 0 and 1 of "data" will be
# 1 and 2 respectively
else:
if data.shape[0] > 3Darray.shape[1]:
"adjust 3Darray.shape[1] so that it equals data[0] by filling with zeroes"
3Darray[num] = data
else:
"adjust data[0] so that it equals 3Darray.shape[1] by filling with zeroes"
3Darray[num] = data
...
If you are counting on having to resize your array, there is very likely not going to be much to be gained by preallocating it. It will probably be simpler to store your arrays in a list, then figure out the size of the array to hold them all, and dump the data into it:
data = []
for num in range(dim3):
data.append(peaksdetect(dataset[num]))
shape = map(max, zip(*(j.shape for j in data)))
shape = (dim3,) + tuple(shape)
data_array = np.zeros(shape, dtype=data[0].dtype)
for j, d in enumerate(data):
data_array[j, :d.shape[0], :d.shape[1]] = d
I am looking for algorithm to solve the following problem :
I have two sets of vectors, and I want to find the matrix that best approximate the transformation from the input vectors to the output vectors.
vectors are 3x1, so matrix is 3x3.
This is the general problem. My particular problem is I have a set of RGB colors, and another set that contains the desired color. I am trying to find an RGB to RGB transformation that would give me colors closer to the desired ones.
There is correspondence between the input and output vectors, so computing an error function that should be minimized is the easy part. But how can I minimize this function ?
This is a classic linear algebra problem, the key phrase to search on is "multiple linear regression".
I've had to code some variation of this many times over the years. For example, code to calibrate a digitizer tablet or stylus touch-screen uses the same math.
Here's the math:
Let p be an input vector and q the corresponding output vector.
The transformation you want is a 3x3 matrix; call it A.
For a single input and output vector p and q, there is an error vector e
e = q - A x p
The square of the magnitude of the error is a scalar value:
eT x e = (q - A x p)T x (q - A x p)
(where the T operator is transpose).
What you really want to minimize is the sum of e values over the sets:
E = sum (e)
This minimum satisfies the matrix equation D = 0 where
D(i,j) = the partial derivative of E with respect to A(i,j)
Say you have N input and output vectors.
Your set of input 3-vectors is a 3xN matrix; call this matrix P.
The ith column of P is the ith input vector.
So is the set of output 3-vectors; call this matrix Q.
When you grind thru all of the algebra, the solution is
A = Q x PT x (P x PT) ^-1
(where ^-1 is the inverse operator -- sorry about no superscripts or subscripts)
Here's the algorithm:
Create the 3xN matrix P from the set of input vectors.
Create the 3xN matrix Q from the set of output vectors.
Matrix Multiply R = P x transpose (P)
Compute the inverseof R
Matrix Multiply A = Q x transpose(P) x inverse (R)
using the matrix multiplication and matrix inversion routines of your linear algebra library of choice.
However, a 3x3 affine transform matrix is capable of scaling and rotating the input vectors, but not doing any translation! This might not be general enough for your problem. It's usually a good idea to append a "1" on the end of each of the 3-vectors to make then a 4-vector, and look for the best 3x4 transform matrix that minimizes the error. This can't hurt; it can only lead to a better fit of the data.
You don't specify a language, but here's how I would approach the problem in Matlab.
v1 is a 3xn matrix, containing your input colors in vertical vectors
v2 is also a 3xn matrix containing your output colors
You want to solve the system
M*v1 = v2
M = v2*inv(v1)
However, v1 is not directly invertible, since it's not a square matrix. Matlab will solve this automatically with the mrdivide operation (M = v2/v1), where M is the best fit solution.
eg:
>> v1 = rand(3,10);
>> M = rand(3,3);
>> v2 = M * v1;
>> v2/v1 - M
ans =
1.0e-15 *
0.4510 0.4441 -0.5551
0.2220 0.1388 -0.3331
0.4441 0.2220 -0.4441
>> (v2 + randn(size(v2))*0.1)/v1 - M
ans =
0.0598 -0.1961 0.0931
-0.1684 0.0509 0.1465
-0.0931 -0.0009 0.0213
This gives a more language-agnostic solution on how to solve the problem.
Some linear algebra should be enough :
Write the average squared difference between inputs and outputs ( the sum of the squares of each difference between each input and output value ). I assume this as definition of "best approximate"
This is a quadratic function of your 9 unknown matrix coefficients.
To minimize it, derive it with respect to each of them.
You will get a linear system of 9 equations you have to solve to get the solution ( unique or a space variety depending on the input set )
When the difference function is not quadratic, you can do the same but you have to use an iterative method to solve the equation system.
This answer is better for beginners in my opinion:
Have the following scenario:
We don't know the matrix M, but we know the vector In and a corresponding output vector On. n can range from 3 and up.
If we had 3 input vectors and 3 output vectors (for 3x3 matrix), we could precisely compute the coefficients αr;c. This way we would have a fully specified system.
But we have more than 3 vectors and thus we have an overdetermined system of equations.
Let's write down these equations. Say that we have these vectors:
We know, that to get the vector On, we must perform matrix multiplication with vector In.In other words: M · I̅n = O̅n
If we expand this operation, we get (normal equations):
We do not know the alphas, but we know all the rest. In fact, there are 9 unknowns, but 12 equations. This is why the system is overdetermined. There are more equations than unknowns. We will approximate the unknowns using all the equations, and we will use the sum of squares to aggregate more equations into less unknowns.
So we will combine the above equations into a matrix form:
And with some least squares algebra magic (regression), we can solve for b̅:
This is what is happening behind that formula:
Transposing a matrix and multiplying it with its non-transposed part creates a square matrix, reduced to lower dimension ([12x9] · [9x12] = [9x9]).
Inverse of this result allows us to solve for b̅.
Multiplying vector y̅ with transposed x reduces the y̅ vector into lower [1x9] dimension. Then, by multiplying [9x9] inverse with [1x9] vector we solved the system for b̅.
Now, we take the [1x9] result vector and create a matrix from it. This is our approximated transformation matrix.
A python code:
import numpy as np
import numpy.linalg
INPUTS = [[5,6,2],[1,7,3],[2,6,5],[1,7,5]]
OUTPUTS = [[3,7,1],[3,7,1],[3,7,2],[3,7,2]]
def get_mat(inputs, outputs, entry_len):
n_of_vectors = inputs.__len__()
noe = n_of_vectors*entry_len# Number of equations
#We need to construct the input matrix.
#We need to linearize the matrix. SO we will flatten the matrix array such as [a11, a12, a21, a22]
#So for each row we combine the row's variables with each input vector.
X_mat = []
for in_n in range(0, n_of_vectors): #For each input vector
#populate all matrix flattened variables. for 2x2 matrix - 4 variables, for 3x3 - 9 variables and so on.
base = 0
for col_n in range(0, entry_len): #Each original unknown matrix's row must be matched to all entries in the input vector
row = [0 for i in range(0, entry_len ** 2)]
for entry in inputs[in_n]:
row[base] = entry
base+=1
X_mat.append(row)
Y_mat = [item for sublist in outputs for item in sublist]
X_np = np.array(X_mat)
Y_np = np.array([Y_mat]).T
solution = np.dot(np.dot(numpy.linalg.inv(np.dot(X_np.T,X_np)),X_np.T),Y_np)
var_mat = solution.reshape(entry_len, entry_len) #create square matrix
return var_mat
transf_mat = get_mat(INPUTS, OUTPUTS, 3) #3 means 3x3 matrix, and in/out vector size 3
print(transf_mat)
for i in range(0,INPUTS.__len__()):
o = np.dot(transf_mat, np.array([INPUTS[i]]).T)
print(f"{INPUTS[i]} x [M] = {o.T} ({OUTPUTS[i]})")
The output is as such:
[[ 0.13654096 0.35890767 0.09530002]
[ 0.31859558 0.83745124 0.22236671]
[ 0.08322497 -0.0526658 0.4417611 ]]
[5, 6, 2] x [M] = [[3.02675088 7.06241873 0.98365224]] ([3, 7, 1])
[1, 7, 3] x [M] = [[2.93479472 6.84785436 1.03984767]] ([3, 7, 1])
[2, 6, 5] x [M] = [[2.90302805 6.77373212 2.05926064]] ([3, 7, 2])
[1, 7, 5] x [M] = [[3.12539476 7.29258778 1.92336987]] ([3, 7, 2])
You can see, that it took all the specified inputs, got the transformed outputs and matched the outputs to the reference vectors. The results are not precise, since we have an approximation from the overspecified system. If we used INPUT and OUTPUT with only 3 vectors, the result would be exact.