I have a for loop that gives me an output of 16 x 8 2D arrays per entry in the loop. I want to stack all of these 2D arrays along the z-axis in a 3D array. This way, I can determine the variance over the z-axis. I have tried multiple commands, such as np.dstack, matrix3D[p,:,:] = ... and np.newaxis both in- and outside the loop. However, the closest I've come to my desired output is just a repetition of the last array stacked on top of each other. Also the dimensions were way off. I need to keep the original 16 x 8 format. By now I'm in a bit too deep and could use some nudge in the right direction!
My code:
excludedElectrodes = [1,a.numberOfColumnsInArray,a.numberOfElectrodes-a.numberOfColumnsInArray+1,a.numberOfElectrodes]
matrixEA = np.full([a.numberOfRowsInArray, a.numberOfColumnsInArray], np.nan)
for iElectrode in range(a.numberOfElectrodes):
if a.numberOfDeflectionsPerElectrode[iElectrode] != 0:
matrixEA[iElectrode // a.numberOfColumnsInArray][iElectrode % a.numberOfColumnsInArray] = 0
for iElectrode in range (a.numberOfElectrodes):
if iElectrode+1 not in excludedElectrodes:
"""Preprocessing"""
# Loop over heartbeats
for p in range (1,len(iLAT)):
# Calculate parameters, store them in right row-col combo (electrode number)
matrixEA[iElectrode // a.numberOfColumnsInArray][iElectrode % a.numberOfColumnsInArray] = (np.trapz(abs(correctedElectrogram[limitA[0]:limitB[0]]-totalBaseline[limitA[0]:limitB[0]]))/(1000))
# Stack all matrixEA arrays along z axis
matrix3D = np.dstack(matrixEA)
This example snippet does what you want, although I suspect your errors have to do more with things not relative to the concatenate part. Here, we use the None keyword in the array to create a new empty dimension (along which we concatenate the 2D arrays).
import numpy as np
# Function does create a dummy (16,8) array
def foo(a):
return np.random.random((16,8)) + a
arrays2D = []
# Your loop
for i in range(10):
# Calculate your (16,8) array
f = foo(i)
# And append it to the list
arrays2D.append(f)
# Stack arrays along new dimension
array3D = np.concatenate([i[...,None] for i in arrays2D], axis = -1)
Basically, I am looking for the numpy primitives which will accomplish the following for loop impementation:
# for a matrix M
argmaxes = np.argmax(M,axis=1)
for i,arg in enumerate(argmaxes):
M[i,arg:] = M[i,arg]
Is there a numpy-ish way to accomplish this?
The following doesn't use for loop, but it creates several intermediate arrays of the same shape as M. So I'm not sure how efficient it is over loop:
maxes = np.max(M, axis=1)
M = np.where(np.arange(M.shape[1]) > argmaxes[:,None],
maxes[:,None],
M)
I'm trying to build a matrix one row at a time.
import numpy as np
f = np.matrix([])
f = np.vstack([ f, np.matrix([1]) ])
This is the error message.
ValueError: all the input array dimensions for the concatenation axis must match exactly, but along dimension 1, the array at index 0 has size 0 and the array at index 1 has size 1
As you can see, np.matrix([]) is NOT an empty list. I'm going to have to do this some other way. But what? I'd rather not do an ugly workaround kludge.
you have to pass some dimension to the initial matrix. Either fill it with some zeros or use np.empty():
f = np.empty(shape = [1,1])
f = np.vstack([f,np.matrix([1])])
you can use np.hstack instead for the first case, then use vstack iteratively.
arr = np.array([])
arr = np.hstack((arr, np.array([1,1,1])))
arr = np.vstack((arr, np.array([2,2,2])))
Now you can convert into a matrix.
mat = np.asmatrix(arr)
Good grief. It appears there is no way to do what I want. Kludgetown it is. I'll build an array with a bogus first entry, then when I'm done make a copy without the bogosity.
For each location in the result matrix, instead of storing the dot product of the corresponding row and column in the argument matrices, I would like like to store the element wise product, which will be a vector extending into a third dimension.
One idea would be to convert the argument matrices to vectors with vector entries, and then take their outer product, but I'm not sure how to do this either.
EDIT:
I figured it out before I saw there was a reply. Here is my solution:
def newdot(A, B):
A = A.reshape((1,) + A.shape)
B = B.reshape((1,) + B.shape)
A = A.transpose(2, 1, 0)
B = B.transpose(1, 0, 2)
return A * B
What I am doing is taking apart each row and column pair that will have their outer product taken, and forming two lists of them, which then get their contents matrix multiplied together in parallel.
It's a little convoluted (and difficult to explain) but this function should get you what you're looking for:
def f(m1, m2):
return (m2.A.T * m1.A.reshape(m1.shape[0],1,m1.shape[1]))
m3 = m1 * m2
m3_el = f(m1, m2)
m3[i,j] == sum(m3_el[i,j,:])
m3 == m3_el.sum(2)
The basic idea is to turn the matrices into arrays and do element-by-element multiplication. One of the arrays gets reshaped to have a size of one in its middle dimension, and array broadcasting rules expand this dimension out to match the height of the other array.
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.