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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 a ndarray (y) including the existing labels. These labels are in binary 0 or 1.
I have a second ndarray (X) representing the training set.
How can extract from X and y all the data corresponding to label 0. So, the new data will be:
the original data: X, y
and another extracted X1 with its corresponding y1.
You can use:
X1, y1 = X[y==0], y[y==0]
Let's say I have x-y data samples sorted by x-value. I'm going to use Pandas as example, but I would be perfectly happy with a Numpy/Scipy-only solution, of course.
In [24]: pd.set_option('display.max_rows', 10)
In [25]: df = pd.DataFrame(np.random.randn(100, 2), columns=['x', 'y'])
In [26]: df = df.sort('x')
In [27]: df
Out[27]:
x y
13 -3.403818 0.717744
49 -2.688876 1.936267
74 -2.388332 -0.121599
52 -2.185848 0.617896
90 -2.155343 -1.132673
.. ... ...
65 1.736506 -0.170502
0 1.770901 0.520490
60 1.878376 0.206113
63 2.263602 1.112115
33 2.384195 -1.877502
[100 rows x 2 columns]
Now, I want to kind of "window" it or "discretize" it and get statistics on each window. But I don't want to do the Pandas moving-window functions because they define windows by rows. I want to define windows by a span of x-values, thus "x-value-window". Specifically, let's define each x-value-window with 2 parameters:
center x-value of each window
in this example, let's say I want x = 0.0 + 0.4 * k for all positive or negative k
thus -3.2, -2.8, -2.4, ..., 1.6, 2.0, 2.4
width of each window
in this example, let's say I want W = 0.5
thus, the example windows will be [-3.2-0.25, -3.2+0.25], [-2.8-0.25, -2.8+0.25], ..., [2.4-0.25, 2.4+0.25]
note that the windows overlap, which is intended
Having thus defined the windows, I would like to ask if there's a function that will produce the following data frame (or numpy array):
x y
-3.2 mean of y-values in x-value-window centered at -3.2
-2.8 mean of y-values in x-value-window centered at -2.8
-2.4 mean of y-values in x-value-window centered at -2.4
... ...
1.6 mean of y-values in x-value-window centered at 1.6
2.0 mean of y-values in x-value-window centered at 2.0
2.4 mean of y-values in x-value-window centered at 2.4
Is there anything that will do this for me? Or do I have to totally roll my own (and probably in a very slow python loop instead of fast numpy or pandas code)?
Extra 1: It would be even better if there's support for weighted windows (such as supported by Pandas's rolling_window function) but of course the weights in this case would not be based on how far the sample's row is from the center row of the window, but rather, how far the sample's x-value is from the center of the x-value-window.
Extra 2: It would be nice if there's support for statistics other than mean on the x-value-windows, e.g. (a) variance of the y-values in each x-value-window or (b) count of the number of samples falling within each x-value-window.
I first create a range of x values centered at zero. This range is wide enough so that then min value minus the width and the max value plus the width will capture all x values.
I then iterate through this range of x values which have k as the step size. At each point, I use loc to capture y values located at the selected x value plus and minus the width. The mean of these selected values are then calculated. These values are used to create the result dataframe.
import math
import numpy as np
import pandas as pd
k = .4
w = .5
np.random.seed(0)
df = pd.DataFrame(np.random.randn(100, 2), columns=['x', 'y'])
x_range = np.arange(math.floor((df.x.min() + w) / k) * k,
k * (math.ceil((df.x.max() - w) / k) + 1), k)
result = pd.DataFrame((df.loc[df.x.between(x - w, x + w), 'y'].mean() for x in x_range),
index=x_range, columns=['y_mean'])
result.index.name = 'centered_x'
>>> result
y_mean
centered_x
-2.400000e+00 0.653619
-2.000000e+00 0.733606
-1.600000e+00 0.576594
-1.200000e+00 0.150462
-8.000000e-01 0.065884
-4.000000e-01 0.022925
-8.881784e-16 0.211693
4.000000e-01 0.057527
8.000000e-01 -0.141970
1.200000e+00 0.233695
1.600000e+00 0.203570
2.000000e+00 0.306409
2.400000e+00 0.576789
In the case of a matrix mat n x n, i can do the following
sym = 0.5 * (mat + mat.T)
the operation gives the desired result sym[i,j] = sym[j,i]
Suppose we have a 3D array ndarr[i,j,k], where i,j,k 0,1,...n,
then ndarr is n x n x n. The idea is to obtain the following "symmetric" form
nsym[i,j,k] = nsym[j,i,k] using ndarr. I tried this:
import numpy as np
# Generate some random matrix, n = 5
ndarr = np.random.beta(0.1,1,(5,5,5))
# First attempt to symmetrize
sym1 = np.array([0.5*(ndarr[:,:,k]+ndarr[:,:,k].T) for k in range(5)])
The problem here is that sym1[i,j,k] != sym1[j,i,k] as it is required. In fact I obtain sym1[i,j,k] = sym1[i,k,j], symmetric under the exchange of the last two symbols!
# Second attempt
sym2 = 0.5*(ndarr+ndarr.T)
Same problem here and sym2 is symmetric with respect the second index sym2[i,j,k]=sym2[k,j,i].
To resume, the goal is to find a symmetric form for a 3D array with respect to the third index and to preserve the values in the diagonal for the original ndarr[i,i,i].
The problem here is that you're not using the correct transpose:
sym = 0.5 * (ndarr + np.transpose(ndarr, (1, 0, 2)))
By default, np.transpose and the .T property will reverse the order of the axes. In your case, we want to only flip the first two axes: (0,1,2) -> (1,0,2).
EDIT: The reason your first attempt failed is because you were concatenating each symmetrized matrix along the first axis, not the last. It's more clear if you make ndarr with shape (5, 5, 3):
In [16]: sym = np.array([0.5*(ndarr[:,:,k]+ndarr[:,:,k].T) for k in range(3)])
In [17]: sym.shape
Out[17]: (3L, 5L, 5L)
In any case, the version above with np.transpose is cleaner and more efficient.
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.