How do I detect multivariate outliers within large data with more than 50 variables. Do i need to plot all of the variables or do i have to group them based independent and dependent variables or do i need an algorithm for this?
We do have a special type of distance formula that we use to find multivariate outliers. It is called Mahalanobis Distance.
The MD is a metric that establishes the separation between a distribution D and a data point x by generalizing the z-score, the MD indicates how far x is from the D mean in terms of standard deviations.
You can use the below function to find out outliers. It returns the index of outliers.
from scipy.stats import chi2
import scipy as sp
import numpy as np
def mahalanobis_method(df):
#M-Distance
x_minus_mean = df - np.mean(df)
cov = np.cov(df.values.T) #Covariance
inv_covmat = sp.linalg.inv(cov) #Inverse covariance
left_term = np.dot(x_minus_mean, inv_covmat)
mahal = np.dot(left_term, x_minus_mean.T)
md = np.sqrt(mahal.diagonal())
#Flag as outliers
outliers = []
#Cut-off point
C = np.sqrt(chi2.ppf((1-0.001), df=df.shape[1])) #degrees of freedom = number of variables
for i, v in enumerate(md):
if v > C:
outliers.append(i)
else:
continue
return outliers, md
If you want to study more about Mahalanobis Distance and its formula you can read this blog.
So, how to understand the above formula? Let’s take the (x – m)^T . C^(-1) term. (x – m) is essentially the distance of the vector from the mean. We then divide this by the covariance matrix (or multiply by the inverse of the covariance matrix). If you think about it, this is essentially a multivariate equivalent of the regular standardization (z = (x – mu)/sigma).
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 2d point clouds (oldPts and newPts) which I whish to combine. They are mx2 and nx2 numpyinteger arrays with m and n of order 2000. newPts contains many duplicates or near duplicates of oldPts and I need to remove these before combining.
So far I have used the histogram2d function to produce a 2d representation of oldPts (H). I then compare each newPt to an NxN area of H and if it is empty I accept the point. This last part I am currently doing with a python loop which i would like to remove. Can anybody show me how to do this with broadcasting or perhaps suggest a completely different method of going about the problem. the working code is below
npzfile = np.load(path+datasetNo+'\\temp.npz')
arrs = npzfile.files
oldPts = npzfile[arrs[0]]
newPts = npzfile[arrs[1]]
# remove all the negative values
oldPts = oldPts[oldPts.min(axis=1)>=0,:]
newPts = newPts[newPts.min(axis=1)>=0,:]
# round to integers
oldPts = np.around(oldPts).astype(int)
newPts = newPts.astype(int)
# put the oldPts into 2d array
H, xedg,yedg= np.histogram2d(oldPts[:,0],oldPts[:,1],
bins = [xMax,yMax],
range = [[0, xMax], [0, yMax]])
finalNewList = []
N = 5
for pt in newPts:
if not H[max(0,pt[0]-N):min(xMax,pt[0]+N),
max(0,pt[1]- N):min(yMax,pt[1]+N)].any():
finalNewList.append(pt)
finalNew = np.array(finalNewList)
The right way to do this is to use linear algebra to compute the distance between each pair of 2-long vectors, and then accept only the new points that are "different enough" from each old point: using scipy.spatial.distance.cdist:
import numpy as np
oldPts = np.random.randn(1000,2)
newPts = np.random.randn(2000,2)
from scipy.spatial.distance import cdist
dist = cdist(oldPts, newPts)
print(dist.shape) # (1000, 2000)
okIndex = np.max(dist, axis=0) > 5
print(np.sum(okIndex)) # prints 1503 for me
finalNew = newPts[okIndex,:]
print(finalNew.shape) # (1503, 2)
Above I use the Euclidean distance of 5 as the threshold for "too close": any point in newPts that's farther than 5 from all points in oldPts is accepted into finalPts. You will have to look at the range of values in dist to find a good threshold, but your histogram can guide you in picking the best one.
(One good way to visualize dist is to use matplotlib.pyplot.imshow(dist).)
This is a more refined version of what you were doing with the histogram. In fact, you ought to be able to get the exact same answer as the histogram by passing in metric='minkowski', p=1 keyword arguments to cdist, assuming your histogram bin widths are the same in both dimensions, and using 5 again as the threshold.
(PS. If you're interested in another useful function in scipy.spatial.distance, check out my answer that uses pdist to find unique rows/columns in an array.)
I have some timecourse data that visually appears to have differing levels of high frequency fluctuation. I have plotted the timecourse Data A and B below.
I have used numpy FFT to perform Fourier Transformation as follows:
Fs = 1.0; # sampling rate
Ts = 864000; # sampling interval
t = np.arange(0,Ts,Fs) # time vector
n = 864000 # number of datapoints of timecourse data
k = np.arange(n)
T = n/Fs
frq = k/T # two sides frequency range
frq = frq[range(n/2)] # one side frequency range
Y = np.fft.fft(A1)/n # fft computing and normalization of timecourse data (A1)
Y = Y[range(n/2)]
Z = abs(Y)
##### Calculate Mean Frequency ########
Mean_Frequency = sum((frq*Z))/(sum(Z))
################################make sure the first value doesnt create an issue
Freq = frq[1:]
Z = Z[1:]
max_y = max(Z) # Find the maximum y value
mode_Frequency = Freq[Z.argmax()] # Find the x value corresponding to the maximum y value
###################### Plot figures
fig, ax = plt.subplots(2, 1)
fig.suptitle(str(D[k]), fontsize=14, fontweight='bold')
ax[0].plot(t,A1, 'black')
ax[0].set_xlabel('Time')
ax[0].set_ylabel('Amplitude')
ax[1].plot(frq,abs(Y),'r') # plotting the spectrum
ax[1].set_xscale('log')
ax[1].set_xlabel('Freq (Hz)')
ax[1].set_ylabel('|Y(freq)|')
text(2*mode_x, 0.95*max_y, "Mode Frequency (Hz): "+str(mode_x))
text(2*mode_x, 0.85*max_y, "Mean Frequency (Hz): "+str(mean_x))
plt.savefig("/home/phoenix/Desktop/Figures/Figure3/FourierGraphs/"+str(D[k])+".png")
plt.close()
This results look like this:
The timecourse data (black) for A on the left appears to me to have much more high frequency noise than than the data on the right (B).
yet the mean frequency is higher for the data on the left.
Is this because I have performed FFT incorrectly, or because I have calculated mean frequency incorrectly or because I need to use a different method to capture the really low frequency noise in timecourse B?
thanks for your time.
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.