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).
hello I'm new to tensorflow and I'm getting a feel for it. so i was given a task to multiply these 4 matrices. i was able to do that but now I'm being asked to Take the (16,4) outputs from the multiplication of the (16,8) and (8,4) and Apply a Logistics function on all outputs. Then multiply this new matrix of shape (16,4) by the (4,2) matrix. Take these (16,2) outputs and apply a Logistics function on them. Now multiply this new (16,2) matrix by the (2,1) matrix. I'm suppose to be able to do all this with matrix manipulation. I'm kind of confused on how to go about it because i only kind of sort of understand linear regression. i know they are similar but i wouldn't know how to apply it. any tips please. no I'm not asking for someone to finish i just would like a better example than what i was given because i can't figure out how to go about a logistic function using a matrix. this is what i have so far
import tensorflow as ts
import numpy as np
import os
# AWESOME SAUCE WARNING MESSAGE WAS GETTING ANNOYING
os.environ['TF_CPP_MIN_LOG_LEVEL']='2' #to avoid warnings about compilation
# for different matrix asked to multiply with
# use random for random numbers in each matrix
m1 = np.random.rand(16,8)
m2 = np.random.rand(8,4)
m3 = np.random.rand(4,2)
m4 = np.random.rand(2,1)
# using matmul to mulitply could use # or dot() but using tensorflow
c = ts.matmul(m1,m2)
d = ts.matmul(c,m3)
e = ts.matmul(d, m4)
#attempting to create log regression
arf = ts.Variable(m1,name = "ARF")
with ts.Session() as s:
r1 = s.run(c)
print("M1 * M2: \n",r1)
r2 = s.run(d)
print("Result of C * M3: \n ", r2)
r3 = s.run(e)
print("Result of D * M4: \n",r3)
#learned i cant reshape just that easily
#r4 = ts.reshape(m1,(16,4))
#print("Result of New M1: \n", r4)
I think you have the right idea. The logistic function is just 1 / (1 + exp(-z)) where z is the matrix you want to apply it to. With that in mind you can simply do:
logistic = 1 / (1 + ts.exp(-c))
This will apply the formula element-wise to the input. The result is that this:
lg = s.run(logistic)
print("Result of logistic function \n ", lg)
…will print a matrix the same size as c (16,4), where all values are between 0 and 1. You can then go on to the rest of the multiplications the assignment is asking for.
I have 3 sparse tensors of dimensions A = P*N, B = Q*N and C = R*N. What is the efficient way to compute the product matrix A*B*C such that dimension of the product matrix is P*Q*R in tensorflow.?
I have tried with tf.matmul and followed by tf.reshape but it won't give the product matrix with the dimension as specified above.
Thanks.
tf.einsum Should help you. Though I'm not really sure what you mean by A*B*C since dimensions are incompatible for matrix multiplication. Probably something like this:
R = tf.einsum('il,jl,kl->ijk', A, B, C)
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.