M N P Q R
A 1 2 3 4 5
B 2 3 2 5 3
C 5 5 5 3 2
Step 1. Mean Calculation for row
M N P Q R Mean
A 1 2 3 4 5 3
B 2 3 2 5 3 3
C 5 5 5 3 2 4
Step 2. Mean Subtraction for row
M N P Q R Mean
A -2 -1 0 1 2 3
B -1 0 -1 2 0 3
C 1 1 1 -1 -2 4
Step 3. Mean Calculation for col
M N P Q R Mean
A -2 -1 0 1 2 3
B -1 0 -1 2 0 3
C 1 1 1 -1 -2 4
Mean -2/3 0 0 -2/3 0
Step 4. Mean Substraction for col
M N P Q R Mean
A -4/3 -1 0 1/3 2 3
B -1/3 0 -1 4/3 0 3
C 5/3 1 1 -1/3 -2 4
Mean -2/3 0 0 -2/3 0
In [1]:
import numpy as np
print np.mean(np.array([1,2,3,4,5]))
print np.mean(np.array([2,3,2,5,3]))
print np.mean(np.array([5,5,5,3,2]))
print "=============="
a = np.array([-2,-1, 1])
# print np.mean(a)
print a - np.mean(a)
a = np.array([-1, 0, 1])
# print np.mean(a)
print a - np.mean(a)
a = np.array([ 0,-1, 1])
# print np.mean(a)
print a - np.mean(a)
a = np.array([ 1, 2,-1])
# print np.mean(a)
print a - np.mean(a)
a = np.array([ 2, 0,-2])
# print np.mean(a)
print a - np.mean(a)
A 1 0 1 0 1 2 B 1 1 0 0 1 6 C 0 1 0 1 0 2
In [28]:
from numpy.linalg import norm
# cos(A, B)
alphas = [0, 0.5, 1, 2]
for alpha in alphas:
print "=========================="
A = np.array([1,0,1,0,1,2*alpha], dtype="float32")
B = np.array([1,1,0,0,1,6*alpha], dtype="float32")
C = np.array([0,1,0,1,0,2*alpha], dtype="float32")
print "Current Alpha: " + str(alpha)
cos_a_b = float(np.dot(A.T, B)) / float( norm(A, ord=2)* norm(B, ord=2))
print "Consine(A, B) = " + str(cos_a_b)
cos_a_c = float(np.dot(A.T, C)) / float( norm(A, ord=2)* norm(C, ord=2))
print "Consine(A, C) = " + str(cos_a_c)
cos_b_c = float(np.dot(B.T, C)) / float( norm(B, ord=2)* norm(C, ord=2))
print "Consine(B, C) = " + str(cos_b_c)
Note: In this question, all columns will be written in their transposed form, as rows, to make the typography simpler. Matrix M has three rows and two columns, and the columns form an orthonormal basis. One of the columns is [2/7,3/7,6/7]. There are many options for the second column [x,y,z]. Write down those constraints on x, y, and z. Then, identify in the list below the one column that could be [x,y,z]. All components are computed to three decimal places, so the constraints may be satisfied only to a close approximation.
In [22]:
import numpy as np
def test_orth(lst, tested_lst):
col_b = np.array(tested_lst)
col_a = np.array(lst)
return (np.dot(col_b.T, col_a), np.linalg.norm(col_b, ord=2) )
a = [2/7.0, 3/7.0, 6/7.0]
print test_orth(a, [-.702, .117, .702])
print test_orth(a, [-.288, -.490, .772])
print test_orth(a, [.728, .485, -.485])
print test_orth(a, [2.250, -.500, -.750])
Note: In this question, all columns will be written in their transposed form, as rows, to make the typography simpler. Matrix M has three rows and three columns, and the columns form an orthonormal basis. One of the columns is [2/7,3/7,6/7], and another is [6/7, 2/7, -3/7]. Let the third column be [x,y,z]. Since the length of the vector [x,y,z] must be 1, there is a constraint that $x^2+y^2+z^2$ = 1. However, there are other constraints, and these other constraints can be used to deduce facts about the ratios among x, y, and z. Compute these ratios, and then identify one of them in the list below.
2x + 3y + 6z = 0 6x + 2y - 3z = 0
14x + 7y = 0 => y = -2x -4x + 6z = 0 => 3z = 2x => z = 2/3x
Suppose we have three points in a two dimensional space: (1,1), (2,2), and (3,4). We want to perform PCA on these points, so we construct a 2-by-2 matrix whose eigenvectors are the directions that best represent these three points. Construct this matrix and identify, in the list below, one of its elements.
In [29]:
M = np.array([[1,1],
[2,2],
[3,4]])
print np.dot(M.T, M)
In [23]:
def test_orth(lst, tested_lst):
flag = True
col_b = np.array(tested_lst)
col_a = np.array(lst)
return np.dot(col_b.T, col_a)
a = [1, 2, 3]
print test_orth(a, [-1, -2, 0])
print test_orth(a, [-1, -2, -3])
print test_orth(a, [-4, 2, -1])
print test_orth(a, [-1, -1, 1])
print "================="
print test_orth(a, [-1,1,-1])
print test_orth(a, [2,-3, 1])
print test_orth(a, [-4,2,-1])
print test_orth(a, [ 3,0,-1])
In [ ]: