Questions

Question 1 — Using the Excel file dataA.xlsx, which contains a 500x3 data matrix (500 data points with 3 attributes), calculate both the mean and the covariance matrix.

Question 2 — Using the Excel file dataB.xlsx, which contains a 500x10 data matrix (500 data points with 10 attributes), calculate both the mean and the covariance matrix.

Question 3 — The data generated is random and normally distributed with a mean for dataA, dataB and covariance for dataA and dataB given in meanA.xlsx, meanB.xlsx, covarianceA.xlsx and covarianceB.xlsx respectively. Briefly explain why your answers are different from the parameters used to generate the data.

Implementation

Below is the python code for calculating mean and covariance matrix -

import pandas
import numpy
import matplotlib.pyplot

# Read dataA.xlsx and dataB.xlsx from excel file
dataA = pandas.read_excel("dataA.xlsx", header=None)
dataB = pandas.read_excel("dataB.xlsx", header=None)

# Convert data to numpy array
datanpyA = pandas.DataFrame.to_numpy(dataA)
datanpyB = pandas.DataFrame.to_numpy(dataB)

# Plot the data
matplotlib.pyplot.figure()
matplotlib.pyplot.scatter(datanpyA[:,0], datanpyA[:,1], c = 'r', marker = '.')
matplotlib.pyplot.scatter(datanpyB[:,0], datanpyB[:,1], c = 'b', marker = '.')

# Calculate the mean
meanA = numpy.mean(datanpyA,axis = 0)
meanB = numpy.mean(datanpyB,axis = 0)

# Subtract mean from the data
datawithoutmeanA = datanpyA - meanA
datawithoutmeanB = datanpyB - meanB

# Calculate covariance (C=X^T.X/(n-1))
covA = numpy.dot(numpy.transpose(datawithoutmeanA), datawithoutmeanA)/(len(datawithoutmeanA) - 1)
covB = numpy.dot(numpy.transpose(datawithoutmeanB), datawithoutmeanB)/(len(datawithoutmeanB) - 1)


# Print the Mean
numpy.set_printoptions(suppress=True)
print("Question 1 Solution:")
print("Mean A =>\n", meanA)
print("Covariance A =>\n", covA)
print("\n------------------------------\n")
print("Question 2 Solution:")
print("Mean B =>\n", meanB)
print("Covariance B =>\n", covB)
print("\n------------------------------\n")
print("Question 3 Solution")
print("The estimate given in mean and covariance excel file is based on population vs what was calulated using python is just a mean and covariance of a sample(data A and data B excel files). \nSince the data is randomly generated from a multivariate normal distribution using meanA and covarianceA for dataA, and meanB and covarianceB for dataB, \nthe resulting mean and covariance from the generated data clouds will not be the exact values as the given starting mean and covariance matrix but will be pretty close.")

matplotlib.pyplot.show()



#-------------- OUTPUT ----------------#
# Question 1 Solution:
# Mean A =>
#  [0.34750193 1.02563712 0.80122132]
# Covariance A =>
#  [[4.0704887  0.1502016  0.26208365]
#  [0.1502016  2.56307135 0.01468606]
#  [0.26208365 0.01468606 3.18321243]]

# ------------------------------

# Question 2 Solution:
# Mean B =>
#  [9.57062029 6.15014874 8.08016477 9.55989208 8.8040749  2.19491256
#  0.20634971 4.54942571 0.06659806 4.65575632]
# Covariance B =>
#  [[ 9.57410499  0.15742552  0.69100599 -0.04315714 -0.15529541  1.12934141
#    0.02644636 -0.48654602  0.95636371  0.53327821]
#  [ 0.15742552  9.51640579  0.47757067  0.41333501  0.00557376  0.53194456
#    0.11100153  0.17033133  0.84105524  1.33915044]
#  [ 0.69100599  0.47757067  8.65741988 -0.31162145  0.16556618  0.19225256
#    0.18585505  0.4101727   0.22889477 -0.15427328]
#  [-0.04315714  0.41333501 -0.31162145 10.27052739  0.2510052   0.34881198
#    0.68992571  0.32255801  0.72253427  1.0499889 ]
#  [-0.15529541  0.00557376  0.16556618  0.2510052   9.65117562  0.65088712
#    0.15264545 -0.16605455  1.35788702 -0.19805019]
#  [ 1.12934141  0.53194456  0.19225256  0.34881198  0.65088712 10.91504476
#    0.80109036  0.33946519  0.09688857  1.34008328]
#  [ 0.02644636  0.11100153  0.18585505  0.68992571  0.15264545  0.80109036
#    9.26492074 -0.19919067 -0.21481801  0.85962642]
#  [-0.48654602  0.17033133  0.4101727   0.32255801 -0.16605455  0.33946519
#   -0.19919067  9.1616525   0.29534256  0.13637128]
#  [ 0.95636371  0.84105524  0.22889477  0.72253427  1.35788702  0.09688857
#   -0.21481801  0.29534256 10.14780653  1.96290271]
#  [ 0.53327821  1.33915044 -0.15427328  1.0499889  -0.19805019  1.34008328
#    0.85962642  0.13637128  1.96290271 10.60596324]]

# ------------------------------

# Question 3 Solution
# The estimate given in mean and covariance excel file is based on population vs what was calulated using python is just a mean and covariance of a sample(data A and data B excel files). 
# Since the data is randomly generated from a multivariate normal distribution using meanA and covarianceA for dataA, and meanB and covarianceB for dataB, 
# the resulting mean and covariance from the generated data clouds will not be the exact values as the given starting mean and covariance matrix but will be pretty close.

Question

Using the multivariate data in the file fld1.xlsx: (a) determine the discriminant line found by Fishers Linear Discriminant. (b) Plot both the data and the discriminant line on a scatter plot (c) Using this line, determine the class of each of the data points in the dataset, assuming that the threshold is 0 (i.e. positive values are in one class and negative values in the other). (d) Determine what percentage of data points are incorrectly classified. NOTE: The first 2 columns in fld1.xlsx are data columns. The third column is the class to which each data point belongs.

Python Implementation

import pandas as pd
import numpy as np
import matplotlib.pyplot

# read data from excel file fld1 and create a scatterplot
fld1 = pd.read_excel("fld1.xlsx", header=None)
fld1_np = pd.DataFrame.to_numpy(fld1)
output_arr = fld1_np[:,2]
# print(output_arr)

fld1_np_1 = pd.DataFrame.to_numpy(fld1.head(300))
fld1_np_0 = pd.DataFrame.to_numpy(fld1.tail(200))

fld1_class_1 = fld1_np_1[:, :2]
# print(fld1_class_1)

fld1_class_0 = fld1_np_0[:, :2]
#print(fld1_class_0)

data_X = np.concatenate((fld1_class_1,fld1_class_0))
#print(data_X)

matplotlib.pyplot.figure()
matplotlib.pyplot.scatter(fld1_class_1[:,0], fld1_class_1[:,1], c = 'r', marker = '.')
matplotlib.pyplot.scatter(fld1_class_0[:,0], fld1_class_0[:,1], c = 'b', marker = '.')


# Calculate the mean
class1_mean = np.mean(fld1_class_1,axis = 0)
class0_mean = np.mean(fld1_class_0,axis = 0)

# Subtract mean from the data
class1_mc = fld1_class_1 - class1_mean
class0_mc = fld1_class_0 - class0_mean

# Calculate covariance (C=X^T.X/(n-1))
class1_cov = np.dot(class1_mc.T, class1_mc)
class0_cov = np.dot(class0_mc.T, class0_mc)
#print(class1_cov)
#print(class0_cov)

#implement fisher's linear discriminant w = Sw^-1*(u1 - u2)
Sw = class1_cov + class0_cov
w = np.dot(np.linalg.inv(Sw),(class1_mean - class0_mean))

print("Fisher's Linear Discriminant point is: \n",w)
print(w[0])
print(w[1])
matplotlib.pyplot.axline((0,0),w,c='black',linestyle='--')

# calc slope and y-intercept to create a discriminant line
thresh = 0
slope_1 = -w[0]/w[1]
y_intercept = thresh/w[1]
print("y-intercept is ", y_intercept)

matplotlib.pyplot.axline((0,y_intercept),slope = slope_1,c='green',linestyle='--')

# prediction and error calculation
prediction = (np.sign(np.dot(w,data_X.T) + thresh) + 1)/2
error = np.sum(abs(prediction - output_arr))
print("\nnumber of errors = ",error)
print("\npercentage of errors = ",(error/500)*100,"%")

# Q = np.squeeze(data_X[error])

# matplotlib.pyplot.scatter(Q[:,0],Q[:,1], c = 'g', marker = 'o')

matplotlib.pyplot.show()

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Bite-sized problem solving

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40 new opportunities

The 40 new opportunities technique inspired from the problem-solving methodology known as TRIZ, or Theory of Inventive Problem Solving is the roadmap for a business could use to continually reinvent its goods and services. A Russian patent examiner who examined 400,000 patent applications in search of similarities between the problems and the inventors’ approaches to solving them created TRIZ and concluded that each discovery might be linked to one of 40 principles.

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Turning the impossible into possible

The underlying denominator for businesses where innovation is preached but not done is that the cultures are too negative for large ideas to even be floated. From Impossible to Possible is a framework that aims to get the issues outlined and as a group think critically about how to resolve them. From the 3 categories of impossibilities — industry, customer, or internal impossibility, decide on 1 of the categories and brainstorm as a small group as many impossibilities for 15 minutes as possible. The idea is to outline ways to make the coworkers’ seemingly hard tasks possible for the next 15 minutes of discussion. By describing what cannot occur, a path to what can is opened, and by bringing in other viewpoints to the problem-solving process, norms can be broken, and employee morale is raised.

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