assign.1.python.txt

# SUBJECT 3
import pandas as pd
import sklearn
## import data set from ecxel
df = pd.read_csv("C:/Users/magvo/Downloads/winequality/white.csv"
,sep=";")
print(df)
from sklearn.preprocessing import StandardScaler
features= ['fixed acidity',"volatile acidity","citric acid",
"residual sugar","chlorides","free sulfur dioxide",
"total sulfur dioxide","density","pH","sulphates","alcohol",
"quality"]
x = df.loc[:, features].values
x = StandardScaler().fit_transform(x)
##pca
from sklearn.decomposition import PCA
pca = PCA(n_components=4)##four new axis
principalComponents = pca.fit_transform(x)
principalDf = pd.DataFrame(data=principalComponents, columns = 
['principal component 1', 'principal component 2',
'principal component 3',
'principal component 4'])
print (principalDf)##new variables 

## from this scores we can compute the percentage of variance by adding one more component
eigenvalues = pca.explained_variance_ 
print(eigenvalues) ## eigenvalues
print("eigen vectors:")
print(pca.components_) ## eigenvectors
                 
#SUBJECT 4
#4.1
import numpy as np
##4.a.python
p1=np.array([1,2,3,4,5,6] )
p2=np.array([1,2,3,4,5,6])
d=np.linalg.norm(p2-p1)
print(d)
#The result is 0 that means we have similar vectors.

##4.b. python
p1=np.array([-0.5, 1, 7.3, 7, 9.4, -8.2, 9, -6, -6.3] )
p2=np.array([0.5, -1, -7.3, -7, -9.4, 8.2, -9, 6, 6.3])
d=np.linalg.norm(p2-p1)
print(d)
#The result is 40.78382, we have big distance between the two vectors.

##4.c.python
p1=np.array([-0.5, 1, 7.3, 7, 9.4, -8.2] )
p2=np.array([1.25, 9.02, -7.3, -7, 5, 1.3])
d=np.linalg.norm(p2-p1)
print(d)
#The result is 24.21059, we have big distance between the two vectors.

##4.d.python
p1=np.array([0, 0, 0.2] )
p2=np.array([0.2, 0.2, 0])
d=np.linalg.norm(p2-p1)
print(d)
#The result is 0.3464102, it is close to zero so we have distance but not so big.
#4.2
import pandas as pd
minutes=(25000,42000,55000,27000,58000)##vector for minutes
sms=(14,17,22,13,21)##vector for sms
mb=(7,9,5,11,13)##vector for mbs

comp=10000000000000000000##we set comp as a big number to compare 
the euclidean distances and keep the smaller
##loop that computes for every individual the dist from 
the 3rd
for i in range(5):
 eudist=((minutes[i]-minutes[4])**2+(sms[i]-sms[4])**2
 +(mb[i]-mb[4])**2)**0.5
 print(eudist)
 if eudist<comp and eudist!=0:##keep the distance if it 
 is smaller than the previous one and not zero to exclude the third 
 individual comparation with him self
        comp=eudist
        pos=i+1

print(comp)
print(pos)##the position of the smallest euclidean distance

##SUBJECT 5
 
#5.a. define two lists or array
x = np.array([9.32, -8.3, 0.2 ])
y = np.array([-5.3, 8.2, 7])
 
print("x:", x)
print("y:", y)
 
# compute cosine similarity
cosine = np.dot(x,y)/(norm(x)*norm(y))
print("Cosine Similarity:", cosine)

#compute distance
distance=(1-cosine)
print("Distance:",distance)
#The CosineSimilarity is -0.7739558 and the distance is 1.7739558.

#5.b. define two lists or array
x = np.array([(6.5, 1.3, 0.3, 16, 2.4, -5.2, 2, -6, -6.3) ])
y = np.array([0.5, -1, -7.3, -7, -9.4, 8.2, -9, 6, 6.3])

print("x:", x)
print("y:", y)

# compute cosine similarity
cosine = np.dot(x,y)/(norm(x)*norm(y))
print("Cosine Similarity:", cosine)

#compute distance
distance=(1-cosine)
print("Distance:",distance)
#The CosineSimilarity is -0.6546319 and the distance is 1.6546319.

#5.c. define two lists or array
x = np.array([-0.5, 1, 7.3, 7, 9.4, -8.2 ])
y = np.array([1.25, 9.02, -7.3, -7, 15, 12.3])
 
print("x:", x)
print("y:", y)
 
# compute cosine similarity
cosine = np.dot(x,y)/(norm(x)*norm(y))
print("Cosine Similarity:", cosine)

#compute distance
distance=(1-cosine)
print("Distance:",distance)
#The CosineSimilarity is -0.140921 and the distance is 1.140921.

#5.d. define two lists or array
x = np.array([2, 8, 5.2 ])
y = np.array([2, 8, 5.2])
 
print("x:", x)
print("y:", y)
 
# compute cosine similarity
cosine = np.dot(x,y)/(norm(x)*norm(y))
print("Cosine Similarity:", cosine)

#compute distance
distance=(1-cosine)
print("Distance:",distance)
#The Cosine Similarity is 1 and the distance is 0 , we have similar vectors.

#SUBJECT 6
#6.a
def jaccard(list1, list2):
    intersection = len(list(set(list1).intersection(list2)))
    union = (len(list1) + len(list2)) - intersection
    return float(intersection) / union
x =("Green", "Potato", "Ford") 
y=("Tyrian purple", "Pasta", "Opel")

xy = jaccard(x,y)
nominalDistance= 1-xy
print(x)
print(y)
print(xy)
print("Nominal distance:" ,nominalDistance)
##The Nominal distance is 1, that means the categorical vectors are dissimilar.

#6.b
def jaccard(list1, list2):
    intersection = len(list(set(list1).intersection(list2)))
    union = (len(list1) + len(list2)) - intersection
    return float(intersection) / union
x=("Eagle", "Ronaldo", "Real madrid", "Prussian blue", "Michael Bay")
y=("Eagle", "Ronaldo", "Real madrid", "Prussian blue", "Michael Bay")
xy = jaccard(x,y)
nominalDistance= 1-xy
print(x)
print(y)
print(xy)
print("Nominal distance:" ,nominalDistance)
##The Nominal distance is 0, that meansthe categorical vectors are simillar. 

#6.c
def jaccard(list1, list2):
    intersection = len(list(set(list1).intersection(list2)))
    union = (len(list1) + len(list2)) - intersection
    return float(intersection) / union
x =("Werner Herzog", "Aquirre, the wrath of God", "Audi",
"Spanish red")
y=("Martin Scorsese","Taxi driver" , "Toyota", "Spanish red")

xy = jaccard(x,y)
nominalDistance= 1-xy
print(x)
print(y)
print(xy)
print("Nominal distance:" ,nominalDistance)
## The Nominal distance is 0.8571428571428572.