Quiz 5: Recomendation Systems

Question 1

Let’s use the content-based method to make some recommendations. The last 5 things I watched on Netflix and their attributes were:

row_names <- c("FUNNY", "ROMANTIC", "SUSPENSE", "DARK", "RATING")
col_names <- c("SHARP OBJECTS", "ARRESTED DEV.", "ARBITRAGE", "MARGIN CALL", "BOJACK", "ORPHAN BLACK", "HINTERLAND")

item <- matrix( c(0, 1, 0, 1, 1, 1, 0,
                  1, 1, 0, 0, 1, 0, 0,
                  1, 1, 1, 0, 1, 0, 1,
                  1, 0, 1, 1, 0, 1, 1,
                  4, 3, 4, 5, 3, NA, NA), byrow = TRUE, nrow = 5, ncol = 7,
                dimnames = list(row_names, col_names))
item %>% 
  kbl() %>%
  kable_styling()

	SHARP OBJECTS	ARRESTED DEV.	ARBITRAGE	MARGIN CALL	BOJACK	ORPHAN BLACK	HINTERLAND
FUNNY	0	1	0	1	1	1	0
ROMANTIC	1	1	0	0	1	0	0
SUSPENSE	1	1	1	0	1	0	1
DARK	1	0	1	1	0	1	1
RATING	4	3	4	5	3	NA	NA

How similar are my preferences to Orphan Black? R: 0.36

Calculate the cosine similarity and report to 2 decimal points (e.g. 0.12)

First, we normalize the ratings to the mean:

rating <- matrix(c(4, 3, 4, 5, 3), nrow=1,ncol=5)
rating_m <- rating-mean(rating)
rating_m %>% 
  kbl() %>%
  kable_styling()

0.2

-0.8

0.2

1.2

-0.8

We now estimate how much the ratings changes with attributes:

user <- item[1:4,1:5] %*% t(rating_m)/rowSums(item[1:4,1:5])
user %>% 
  kbl() %>%
  kable_styling()

FUNNY	-0.133
ROMANTIC	-0.467
SUSPENSE	-0.300
DARK	0.533

Finally, we take the similarity with respect to Orphan Black:

new_item <- item[1:4,6:7]

CS <- t(new_item) %*% user / sqrt( colSums(new_item^2) )*sqrt( sum(user^2) )
round(CS,2) %>% 
  kbl() %>%
  kable_styling()

ORPHAN BLACK	0.22
HINTERLAND	0.13

Question 2

How similar are my preferences to Hinterland? R: 0.21

Calculate the cosine similarity and report to 2 decimal points (e.g. 0.12).

CS <- t(new_item) %*% user / sqrt( colSums(new_item^2) )*sqrt( sum(user^2) )
CS %>% 
  kbl() %>%
  kable_styling()

ORPHAN BLACK	0.221
HINTERLAND	0.129

Question 3

What should Netflix recommend to me, based on the content-based method?

## Orphan Black, because is closer to 1

Question 4

Now, let’s do the same exercise through the lens of collaborative filtering (on users, not items).

row_names<-c("George", "Adam", "Ben", "Cam", "Dan")
col_names<-c("Sharp Obj", "Arrested Dev", "Arbitrage", "Margin C", "Bojack", "Orphan B", "Hinterland")

util <- matrix(c(4,3,4,5,3,NA,NA,
                 4,3,4,4,3,NA,NA,
                 3,4,3,1,3,5,NA,
                 4,4,4,4,4,2,4,
                 2,1,2,3,1,NA,3),
               byrow = TRUE, nrow = 5, ncol = 7, 
               dimnames=list(row_names,col_names))
util %>% 
  kbl() %>%
  kable_styling()

	Sharp Obj	Arrested Dev	Arbitrage	Margin C	Bojack	Orphan B	Hinterland
George	4	3	4	5	3	NA	NA
Adam	4	3	4	4	3	NA	NA
Ben	3	4	3	1	3	5	NA
Cam	4	4	4	4	4	2	4
Dan	2	1	2	3	1	NA	3

Imagine we want to again make a recommendation as to which movie George should see, based on his ratings and those of Adam, Ben, Cam, and Dan:

Who is most useful in predicting George’s rating for Orphan Black?

We take the relevant users:

m <- cor( t(util), use="pairwise.complete.obs")

## Warning in cor(t(util), use = "pairwise.complete.obs"): the standard deviation is zero

# The relevant row
users_corr <- m[row=c("Adam","Ben","Cam", "Dan"), col=c("Adam")]
users_corr %>% 
  kbl() %>%
  kable_styling()

	x
Adam	1.000
Ben	-0.583
Cam	NA
Dan	0.873

We normalize the ratings:

util_n <- util - rowMeans(util, na.rm=TRUE) #normalize

# The Movies we care:
movies_corr <- util_n[row=c("Adam","Ben","Cam", "Dan"), col=c("Orphan B", "Hinterland")]
movies_corr %>% 
  kbl() %>%
  kable_styling()

	Orphan B	Hinterland
Adam	NA	NA
Ben	1.83	NA
Cam	-1.71	0.286
Dan	NA	1.000

## The most useful user for Orphan Black is Ben

Question 5

Who is most useful in predicting George’s rating for Hinterland?

movies_corr %>% 
  kbl() %>%
  kable_styling()

	Orphan B	Hinterland
Adam	NA	NA
Ben	1.83	NA
Cam	-1.71	0.286
Dan	NA	1.000

## The most useful user for Hinterland is Dan

Question 6

According to the collaborative filtering method, should Netflix better recommend Orphan Black or Hinterland?

Finally, we predict:

predm <- users_corr * movies_corr
predm %>% 
  kbl() %>%
  kable_styling()

	Orphan B	Hinterland
Adam	NA	NA
Ben	-1.07	NA
Cam	NA	NA
Dan	NA	0.873

# Take the average
pred <- colMeans(predm, na.rm=TRUE)
pred %>% 
  kbl() %>%
  kable_styling()

	x
Orphan B	-1.069
Hinterland	0.873

predm<-m[row=c("Ben","Dan"),col=c("George")]*util_n[row=c("Ben","Dan"),col=c("Orphan B", "Hinterland")]
predm %>% 
  kbl() %>%
  kable_styling()

	Orphan B	Hinterland
Ben	-1.6	NA
Dan	NA	1

# Take the average
pred <- colMeans(predm, na.rm=TRUE)
pred %>% 
  kbl() %>%
  kable_styling()

	x
Orphan B	-1.6
Hinterland	1.0