07 problem-solving

Stat405 Problem solving

Hadley Wickham
Monday, 13 September 2010

1. Projects
2. Saving data
3. Slot machine question
4. Basic strategy
5. Turning ideas into code


Projects

By Thursday, each group should send
Hadley an email with
1. the group name
2. the group members


Saving
data

Recall

To load data,
slots <- read.csv(“slots.csv”)


Your turn

Guess the name of the function you might
use to write an R object back to a csv ﬁle
on disk. Use it to save slots to
slots-2.csv.
What happens if you now read in
slots.csv with read.csv?


write.csv(slots, "slots-2.csv")
slots2 <- read.csv("slots-2.csv")

head(slots)
head(slots2)

str(slots)
str(slots2)

# Better, but still loses factor levels
write.csv(slots, file = "slots-3.csv", row.names = F)
slots3 <- read.csv("slots-3.csv")


Saving data

# For long-term storage
write.csv(slots, file = "slots.csv",
row.names = FALSE)

# For short-term caching
# Preserves factors etc.
save(slots, file = "slots.rdata")


.csv .rdata

read.csv() load()
write.csv( row.names =
FALSE) save()

Only data frames Any R object
Can be read by any
program
Only by R
Short term caching of
Long term storage expensive computations


Compression
Easy to store compressed ﬁles to save
space:
write.csv(slots,
file = bzfile("slots.csv.bz2"),
row.names = FALSE)
slots4 <- read.csv("slots.csv.bz2")
Files stored with save() are automatically
compressed.


Slot
machine
question

Slots
Casino claims that slot machines have
prize payout of 92%. Is this claim true?
mean(slots$prize)
t.test(slots$prize, mu = 0.92)
qplot(prize, data = slots, binwidth = 1)
Can we do better?


Doing that lots of times, and calculating
the payoff should give us a better idea of
the payoff. But ﬁrst we need some way to
calculate the prize from the windows.

Solution: Write a function


Strategy
1. Break complex tasks into smaller parts
2. Use words to describe how each part
should work
3. Translate words to R
4. When all parts work, combine into a
function (next class)


DD DD DD 800
windows <- c("7", "C", "C")
7 7 7 80
# How can we calculate the
BBB BBB BBB 40
# payoff?
BB BB BB 25
B B B 10
C C C 10
Any bar Any bar Any bar 5
C C * 5
C * C 5
C * * 2
DD doubles any winning
* C * 2 combination. Two DD
* * C 2 quadruples. DD is wild.

Your turn

We can simplify this table into 3 basic
cases of prizes. What are they? Take 3
minutes to brainstorm with a partner.


Cases

1. All windows have same value
2. A bar (B, BB, or BBB) in every window
3. Cherries and diamonds
4. (No prize)


Same values


Same values

1. Check whether all windows are the
same. How?


Same values

same. How?
2. If so, look up prize value. How?


Same values

same. How?
2. If so, look up prize value. How?

With a partner, brainstorm
for 2 minutes on how to
solve one of these problems

# Same value
same <- length(unique(windows)) == 1

# OR
same <- windows[1] == windows[2] &&
windows[2] == windows[3]

if (same) {
# Lookup value
}


&&, || vs. &, |
Use && and || to combine sub-conditions and
return a single TRUE or FALSE
Not & and | - these return vectors when
given vectors
&& and || are “short-circuiting”: they do the
minimum amount of work


If
if (condition) {
expression
}
Condition should be a logical vector of
length 1


if (TRUE) {
# This will be run
}

if (FALSE) {
# This won't be run
} else {
# This will be
}

# Single line form: (not recommended)
if (TRUE) print("True!)
if (FALSE) print("True!)


if (TRUE) {
# This will be run
}

if (FALSE) {
# This won't be run
} else {
# This will be
}
Note indenting.
Very important!
# Single line form: (not recommended)
if (TRUE) print("True!)
if (FALSE) print("True!)


x <- 5
if (x < 5) print("x < 5")
if (x == 5) print("x == 5")

x <- 1:5
if (x < 3) print("What should happen here?")

if (x[1] < x[2]) print("x1 < x2")
if (x[1] < x[2] && x[2] < x[3]) print("Asc")
if (x[1] < x[2] || x[2] < x[3]) print("Asc")


if (window[1] == "DD") {
prize <- 800
} else if (windows[1] == "7") {
prize <- 80
} else if (windows[1] == "BBB") ...

# Or use subsetting
c("DD" = 800, "7" = 80, "BBB" = 40)
c("DD" = 800, "7" = 80, "BBB" = 40)["BBB"]
c("DD" = 800, "7" = 80, "BBB" = 40)["0"]
c("DD" = 800, "7" = 80, "BBB" = 40)[window[1]]


Your turn

Complete the previous code so that if all
the values in win are the same, then prize
variable will be set to the correct amount.


All bars

How can we determine if all of the
windows are B, BB, or BBB?
(windows[1] == "B" ||
windows[1] == "BB" ||
windows[1] === "BBB") && ... ?


All bars

How can we determine if all of the
windows are B, BB, or BBB?
(windows[1] == "B" ||
windows[1] == "BB" ||
windows[1] === "BBB") && ... ?

Take 1 minute to brainstorm
possible solutions

windows[1] %in% c("B", "BB", "BBB")
windows %in% c("B", "BB", "BBB")

allbars <- windows %in% c("B", "BB", "BBB")
allbars[1] & allbars[2] & allbars[3]
all(allbars)

# See also ?any for the complement


Your turn

Complete the previous code so that the
correct value of prize is set if all the
windows are the same, or they are all
bars


payoffs <- c("DD" = 800, "7" = 80, "BBB" = 40,
"BB" = 25, "B" = 10, "C" = 10, "0" = 0)

allbars <- all(windows %in% c("B", "BB", "BBB"))

if (same) {
prize <- payoffs[windows[1]]
} else if (allbars) {
prize <- 5
}


Cherries

Need numbers of cherries, and numbers
of diamonds (hint: use sum)
Then need to look up values (like for the
ﬁrst case) and multiply together


cherries <- sum(windows == "C")
diamonds <- sum(windows == "DD")

c(0, 2, 5)[cherries + 1] *
c(1, 2, 4)[diamonds + 1]


payoffs <- c("DD" = 800, "7" = 80, "BBB" = 40,
"BB" = 25, "B" = 10, "C" = 10, "0" = 0)

allbars <- all(windows %in% c("B", "BB", "BBB"))

if (same) {
prize <- payoffs[windows[1]]
} else if (allbars) {
prize <- 5
} else {
cherries <- sum(windows == "C")
diamonds <- sum(windows == "DD")

prize <- c(0, 2, 5)[cherries + 1] *
c(1, 2, 4)[diamonds + 1]
}


Writing a function

Now we need to wrap up this code in to a
reusable fashion. We need a function
Have used functions a lot, next time we’ll
learn how to write one.


07 problem-solving

More Related Content

Similar to 07 problem-solving (20)

More from Hadley Wickham (19)

07 problem-solving