Plan for this week
Last week:
- user-defined data types
- and how to manipulate them using pattern matching and recursion
- how to make recursive functions more efficient with tail recursion
This week:
code reuse with higher-order functions (HOFs)
some useful HOFs:
map
,filter
, andfold
Recursion is good…
Recursive code mirrors recursive data
- Base constructor -> Base case
- Inductive constructor -> Inductive case (with recursive call)
But it can get kinda repetitive!
Example: evens
Let’s write a function evens
:
Example: four-letter words
Let’s write a function fourChars
:
Yikes, Most Code is the Same
Lets rename the functions to foo
:
foo [] = []
foo (x:xs)
| x mod 2 == 0 = x : foo xs
| otherwise = foo xs
foo [] = []
foo (x:xs)
| length x == 4 = x : foo xs
| otherwise = foo xs
Only difference is condition
x mod 2 == 0
vslength x == 4
Moral of the day
D.R.Y. Don’t Repeat Yourself!
Can we
- reuse the general pattern and
- substitute in the custom condition?
HOFs to the rescue!
General Pattern
- expressed as a higher-order function
- takes customizable operations as arguments
Specific Operation
- passed in as an argument to the HOF
The “filter” pattern
General Pattern
- HOF
filter
- Recursively traverse list and pick out elements that satisfy a predicate
Specific Operations
- Predicates
isEven
andisFour
Avoid duplicating code!
Let’s talk about types
-- evens [1,2,3,4] ==> [2,4]
evens :: [Int] -> [Int]
evens xs = filter isEven xs
where
isEven :: Int -> Bool
isEven x = x `mod` 2 == 0
-- fourChars ["i","must","do","work"] ==> ["must","work"]
fourChars :: [String] -> [String]
fourChars xs = filter isFour xs
where
isFour :: String -> Bool
isFour x = length x == 4
So what’s the type of filter
?
filter :: (Int -> Bool) -> [Int] -> [Int] -- ???
filter :: (String -> Bool) -> [String] -> [String] -- ???
It does not care what the list elements are
- as long as the predicate can handle them
It’s type is polymorphic (generic) in the type of list elements
-- For any type `a`
-- if you give me a predicate on `a`s
-- and a list of `a`s,
-- I'll give you back a list of `a`s
filter :: (a -> Bool) -> [a] -> [a]
Example: all caps
Lets write a function shout
:
Example: squares
Lets write a function squares
:
Yikes, Most Code is the Same
Lets rename the functions to foo
:
-- shout
foo [] = []
foo (x:xs) = toUpper x : foo xs
-- squares
foo [] = []
foo (x:xs) = (x * x) : foo xs
Lets refactor into the common pattern
The “map” pattern
General Pattern
- HOF
map
- Apply a transformation
f
to each element of a list
Specific Operations
- Transformations
toUpper
and\x -> x * x
Lets refactor shout
and squares
QUIZ
What is the type of map
?
(A) (Char -> Char) -> [Char] -> [Char]
(B) (Int -> Int) -> [Int] -> [Int]
(C) (a -> a) -> [a] -> [a]
(D) (a -> b) -> [a] -> [b]
(E) (a -> b) -> [c] -> [d]
-- For any types `a` and `b`
-- if you give me a transformation from `a` to `b`
-- and a list of `a`s,
-- I'll give you back a list of `b`s
map :: (a -> b) -> [a] -> [b]
Type says it all!
The only meaningful thing a function of this type can do is apply its first argument to elements of the list
Hoogle it!
Things to try at home:
can you write a function
map' :: (a -> b) -> [a] -> [b]
whose behavior is different frommap
?can you write a function
map' :: (a -> b) -> [a] -> [b]
such thatmap' f xs
returns a list whose elements are not inmap f xs
?
QUIZ
What is the value of quiz
?
(A) [2, 4, 6]
(B) [3, 5]
(C) Syntax Error
(D) Type Error
(E) None of the above
Don’t Repeat Yourself
Benefits of factoring code with HOFs:
Reuse iteration pattern
think in terms of standard patterns
less to write
easier to communicate
Avoid bugs due to repetition
Recall: length of a list
Recall: summing a list
Example: string concatenation
Let’s write a function cat
:
-- cat [] ==> ""
-- cat ["carne","asada","torta"] ==> "carneasadatorta"
cat :: [String] -> String
cat [] = ...
cat (x:xs) = ...
Can you spot the pattern?
-- len
foo [] = 0
foo (x:xs) = 1 + foo xs
-- sum
foo [] = 0
foo (x:xs) = x + foo xs
-- cat
foo [] = ""
foo (x:xs) = x ++ foo xs
The “fold-right” pattern
General Pattern
- Recurse on tail
- Combine result with the head using some binary operation
Let’s refactor sum
, len
and cat
:
Factor the recursion out!
You can write it more clearly as
QUIZ
What does this evaluate to?
(A) Type error
(B) [1,2,3]
(C) [3,2,1]
(D) [[3],[2],[1]]
(E) [[1],[2],[3]]
foldr f b [] = b
foldr f b (x:xs) = f x (foldr f b xs)
foldr (:) [] [1,2,3]
==> (:) 1 (foldr (:) [] [2, 3])
==> (:) 1 ((:) 2 (foldr (:) [] [3]))
==> (:) 1 ((:) 2 ((:) 3 (foldr (:) [] [])))
==> (:) 1 ((:) 2 ((:) 3 []))
== 1 : (2 : (3 : []))
== [1,2,3]
The “fold-right” pattern
foldr f b [x1, x2, x3, x4]
==> f x1 (foldr f b [x2, x3, x4])
==> f x1 (f x2 (foldr f b [x3, x4]))
==> f x1 (f x2 (f x3 (foldr f b [x4])))
==> f x1 (f x2 (f x3 (f x4 (foldr f b []))))
==> f x1 (f x2 (f x3 (f x4 b)))
Accumulate the values from the right
For example:
foldr (+) 0 [1, 2, 3, 4]
==> 1 + (foldr (+) 1 [2, 3, 4])
==> 1 + (2 + (foldr (+) 0 [3, 4]))
==> 1 + (2 + (3 + (foldr (+) 0 [4])))
==> 1 + (2 + (3 + (4 + (foldr (+) 0 []))))
==> 1 + (2 + (3 + (4 + 0)))
QUIZ
What is the most general type of foldr
?
(A) (a -> a -> a) -> a -> [a] -> a
(B) (a -> a -> b) -> a -> [a] -> b
(C) (a -> b -> a) -> b -> [a] -> b
(D) (a -> b -> b) -> b -> [a] -> b
(E) (b -> a -> b) -> b -> [a] -> b
Is foldr
tail recursive?
What about tail-recursive versions?
Let’s write tail-recursive sum
!
Lets run sumTR
to see how it works
sumTR [1,2,3]
==> helper 0 [1,2,3]
==> helper 1 [2,3] -- 0 + 1 ==> 1
==> helper 3 [3] -- 1 + 2 ==> 3
==> helper 6 [] -- 3 + 3 ==> 6
==> 6
Note: helper
directly returns the result of recursive call!
Let’s write tail-recursive cat
!
Lets run catTR
to see how it works
catTR ["carne", "asada", "torta"]
==> helper "" ["carne", "asada", "torta"]
==> helper "carne" ["asada", "torta"]
==> helper "carneasada" ["torta"]
==> helper "carneasadatorta" []
==> "carneasadatorta"
Note: helper
directly returns the result of recursive call!
Can you spot the pattern?
-- sumTR
foo xs = helper 0 xs
where
helper acc [] = acc
helper acc (x:xs) = helper (acc + x) xs
-- catTR
foo xs = helper "" xs
where
helper acc [] = acc
helper acc (x:xs) = helper (acc ++ x) xs
The “fold-left” pattern
General Pattern
- Use a helper function with an extra accumulator argument
- To compute new accumulator, combine current accumulator with the head using some binary operation
Let’s refactor sumTR
and catTR
:
Factor the tail-recursion out!
QUIZ
What does this evaluate to?
foldl f b xs = helper b xs
where
helper acc [] = acc
helper acc (x:xs) = helper acc' xs
where acc' = f x acc
quiz = foldl (:) [] [1,2,3]
[] [1,2,3,4,5,6,7]
=> 1:[] [2,3,4,5,6,7]
=> 2:1:[] [3,4,5,6,7]
=> 3:2:1:[] [4,5,6,7]
=> 4:3:2:1:[] [5,6,7]
=> 5:4:3:2:1:[] [6,7]
=> 7:6:5:4:3:2:1:[] []
foldl f b (x1: x2: x3 : [])
==> helper b (x1: x2: x3 : [])
==> helper (f x1 b) (x2: x3 : [])
==> helper (f x2 (f x1 b)) (x3 : [])
==> helper (f x3 (f x2 (f x1 b))) []
==> ( x3 : (x2 : (x1 : [])))
(A) Type error
(B) [1,2,3]
———–
(C) [3,2,1]
———–
(D) [[3],[2],[1]]
(E) [[1],[2],[3]]
QUIZ
What does this evaluate to?
foldl f b xs = helper b xs
where
helper acc [] = acc
helper acc (x:xs) = helper (f acc x) xs
quiz = foldl (\xs x -> x : xs) [] [1,2,3]
(A) Type error
(B) [1,2,3]
(C) [3,2,1]
(D) [[3],[2],[1]]
(E) [[1],[2],[3]]
The “fold-left” pattern
foldl f b [x1, x2, x3, x4]
==> helper b [x1, x2, x3, x4]
==> helper (f b x1) [x2, x3, x4]
==> helper (f (f b x1) x2) [x3, x4]
==> helper (f (f (f b x1) x2) x3) [x4]
==> helper (f (f (f (f b x1) x2) x3) x4) []
==> (f (f (f (f b x1) x2) x3) x4)
Accumulate the values from the left
For example:
foldl (+) 0 [1, 2, 3, 4]
==> helper 0 [1, 2, 3, 4]
==> helper (0 + 1) [2, 3, 4]
==> helper ((0 + 1) + 2) [3, 4]
==> helper (((0 + 1) + 2) + 3) [4]
==> helper ((((0 + 1) + 2) + 3) + 4) []
==> ((((0 + 1) + 2) + 3) + 4)
Left vs. Right
foldl f b [x1, x2, x3] ==> f (f (f b x1) x2) x3 -- Left
foldr f b [x1, x2, x3] ==> f x1 (f x2 (f x3 b)) -- Right
For example:
foldl (+) 0 [1, 2, 3] ==> ((0 + 1) + 2) + 3 -- Left
foldr (+) 0 [1, 2, 3] ==> 1 + (2 + (3 + 0)) -- Right
Different types!
Useful HOF: flip
-- you can write
foldl (\xs x -> x : xs) [] [1,2,3]
-- more concisely like so:
foldl (flip (:)) [] [1,2,3]
What is the type of flip
?
Useful HOF: compose
What is the type of (.)
?
Higher Order Functions
Iteration patterns over collections:
- Filter values in a collection given a predicate
- Map (iterate) a given transformation over a collection
- Fold (reduce) a collection into a value, given a binary operation to combine results
Useful helper HOFs:
- Flip the order of function’s (first two) arguments
- Compose two functions
HOFs can be put into libraries to enable modularity
Data structure library implements
map
,filter
,fold
for its collectionsgeneric efficient implementation
generic optimizations:
map f (map g xs) --> map (f.g) xs
Data structure clients use HOFs with specific operations
- no need to know the implementation of the collection
Enabled the “big data” revolution e.g. MapReduce, Spark
That’s all folks!