Vim script has a few very basic functional programming facilities.
It has map() and filter(), but as far as I know it lacks a reduce() function. "Reduce" reduces a collection of values to a single value.
Is there a way to create reduce() or emulate it somehow in Vim script? Is it possible to reduce a list of values, without writing an explicit loop, in a Vim script expression? As an example, is there a way to reduce the first five positive integers over the addition operation, as is par for the course in functional languages?
In JavaScript:
[1, 2, 3, 4, 5].reduce(function(x, y) { return x + y; });
15
In Clojure:
(reduce + (range 1 (inc 5)))
15
In Haskell:
foldl (+) 0 [1..5]
15
In J:
+/>:i.5
15
In Vim script: ...?
For future reference, here are my variations on the theme, inspired by the answers that #MatthewStrawbridge linked.
The expression for the original example problem:
eval(join(range(1, 5), '+'))
A more general solution in the same vein, using Add(), where a is range(1, 5):
eval(repeat('Add(',len(a)-1).a[0].','.join(a[1:],'),').')')
This constructs the string "Add(Add(Add(Add(1,2),3),4),5)", and then evals it. Fun!
Finally, Reduce(), which takes a Funcref and a list, and then reduces it in a loop using Vim's list "destructuring" syntax [x, y; z]. See :h :let-unpack.
function! Reduce(f, list)
let [acc; tail] = a:list
while !empty(tail)
let [head; tail] = tail
let acc = a:f(acc, head)
endwhile
return acc
endfunction
And this is how it's used:
:echo Reduce(function('Add'), range(1, 5))
15
I think you're expected to construct a string and then execute it (which I admit feels a bit clunky). The help (:h E714) gives this example:
:exe 'let sum = ' . join(nrlist, '+')
So in your case, where nrlist is [1, 2, 3, 4, 5], it would construct the string let sum = 1+2+3+4+5 and then execute it.
Alternatively, you can probably code up your own reduce function as there isn't one built in.
Edit:
I found a discussion on the vim_use Google Group (How powerful is language build in vim compare with the language build in emacs?, 25th Jan 2010) about functional programming in Vim, which included a couple of implementations of just such a reduce function.
The first, by Tom Link, is as follows:
function! Reduce(ffn, list) "{{{3
if empty(a:list)
return ''
else
let list = copy(a:list)
let s:acc = remove(list, 0)
let ffn = substitute(a:ffn, '\<v:acc\>', "s:acc", 'g')
for val in list
let s:acc = eval(substitute(ffn, '\<v:val\>', val, 'g'))
endfor
return s:acc
endif
endf
echom Reduce("v:val + v:acc", [1, 2, 3, 4])
echom Reduce("v:val > v:acc ? v:val : v:acc", [1, 2, 3, 4])
echom Reduce("'v:val' < v:acc ? 'v:val' : v:acc", split("characters",
'\zs'))
The second, by Antony Scriven, is as follows:
fun Reduce(funcname, list)
let F = function(a:funcname)
let acc = a:list[0]
for value in a:list[1:]
let acc = F(acc, value)
endfor
return acc
endfun
fun Add(a,b)
return a:a + a:b
endfun
fun Max(a,b)
return a:a > a:b ? a:a : a:b
endfun
fun Min(a,b)
return a:a < a:b ? a:a : a:b
endfun
let list = [1,2,3,4,5]
echo Reduce('Add', list)
echo Reduce('Max', list)
echo Reduce('Min', list)
It's sad that it doesn't have a reduce function, here is mine
function s:reduce(acc, fn, args) abort
let acc = a:acc
for item in a:args
let acc = a:fn(a:acc, item)
endfor
return acc
endfunc
Here is a sum function defined on terms of reduce
let Sum = {... -> s:reduce(0, {acc, arg -> acc + arg}, a:000)}
Related
Consider a function g(x):
g(x) = x == 0 ? 0 : g(g(x - 1))
I don't think this is tail recursive, since the call to g(x) can be broken down into two parts:
let y = g(x - 1);
return g(y);
how to convert this to tail recursion?
continuation-passing style
You can convert from direct style to continuation-passing style -
g'(x,return) =
x == 0
? return(0)
: g'(x - 1, x' => # tail
g'(x', return)) # tail
Now write g to call g' with the default continuation -
g(x) = g'(x, x' => x')
Depending on the language you use, you may have to rename return to something else. k is a popular choice.
another example
We can see this technique applied to other problems like fib -
# direct style
fib(x) =
x < 2
? x
: fib(x - 1) + fib(x - 2) # "+" has tail position; not fib
# continuation-passing style
fib'(x, return) =
x < 2
? return(x)
: fib'(x - 1, x' => # tail| first compute x - 1
fib(x - 2, x'' => # tail| then compute x - 2
return(x' + x''))) # tail| then return sum
fib(x) = fib'(x, z => z)
other uses
Continuation-passing style is useful in other ways too. For example, it can be used to provide branching or early-exit behavior in this search program -
search(t, index, match, ifFound, notFound) =
i >= length(t)
? notFound()
: t[index] == match
? ifFound(index)
: search(t, index + 1, match, ifFound, notFound)
We can call search with continuations for each possible outcome -
search(["bird", "cat", "dog"],
0,
"cat",
matchedIndex => print("found at: " + matchedIndex),
() => print("not found")
)
how to
A function written in continuation-passing style takes an extra argument: an explicit "continuation"; i.e., a function of one argument — wikipedia
(* direct style *)
add(a, b) = ...
(* continuation-passing style takes extra argument *)
add(a, b, k) = ...
When the CPS function has computed its result value, it "returns" it by calling the continuation function with this value as the argument.
(* direct style *)
add(a, b) =
a + b
(* continuation-passing style "returns" by calling continuation *)
add(a, b, k) =
k(a + b) (* call k with the return value *)
That means that when invoking a CPS function, the calling function is required to supply a procedure to be invoked with the subroutine's "return" value.
(* direct style *)
print(add(5, 3)) (* 8 *)
(* continuation-passing style *)
add(5, 3, print) (* 8 *)
Expressing code in this form makes a number of things explicit which are implicit in direct style. These include: procedure returns, which become apparent as calls to a continuation; intermediate values, which are all given names; order of argument evaluation, which is made explicit; and tail calls, which simply call a procedure with the same continuation, unmodified, that was passed to the caller.
you've probably used continuations before
If you've ever run into someone saying "callback", what they really mean is continuation.
"When the button is clicked, continue the program with event => ..." -
document.querySelector("#foo").addEventListener("click", event => {
console.log("this code runs inside a continuation!")
})
<button id="foo">click me</button>
"When the file contents are read, continue the program with (err, data) => ..." -
import { readFile } from 'fs';
readFile('/etc/passwd', (err, data) => {
if (err) throw err;
console.log(data);
});
I have the following question "Given a list of integer pairs, write a function to return a list of even numbers in that list in sml".
this is what I've achieved so far
val x = [(6, 2), (3, 4), (5, 6), (7, 8), (9, 10)];
fun isEven(num : int) =
if num mod 2 = 0 then num else 0;
fun evenNumbers(list : (int * int) list) =
if null list then [] else
if isEven(#1 (hd list)) <> 0
then if isEven(#2 (hd list)) <> 0
then #1 (hd list) :: #1 (hd list) :: evenNumbers(tl list)
else []
else if isEven(#2 (hd list)) <> 0
then #1 (hd list) :: evenNumbers(tl list)
else [];
evenNumbers(x);
the result should be like this [6,2,4,6,8,10]
any help would be appreciated.
I see two obvious problems.
If both the first and second number are even, you do
#1 (hd list) :: #1 (hd list) :: evenNumbers(tl list)
which adds the first number twice and ignores the second.
If the first number is odd and the second even, you do
#1 (hd list) :: evenNumbers(tl list)
which adds the number that you know is odd and ignores the one you know is even.
Programming with selectors and conditionals gets complicated very quickly (as you've noticed).
With pattern matching, you could write
fun evenNumbers [] = []
| evenNumber ((x,y)::xys) = ...
and reduce the risk of using the wrong selector.
However, this still makes for complicated logic, and there is a better way.
Consider the simpler problem of filtering the odd numbers out of a list of numbers, not pairs.
If you transform the input into such a list, you only need to solve that simpler problem (and there's a fair chance that you've already solved something very similar in a previous exercise).
Exercise: implement this transformation. Its type will be ('a * 'a) list -> 'a list.
Also, your isEven is more useful if it produces a truth value (if you ask someone, "is 36 even?", "36" is a very strange answer).
fun isEven x = x mod 2 = 0
Now, evenNumbers can be implemented as "just" a combination of other, more general, functions.
So running your current code,
- evenNumbers [(6, 2), (3, 4), (5, 6), (7, 8), (9, 10)];
val it = [6,6,3,5,7,9] : int list
suggests that you're not catching all even numbers, and that you're catching some odd numbers.
The function isEven sounds very much like you want to have the type int -> bool like so:
fun isEven n =
n mod 2 = 0
Instead of addressing the logic error of your current solution, I would like to propose a syntactically much simpler approach which is to use pattern matching and fewer explicit type annotations. One basis for such a solution could look like:
fun evenNumbers [] = ...
| evenNumbers ((x,y)::pairs) = ...
Using pattern matching is an alternative to if-then-else: the [] pattern is equivalent to if null list ... and the (x,y)::pairs pattern matches when the input list is non-empty (holds at least one element, being (x,y). At the same time, it deconstructs this one element into its parts, x and y. So in the second function body you can express isEven x and isEven y.
As there is a total of four combinations of whether x and y are even or not, this could easily end up with a similarly complicated nest of if-then-else's. For this I might do either one of two things:
Use case-of (and call evenNumbers recursively on pairs):
fun evenNumbers [] = ...
| evenNumbers ((x,y)::pairs) =
case (isEven x, isEven y) of
... => ...
| ... => ...
Flatten the list of pairs into a list of integers and filter it:
fun flatten [] = ...
| flatten ((x,y)::pairs) = ...
val evenNumbers pairs = ...
I am absolute OCaml beginner. I want to create a function that repeats characters 20 times.
This is the function, but it does not work because of an error.
let string20 s =
let n = 20 in
s ^ string20 s (n - 1);;
string20 "u";;
I want to run like this
# string20 "u"
- : string = "uuuuuuuuuuuuuuuuuuuu"
Your function string20 takes one parameter but you are calling it recursively with 2 parameters.
The basic ideas are in there, but not quite in the right form. One way to proceed is to separate out the 2-parameter function as a separate "helper" function. As #PierreG points out, you'll need to delcare the helper function as a recursive function.
let rec string n s =
if n = 0 then "" else s ^ string (n - 1) s
let string20 = string 20
It is a common pattern to separate a function into a "fixed" part and inductive part. In this case, a nested helper function is needed to do the real recursive work in a new scope while we want to fix an input string s as a constant so we can use to append to s2. s2 is an accumulator that build up the train of strings over time while c is an inductor counting down to 1 toward the base case.
let repeat s n =
let rec helper s1 n1 =
if n1 = 0 then s1 else helper (s1 ^ s) (n1 - 1)
in helper "" n
A non-tail call versions is more straightforward since you won't need a helper function at all:
let rec repeat s n =
if n = 0 then "" else s ^ repeat s (n - 1)
On the side note, one very fun thing about a functional language with first-class functions like Ocaml is currying (or partial application). In this case you can create a function named repeat that takes two arguments n of type int and s of type string as above and partially apply it to either n or s like this:
# (* top-level *)
# let repeat_foo = repeat "foo";;
# repeat_foo 5;;
- : bytes = "foofoofoofoofoo" (* top-level output *)
if the n argument was labeled as below:
let rec repeat ?(n = 0) s =
if n = 0 then "" else s ^ repeat s (n - 1)
The order of application can be exploited, making the function more flexible:
# (* top-level *)
# let repeat_10 = repeat ~n:10;;
# repeat_10 "foo";;
- : bytes = "foofoofoofoofoofoofoofoofoofoo" (* top-level output *)
See my post Currying Exercise in JavaScript (though it is in JavaScript but pretty simple to follow) and this lambda calculus primer.
Recursive functions in Ocaml are defined with let rec
As pointed out in the comments you've defined your function to take one parameter but you're trying to recursively call with two.
You probably want something like this:
let rec stringn s n =
match n with
1 -> s
| _ -> s ^ stringn s (n - 1)
;;
I'm trying to write some code in a functional paradigm for practice. There is one case I'm having some problems wrapping my head around. I am trying to create an array of 5 unique integers from 1, 100. I have been able to solve this without using functional programming:
let uniqueArray = [];
while (uniqueArray.length< 5) {
const newNumber = getRandom1to100();
if (uniqueArray.indexOf(newNumber) < 0) {
uniqueArray.push(newNumber)
}
}
I have access to lodash so I can use that. I was thinking along the lines of:
const uniqueArray = [
getRandom1to100(),
getRandom1to100(),
getRandom1to100(),
getRandom1to100(),
getRandom1to100()
].map((currentVal, index, array) => {
return array.indexOf(currentVal) > -1 ? getRandom1to100 : currentVal;
});
But this obviously wouldn't work because it will always return true because the index is going to be in the array (with more work I could remove that defect) but more importantly it doesn't check for a second time that all values are unique. However, I'm not quite sure how to functionaly mimic a while loop.
Here's an example in OCaml, the key point is that you use accumulators and recursion.
let make () =
Random.self_init ();
let rec make_list prev current max accum =
let number = Random.int 100 in
if current = max then accum
else begin
if number <> prev
then (number + prev) :: make_list number (current + 1) max accum
else accum
end
in
make_list 0 0 5 [] |> Array.of_list
This won't guarantee that the array will be unique, since its only checking by the previous. You could fix that by hiding a hashtable in the closure between make and make_list and doing a constant time lookup.
Here is a stream-based Python approach.
Python's version of a lazy stream is a generator. They can be produced in various ways, including by something which looks like a function definition but uses the key word yield rather than return. For example:
import random
def randNums(a,b):
while True:
yield random.randint(a,b)
Normally generators are used in for-loops but this last generator has an infinite loop hence would hang if you try to iterate over it. Instead, you can use the built-in function next() to get the next item in the string. It is convenient to write a function which works something like Haskell's take:
def take(n,stream):
items = []
for i in range(n):
try:
items.append(next(stream))
except StopIteration:
return items
return items
In Python StopIteration is raised when a generator is exhausted. If this happens before n items, this code just returns however much has been generated, so perhaps I should call it takeAtMost. If you ditch the error-handling then it will crash if there are not enough items -- which maybe you want. In any event, this is used like:
>>> s = randNums(1,10)
>>> take(5,s)
[6, 6, 8, 7, 2]
of course, this allows for repeats.
To make things unique (and to do so in a functional way) we can write a function which takes a stream as input and returns a stream consisting of unique items as output:
def unique(stream):
def f(s):
items = set()
while True:
try:
x = next(s)
if not x in items:
items.add(x)
yield x
except StopIteration:
raise StopIteration
return f(stream)
this creates an stream in a closure that contains a set which can keep track of items that have been seen, only yielding items which are unique. Here I am passing on any StopIteration exception. If the underlying generator has no more elements then there are no more unique elements. I am not 100% sure if I need to explicitly pass on the exception -- (it might happen automatically) but it seems clean to do so.
Used like this:
>>> take(5,unique(randNums(1,10)))
[7, 2, 5, 1, 6]
take(10,unique(randNums(1,10))) will yield a random permutation of 1-10. take(11,unique(randNums(1,10))) will never terminate.
This is a very good question. It's actually quite common. It's even sometimes asked as an interview question.
Here's my solution to generating 5 integers from 0 to 100.
let rec take lst n =
if n = 0 then []
else
match lst with
| [] -> []
| x :: xs -> x :: take xs (n-1)
let shuffle d =
let nd = List.map (fun c -> (Random.bits (), c)) d in
let sond = List.sort compare nd in
List.map snd sond
let rec range a b =
if a >= b then []
else a :: range (a+1) b;;
let _ =
print_endline
(String.concat "\t" ("5 random integers:" :: List.map string_of_int (take (shuffle (range 0 101)) 5)))
How's this:
const addUnique = (ar) => {
const el = getRandom1to100();
return ar.includes(el) ? ar : ar.concat([el])
}
const uniqueArray = (numberOfElements, baseArray) => {
if (numberOfElements < baseArray.length) throw 'invalid input'
return baseArray.length === numberOfElements ? baseArray : uniqueArray(numberOfElements, addUnique(baseArray))
}
const myArray = uniqueArray(5, [])
For an assignment, i have written the following code in recursion. It takes a list of a vector data type, and a vector and calculates to closeness of the two vectors. This method works fine, but i don't know how to do the recursive version.
let romulus_iter (x:vector list) (vec:vector) =
let vector_close_hash = Hashtbl.create 10 in
let prevkey = ref 10000.0 in (* Define previous key to be a large value since we intially want to set closefactor to prev key*)
if List.length x = 0 then
{a=0.;b=0.}
else
begin
Hashtbl.clear vector_close_hash;
for i = 0 to (List.length x)-1 do
let vecinquestion = {a=(List.nth x i).a;b=(List.nth x i).b} in
let closefactor = vec_close vecinquestion vec in
if (closefactor < !prevkey) then
begin
prevkey := closefactor;
Hashtbl.add vector_close_hash closefactor vecinquestion
end
done;
Hashtbl.find vector_close_hash !prevkey
end;;
The general recursive equivalent of
for i = 0 to (List.length x)-1 do
f (List.nth x i)
done
is this:
let rec loop = function
| x::xs -> f x; loop xs
| [] -> ()
Note that just like a for-loop, this function only returns unit, though you can define a similar recursive function that returns a meaningful value (and in fact that's what most do). You can also use List.iter, which is meant just for this situation where you're applying an impure function that doesn't return anything meaningful to each item in the list:
List.iter f x