I've found a basic example of an f# recursive function that takes a list and returns a list of only even integers. I understand it for the most part, but there's a little i'm confused about.
let numbers = [1..4]
let rec even ls =
match ls with
| [] -> []
|head :: tail when head % 2 = 0 -> head :: even tail
|_::tail -> even tail
The line that matches head confuses me. This is how I read it. Append head to tail when head is even, then call even tail again. Because we appended head to tail, wouldn't that just get caught in a loop of adding head over and over again?
Also, the final line _::tail I assume means "do nothing, recurse again", but
I looked up the operator _ in f# and it says it's a wildcard pattern. Does that essentially mean if not covered by either of my first two matches, do this thing?
Hoping someone can clarify! f# looks like a lot of fun, but coming from an imperative background it's difficult to understand.
This is a pattern matching expression. The way to read each line is this: to the left of the arrow -> is the instruction of how to inspect the data, and to the right of the arrow is the instruction of what to do with the results of inspection.
Applying this to your case, follow the comments (I've inserted some newlines for clarity):
match ls with // Look at `ls`
| [] -> // when `ls` is an empty list,
[] // return an empty list
| head :: tail // when `ls` is a `head` attached to a `tail`,
when head % 2 = 0 -> // AND `head` is an even number,
head :: even tail // call `even tail`, attach `head` to it, and return that result
| _::tail -> // when `ls` is "something" attached to a `tail`,
even tail // call `even tail` and return that result
Note that the very last case will apply in all situations in which the second case applies. This ambiguity is resolved by the order of the cases: the program will attempt to match ("look at", "inspect") the datum according to each case in turn, and will execute code for the first case that matches.
Another subtle point that you seem to be missing: immutability. The list is immutable. You cannot "update" ("change", "alter") the list in place. If you have a list, it will always stay that way, the compiler guarantees it. Instead, the way data evolves through the program is by creating new data based on the old ones. In particular, if you say a :: b, this does not modify the list b, but instead creates a new list, in which a is head and b is tail.
These are pretty basic concepts in F#, and the fact that you're missing them suggests to me that you've only just started looking at the language (and perhaps at functional programming in general). If this is true, I recommend first reading some introductory material to familiarize yourself with basic concepts. My favourite is fsharpforfunandprofit.com, I can't recommend it enough.
Append head to tail when head is even, then call even tail again. Because we appended head to tail, wouldn't that just get caught in a loop of adding head over and over again?
Close but not quite. It's prepending head onto the list returned by recursing even tail. With each recursion the tail value in this pattern matching expression has one less item in it.
Also, the final line _::tail I assume means "do nothing, recurse again", but I looked up the operator _ in F# and it says it's a wildcard pattern. Does that essentially mean if not covered by either of my first two matches, do this thing?
_ can be a wildcard pattern, but in this pattern matching example it simply indicates we don't care about the head value when we destructure the list; we only care about the tail. This clause (because it comes after the clause that tests for evenness) causes non-even head values to be discarded.
You are dealing with immutable data-types (immutable list) here. This means lists are not changed, every operations creates and returns a new list.
Pattern Matching is a way to deconstruct data into multiple pieces. As an example. If you have the list [1;2;3;4] and you write.
let list = [1;2;3;4]
let (head :: tail) = list
Then head is the value 1. tail is the list [2;3;4] and list is still [1;2;3;4].
Let's go through your example step-by-step.
even [1;2;3;4] is called. Then there are three cases with Pattern Matching. The first one checks if [1;2;3;4] is an empty list. Its not so it checks the next one. head :: tail extracts the list into two pieces like above. So head is 1 and tail represents [2;3;4], but when head % 2 = 0 adds a conditional. It checks if head (currently 1) is dividable by two. It isn't, so it goes to the last Pattern Matching.
The last one is _::tail. First it does the exact same as head::tail. It extracts the first value 1 and stores it in the variable _. The reason to use _ as a variable name is to clarify that the first name is never used. You also could change _ to head and the code works the same.
The last Pattern Match matches, and now you have 1 stored in _ and [2;3;4] stored in tail. And all you do is call even tail. Or in that case, you call even [2;3;4] and return the result of this function call as your result.
When even [2;3;4] is called it does the same above.
It checks if its an empty list. It's not. Then it extract the first value 2 in head and [3;4] into tail. It checks the when condition. This time it is true. So now it executes head :: even tail
Or if we replace the values we get 2 :: even [3;4]
:: is a concatenation of a list, but before we can concatenate a list, first the function call even [3;4] needs to return a list. So even [3;4] is called.
This then checks again.
Is [3;4] an empty list. Nope.
Is the head the value 3 dividable by 2. Nope.
Extract 3 into _ and [4] into tail, and call even [4].
even [4] then does the same:
Is [4] an empty list. Nope.
Is the value 4 assigned to head an even number? Yes it is. So 4 :: even [] is called.
even [] then does the same:
Is [] an empty list. Yes it is. So return the empty list. []
Then it goes backwards.
-> means "returns"
even [] -> []
4 :: even [] -> [4]
even [3;4] -> [4]
2 :: even [3;4] -> [2;4]
even [2;3;4] -> [2;4]
even [1;2;3;4] -> [2;4]
I'm trying to implement a sliding window algorithm for matching words in a text file. I come from a procedural background and my first attempt to do this in a functional language like Erlang seems to require time O(n^2) (or even more). How would one do this in a functional language?
-module(test).
-export([readText/1,patternCount/2,main/0]).
readText(FileName) ->
{ok,File} = file:read_file(FileName),
unicode:characters_to_list(File).
patternCount(Text,Pattern) ->
patternCount_(Text,Pattern,string:len(Pattern),0).
patternCount_(Text,Pattern,PatternLength,Count) ->
case string:len(Text) < PatternLength of
true -> Count;
false ->
case string:equal(string:substr(Text,1,PatternLength),Pattern) of
true ->
patternCount_(string:substr(Text,2),Pattern,PatternLength,Count+1);
false ->
patternCount_(string:substr(Text,2),Pattern,PatternLength,Count)
end
end.
main() ->
test:patternCount(test:readText("file.txt"),"hello").
Your question is a bit too broad, since it asks about implementing this algorithm in functional languages but how best to do that is language-dependent. My answer therefore focuses on Erlang, given your example code.
First, note that there's no need to have separate patternCount and patternCount_ functions. Instead, you can just have multiple patternCount functions with different arities as well as multiple clauses of the same arity. First, let's rewrite your functions to take that into account, and also replace calls to string:len/1 with the length/1 built-in function:
patternCount(Text,Pattern) ->
patternCount(Text,Pattern,length(Pattern),0).
patternCount(Text,Pattern,PatternLength,Count) ->
case length(Text) < PatternLength of
true -> Count;
false ->
case string:equal(string:substr(Text,1,PatternLength),Pattern) of
true ->
patternCount(string:substr(Text,2),Pattern,PatternLength,Count+1);
false ->
patternCount(string:substr(Text,2),Pattern,PatternLength,Count)
end
end.
Next, the multi-level indentation in the patternCount/4 function is a "code smell" indicating it can be done better. Let's split that function into multiple clauses:
patternCount(Text,Pattern,PatternLength,Count) when length(Text) < PatternLength ->
Count;
patternCount(Text,Pattern,PatternLength,Count) ->
case string:equal(string:substr(Text,1,PatternLength),Pattern) of
true ->
patternCount(string:substr(Text,2),Pattern,PatternLength,Count+1);
false ->
patternCount(string:substr(Text,2),Pattern,PatternLength,Count)
end.
The first clause uses a guard to detect that no more matches are possible, while the second clause looks for matches. Now let's refactor the second clause to use Erlang's built-in matching. We want to advance through the input text one element at a time, just as the original code does, but we also want to detect matches as we do so. Let's perform the matches in our function head, like this:
patternCount(_Text,[]) -> 0;
patternCount(Text,Pattern) ->
patternCount(Text,Pattern,Pattern,length(Pattern),0).
patternCount(Text,_Pattern,_Pattern,PatternLength,Count) when length(Text) < PatternLength ->
Count;
patternCount(Text,[],Pattern,PatternLength,Count) ->
patternCount(Text,Pattern,Pattern,PatternLength,Count+1);
patternCount([C|TextTail],[C|PatternTail],Pattern,PatternLength,Count) ->
patternCount(TextTail,PatternTail,Pattern,PatternLength,Count);
patternCount([_|TextTail],_,Pattern,PatternLength,Count) ->
patternCount(TextTail,Pattern,Pattern,PatternLength,Count).
First, note that we added a new argument to the bottom four clauses: we now pass Pattern as both the second and third arguments to allow us to use one of them for matching and one of them to maintain the original pattern, as explained more fully below. Note also that we added a new clause at the very top to check for an empty Pattern and just return 0 in that case.
Let's focus only on the bottom three patternCount/5 clauses. These clauses are tried in order at runtime, but let's look at the second of these three clauses first, then the third clause, then the first of the three:
In the second of these three clauses, we write the first and second arguments in [Head|Tail] list notation, which means Head is the first element of the list and Tail is the rest of the list. We use the same variable for the head of both lists, which means that if the first elements of both lists are equal, we have a potential match in progress, so we then recursively call patternCount/5 passing the tails of the lists as the first two arguments. Passing the tails allows us to advance through both the input text and the pattern an element at a time, checking for matching elements.
In the last clause, the heads of the first two arguments do not match; if they did, the runtime would execute the second clause, not this one. This means that our pattern match has failed, and so we no longer care about the first element of the first argument nor about the second argument, and we have to advance through the input text to look for a new match. Note that we write both the head of the input text and the second argument as the _ "don't care" variable, as they are no longer important to us. We recursively call patternCount/5, passing the tail of the input text as the first argument and the full Pattern as the second argument, allowing us to start looking for a new match.
In the first of these three clauses, the second argument is the empty list, which means we've gotten here by successfully matching the full Pattern, element by element. So we recursively call patternCount/5 passing the full Pattern as the second argument to start looking for a new match, and we also increment the match count.
Try it! Here's the full revised module:
-module(test).
-export([read_text/1,pattern_count/2,main/0]).
read_text(FileName) ->
{ok,File} = file:read_file(FileName),
unicode:characters_to_list(File).
pattern_count(_Text,[]) -> 0;
pattern_count(Text,Pattern) ->
pattern_count(Text,Pattern,Pattern,length(Pattern),0).
pattern_count(Text,_Pattern,_Pattern,PatternLength,Count)
when length(Text) < PatternLength ->
Count;
pattern_count(Text,[],Pattern,PatternLength,Count) ->
pattern_count(Text,Pattern,Pattern,PatternLength,Count+1);
pattern_count([C|TextTail],[C|PatternTail],Pattern,PatternLength,Count) ->
pattern_count(TextTail,PatternTail,Pattern,PatternLength,Count);
pattern_count([_|TextTail],_,Pattern,PatternLength,Count) ->
pattern_count(TextTail,Pattern,Pattern,PatternLength,Count).
main() ->
pattern_count(read_text("file.txt"),"hello").
A few final recommendations:
Searching through text element by element is slower than necessary. You should have a look at the Boyer-Moore algorithm and other related algorithms to see ways of advancing through text in larger chunks. For example, Boyer-Moore attempts to match at the end of the pattern first, since if that's not a match, it can advance through the text by as much as the full length of the pattern.
You might want to also looking into using Erlang binaries rather than lists, as they are more compact memory-wise and they allow for matching more than just their first elements. For example, if Text is the input text as a binary and Pattern is the pattern as a binary, and assuming the size of Text is equal to or greater than the size of Pattern, this code attempts to match the whole pattern:
case Text of
<<Pattern:PatternLength/binary, TextTail/binary>> = Text ->
patternCount(TextTail,Pattern,PatternLength,Count+1);
<<_/binary,TextTail/binary>> ->
patternCount(TextTail,Pattern,PatLen,Count)
end.
Note that this code snippet reverts to using patternCount/4 since we no longer need the extra Pattern argument to work through element by element.
As shown in the full revised module, when calling functions in the same module, you don't need the module prefix. See the simplified main/0 function.
As shown in the full revised module, conventional Erlang style does not use mixed case function names like patternCount. Most Erlang programmers would use pattern_count instead.
I find the following behaviour of the R person object rather unexpected:
Let's create a multi-person object:
a = c(person("Huck", "Finn"), person("Tom", "Sawyer"))
Imagine we want to update the given name of one person in the object:
a[[1]]$given <- 'Huckleberry'
Then if we inspect our object, to my surprise we have:
> a
[1] " <> [] ()" "Tom Sawyer"
Where'd Huckleberry Finn go?! (Note that if we try this with just a single person object, it works fine.) Why does this happen?
How can we do the above so that we get the more logical behavior of correcting just the first name?
The syntax you want here is
a <- c(person("Huck", "Finn"), person("Tom", "Sawyer"))
a[1]$given<-"Huckleberry"
a
#[1] "Huckleberry Finn" "Tom Sawyer"
A group of people is still a "person" and it has it's own special indexing function [.person and concat function c.person so it has perhaps different behavior than you were expecting. The problem was that [[ ]] was messing with the underlying hidden list.
Actually, it's interesting because they've overloaded nearly all the indexing methods for person but not the [<- or [[<- and that's really what's causing the error. Because up to here, we're the same
`$<-`(`[`(a,1), "given", "Huckleberry") #works
`$<-`(`[[`(a,1), "given", "Huckleberry") #works
but when we get to
`[<-`(a, 1, `$<-`(`[`(a,1), "given", "Huckleberry")) #works
`[[<-`(a, 1, `$<-`(`[[`(a,1), "given", "Huckleberry")) #no work
we see a difference. The special wrapping/unwrapping that happens during retrieval does not happen during assignment.
So what's going on is that a "person" is always a list of lists. The outer list holds all the people and the inner lists hold the data. You can think of the data like this
x<-list(
list(name="a"),list(name="b")
)
y<-list(
list(name="c")
)
where x is a collection of two people and y is a "single" person. When you do
x[1]<-y
x
you end up with
list(
list(name="c"),list(name="b")
)
since you're replacing a list with a list which is how [ indexing works with lists. But if you try to replace the element at [[1]] with a list of lists, that list will get nested. For example
x[[1]]<-y
x
becomes
x<-list(
list(list(name="c")),list(name="b")
)
And that extra level of nesting is what's confusing R when it goes to print the person in the first position. That first person won't have any named elements at the second level, so when it goes to print, it will return
emptyp <- structure(list(structure(list(), class="person")), class="person")
utils:::format.person(emptyp)
# " <> [] ()"
which gives is the symbols where it's trying to place the name, e-mail address, role, and comment.
Heres a snippet:
translate("a", "4").
translate("m", "/\\/\\").
tol33t([], []).
tol33t([Upper|UpperTail], [Lower|LowerTail]) :-
translate([Upper], [Lower]),
tol33t(UpperTail, LowerTail).
Basically what i want to do is look up in the table for a letter and then get that letter and add it to the new list.
What i have works if its a character, but I'm not sure how to append the new list of characters with the old.
Example input:
l33t("was", L).
It will be put through like this:
l33t([119,97,115], L).
Now that should come back as:
[92,47,92,47]++[52]++[53] or [92,47,92,47,52,53]
Problem is i don't know how to append it like that.
Consider these modifications to tol33t/2:
tol33t([], []).
tol33t([Code|Codes], Remainder) :-
translate([Code], Translation), !,
tol33t(Codes, Rest),
append(Translation, Rest, Remainder).
tol33t([Code|Codes], [Code|Remainder]) :-
tol33t(Codes, Remainder).
The first clause is the base case.
The second clause will succeed iff there is a translation for the current Code via translate/2, as a list of characters of arbitrary length (Translation - note you had [Lower] instead, which restricted results to lists of length 1 only). The cut (!) after the check for a code translation commits to finding the Rest of the solution recursively and then appends the Translation to the front, as the Remainder to return.
The third clause is executed iff there was no translation for the current Code (i.e., the call to translate/2) in the second clause. In this case, no translation for the Code means we just return it as is and compute the rest.
EDIT:
If you don't have cut (!), the second and third clauses can be combined to become:
tol33t([Code|Codes], Remainder) :-
tol33t(Codes, Rest),
(translate([Code], Translation) ->
append(Translation, Rest, Remainder)
; Remainder = [Code|Rest]
).
This (unoptimized) version checks, at every Code in the character list, if there is a translate/2 that suceeds; if so, the Translation is appended to the Rest, else the Code is passed through unchanged. Note that this has the same semantics as the implementation above, in that solutions are commited to (i.e., simulating a cut !) if the antecedent to -> (translate/2) succeeds. Note that the cut in both implementations is strictly necessary; without it, the program will backtrack to find solutions where Code bindings are not translated where there exists an applicable translate/2 clause.