If I have a function
let rec function n =
if n<0 then []
else n-2 # function n-2 ;;
I get an error saying that the expression function n-2 is a list of int but it is expecting an int.
How do I concatenate the values to return all the n-2 values above zero as a list?
I cannot use the List module to fold.
Thanks
Your title asks how to concatenate lists, but your question seems rather different.
To concatenate lists, you can use the # operator. In many cases, code that depends on this operator is slower than it needs to be (something to keep in mind for later :-).
Here are some things I see wrong with the code you give:
a. You can't name a function function, because function is a keyword in OCaml.
b. If you use the # operator, you should have lists on both sides of it. As near as I can see, the thing on the left in your code is not a list.
c. Function calls have higher precedence than infix operators. So myfun n - 2 is parsed as (myfun n) - 2. You probably want something closer to myfun (n - 2).
Even with these changes, your code seems to generate a list of integers that are 2 apart, which isn't what you say you want. However, I can't understand what the function is actually supposed to return.
It seems like you are not concatenating lists, but concatenating ints instead. This is done by the :: operator. So your code would look like:
else (n-2)::(fun (n-2))
Although I could see this function possibly not producing the desired output if you put in negative numbers. For example if you pass through n = 1, n-2 will evaluate to -1 which is less than zero.
Related
i am a beginner in ocaml and I am stuck in my project.
I would like to count the number of elements of a list contained in a list.
Then test if the list contains odd or even lists.
let listoflists = [[1;2] ; [3;4;5;6] ; [7;8;9]]
output
l1 = even
l2 = even
l3 = odd
The problem is that :
List.tl listoflists
Gives the length of the rest of the list
so 2
-> how can I calculate the length of the lists one by one ?
-> Or how could I get the lists and put them one by one in a variable ?
for the odd/even function, I have already done it !
Tell me if I'm not clear
and thank you for your help .
Unfortunately it's not really possible to help you very much because your question is unclear. Since this is obviously a homework problem I'll just make a few comments.
Since you talk about putting values in variables you seem to have some programming experience. But you should know that OCaml code tends to work with immutable variables and values, which means you have to look at things differently. You can have variables, but they will usually be represented as function parameters (which indeed take different values at different times).
If you have no experience at all with OCaml it is probably worth working through a tutorial. The OCaml.org website recommends the first 6 chapters of the OCaml manual here. In the long run this will probably get you up to speed faster than asking questions here.
You ask how to do a calculation on each list in a list of lists. But you don't say what the answer is supposed to look like. If you want separate answers, one for each sublist, the function to use is List.map. If instead you want one cumulative answer calculated from all the sublists, you want a fold function (like List.fold_left).
You say that List.tl calculates the length of a list, or at least that's what you seem to be saying. But of course that's not the case, List.tl returns all but the first element of a list. The length of a list is calculated by List.length.
If you give a clearer definition of your problem and particularly the desired output you will get better help here.
Use List.iter f xs to apply function f to each element of the list xs.
Use List.length to compute the length of each list.
Even numbers are integrally divisible by two, so if you divide an even number by two the remainder will be zero. Use the mod operator to get the remainder of the division. Alternatively, you can rely on the fact that in the binary representation the odd numbers always end with 1 so you can use land (logical and) to test the least significant bit.
If you need to refer to the position of the list element, use List.iteri f xs. The List.iteri function will apply f to two arguments, the first will be the position of the element (starting from 0) and the second will be the element itself.
I need writing a function which takes as input
a = [12,39,48,36]
and produces as output
b=[4,4,4,13,13,13,16,16,16,12,12,12]
where the idea is to repeat one element three times or two times (this should be variable) and divided by 2 or 3.
I tried doing this:
c=[12,39,48,36]
a=size(c)
for i in a
repeat(c[i]/3,3)
end
You need to vectorize the division operator with a dot ..
Additionally I understand that you want results to be Int - you can vectorizing casting to Int too:
repeat(Int.(a./3), inner=3)
Przemyslaw's answer, repeat(Int.(a./3), inner=3), is excellent and is how you should write your code for conciseness and clarity. Let me in this answer analyze your attempted solution and offer a revised solution which preserves your intent. (I find that this is often useful for educational purposes).
Your code is:
c = [12,39,48,36]
a = size(c)
for i in a
repeat(c[i]/3, 3)
end
The immediate fix is:
c = [12,39,48,36]
output = Int[]
for x in c
append!(output, fill(x/3, 3))
end
Here are the changes I made:
You need an array to actually store the output. The repeat function, which you use in your loop, would produce a result, but this result would be thrown away! Instead, we define an initially empty output = Int[] and then append! each repeated block.
Your for loop specification is iterating over a size tuple (4,), which generates just a single number 4. (Probably, you misunderstand the purpose of the size function: it is primarily useful for multidimensional arrays.) To fix it, you could do a = 1:length(c) instead of a = size(c). But you don't actually need the index i, you only require the elements x of c directly, so we can simplify the loop to just for x in c.
Finally, repeat is designed for arrays. It does not work for a single scalar (this is probably the error you are seeing); you can use the more appropriate fill(scalar, n) to get [scalar, ..., scalar].
Is it possible to write functions that returns functions whose signature depends on the arguments to the builder function?
Specifically I am refining an implementation of primitive recursion I wrote. I want to have factory like functions that for numeric parameters generate a function that works on tuples or lists of the length equal to the arguments passed for the numeric parameters. Currently I am handling cases like pr 1 2 [0,5,13], which is an invalid statement from the perspective of primitive recursion, at runtime through Either:
pr :: (Show a) => Int -> Int -> [a] -> Either String a
pr i k args
| 1 <= i && i <= k && length args == k = Right $ args!!(i-1)
| i <= 0 = Left "first argument of pr needs to be greater or equal to 1"
| k < i = Left "first argument of pr needs to be lesser or equal to the second argument"
| length args /= k = Left $ "pr expected "++(show k)++" arguments, got "++(show $ length args)++": pr "++(concat[show i, " ", show k, " ", show args])
But I would like to somehow catch that case at compile time, as from the perspective of the formal system I want to implement with this, this is a compile time error -- passing more arguments to a function than its domain specifies.
Is this somehow possible and if not what would be the correct approach to get compile time errors for what should be invalid statements.
What you want is a sized vector. It is like a list but in addition to the type of its elements, it is also parametrised by type level natural numbers.
sized-vector package on Hackage is what you need. As it happens, the function you're trying to implement is the last function in this library.
Note that every time you call last you will have to prove the compiler that its argument vector is of size at least 1. You can do this by constructing the vector in the source code (for example, the compiler will understand 1 :- 2 :- Nil is of size 2) or if the vector is obtained at runtime perhaps by conversion from a list, you'll have to write a function that either gives a run time error if it has no elements or constructs a vector of size at least one i.e. have the type level size S n for some n.
If you're not familiar with dependently typed programming (a paradigm that includes this and much much more) I suggest you look through some tutorials first. For example, this post is a good example that includes how to implement vectors from scratch and write functions for them.
A word of caution, learning and using dependently typed programming is exciting, addictive, but also time consuming. So if you want to focus on the task at hand, you might like to live with runtime checks for now.
I'm a newbie prolog programmer, and for an assignment, I have to have a basic program that succeeds if and only if list X is a list of two elements, with the first as the same as the second.
From my view of prolog, programs seem to be pretty small, so I typed this in:
firstPair(x,x).
When I run it under swipl, I get this as output:
Syntax error: Operator expected
Is there something more that needs to be done? I thought that if I executed this with say, firstPair(1,2). this would be all it would need to know that it is false.
First, lowercase x is not a variable, it's an atom. Make x uppercase to fix the problem:
firstPair(X,X).
Second, you do not type this into the interpreter. Rather, you write it into a file firstPair.pl, and then read that file into Prolog.
At the command prompt, type this:
['firstPair.pl'].
Press enter. Now you can use your firstPair/2 rule.
Finally, since the assignment talks about lists, I think the instructor wanted you to write firstPair/1, not firstPair/2:
firstPair([X,X]).
Your program/fact
firstPair(X,X).
will succeed if the two arguments given it can be unified, whether they are lists, atoms, variables, etc. To meet your specification, a
program that succeeds if and only if list X is a list of two elements,
with the first as the same as the second.
You need something like this:
list_of_two_elements( [X,X] ).
This will succeed if passed a single term that is (or can be unified with) a list of two elements that are, or can be made through unification, identical. For instance, all of the following will succeed:
list_of_two_elements( X ).
on success, the variable X will be unified with a list of two elements containing the same unbound variable, something like [V1,V1].
list_of_two_elements( [1,1] ).
list_of-two_elements( [1,X] ). (on success, X here will have been unified with the integer 1.)
So I'm reading http://learnyouahaskell.com/starting-out as it explains lists, and using ghci on Vista 64. It says that [2,4..20] steps by 2 from 4 to 20. This works. It says [20,19..1] goes from 20 to 1, but doesn't explain. I take it that the first number is NOT the step, the step is the difference between the 1st and 2nd number. This is confirmed by [4,4..20] which hangs (no error message, must kill console). This is unlike operators like !! and take which check the index's range and give an error message.
My question is: is this a bug on Vista port or is that the way it's supposed to be?
[x,y..z] does indeed step from x to z by step y-x. When y-x is 0 this leads to an infinite list. This is intended behavior.
Note that if you use the list in an expression like take 20 [2,2..20], ghci won't try to print the whole list (which is impossible with infinite lists of course) and it won't "hang".
Quoting this book:
[n,p..m] is the list of numbers from n to m in steps of p-n.
Your list [4,4..20] "hangs", because you have a step of 4-4=0, so it's an infinite list containing only the number 4 ([4, 4, 4, 4...]).
Haskell allows infinite lists and as the Haskell is the "lazy evaluation language", meaning it will only compute what is necessary to give you the result, so the infinite structures are allowed in Haskell.
In Haskell you could compute something like "head[1..]". This is because Haskell only calculates what is required for the result. So in the example above it would generate only the first element of the infinite list (number 1) and head would return you this element (number 1).
So, in that case program will terminate! However, if you calculate [1..] (infinite list) program won't terminate. Same applies to your example, you created an infinite list and there is no way of terminating it.
That syntax basically is derived from listing the whole list. [1,3,5,7,9,11,13,15,17,19] for example can be shortened by simply omitting the obvious parts. So you could say, if I specify the first two elements, it is clear how it would continue. So the above list equals to [1,3..19].
It's worth noting that the .. syntax in lists desugars to the enumFrom functions given by the Enum typeclass:
http://hackage.haskell.org/packages/archive/base/latest/doc/html/Prelude.html#t:Enum