I'm very new to haskell programing and have alot of difficulties quite sometime. The task I'm given here is that a have a list of numbers it then compares to another list of numbers and returns three numbers as feedback based on which interpretation on other numbers could be filtered.
eg.
[4,9],[7,9],[10,18],[2,9] is my list
it should two separate lists now [4,9] and [7,9] it should give a feedback (1,1,0)
first function should check whether two lists have the same value if they have then it should return value 2, if there is only one value then it should return 1, no values then it returns 0
i tried to do this with elem method but im not able to succeed . So the answer for this must be 1 since it has 9 on both lists.
second function checks for the lowest value. It should only compare from the second element to the first. from the same example above [4,9] and [7,9] it first finds out the lowest value in [7,9] which is 7 and then checks for any value lower than 7 in [4,9], it there then it returns a feedback 1 else its 0
third function is the same as the second except it checks for the highest value
eg . [[3,13],[10,9],[5,7]] we take first two elements of the list
[3,13] and [10,9], now we check the highest number in [10,9] which is 10 and check for that value in [3,13] here the answer must be 1 since 13 is higher than 10 else it is 0
please help in declaring these functions
would be much obliged
Your question is a bit confusing without seeing any sample code. I think I understand how your first function needs to work:
It seems that you are using a list of lists, [[Int]] but all of your lists only have two items. It would be easier to use a pair [(Int,Int)]
Then your first function could be written with nested if statements:
firstTest (a1,b2) (a2,b2) = if a1 == a2 && b1 == b2 then 2 else if a1 == a2 || b1 == b2 then 1 else 0
If it can't be a pair and it must be a list of lists then it might be easier to reify your problem and turn the two lists into Sets using Data.Set. Comparing two sets is far more efficient than comparing two lists - and Data.Set provides us the useful isSubsetOf function. You can transform a list into a set using the fromList function.
import qualified Data.Set as S
firstTest' :: S.Set Int -> S.Set Int -> Int
firstTest' sas sbs = if sas == sbs then 2 else if S.isSubsetOf sas sbs then 1 else 0
Comparing adjacent items in a list is a bit challenging. Take a look at my answer to this question: https://stackoverflow.com/a/25777940/3792504
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.
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 want to match numbers that are present in two arrays (not of equal length) and output them to another array if there is a match. The numbers are floating point.
Currently I have a working program in Python but it is slow when I run it for large datasets. What I've done is two nested for loops.
The first nested for loop runs through array1 and checks if any numbers from array2 are in array 1. If there is a match, I write it to an array called arrayMatch1.
I then check array2 and see if there is a match with arrayMatch1. And output the final result to arrayFinal.
arrayFinal will have all numbers that exist within both array1, array2.
My problem:
Two nested for loops give me a complexity of O(n^2). This method works fine for data sets under an array length of 25000 but slows down significant if greater. How can I make it more efficient. The numbers are floating point and always are in this format ######.###
I want to speed up my program but keep using Python because of the simplicity. Are there better ways to find matches between two arrays?
Why not just find the interesection of two lists?
a = [1,2,3,4.3,5.7,9,11,15]
b = [4.3,5.7,6.3,7.9,8.1]
def intersect(a, b):
return list(set(a) & set(b))
print intersect(a, b)
Output:
[5.7, 4.3]
Gotten from this question.
So what you're basically trying to do is find intersection(logically correct term) of 2 list.
First you need to eliminate the duplicate form the list itself, set is great way to do that, then you can just & those lists and you will be good to go.
a = [23.3213,23.123,43.213,12.234] #List First
b = [12.234,23.345,34.224] #List Second
def intersect(a, b):
return list(set(a) & set(b))
print intersect(a, b)
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.
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