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Given a list of lists of integers, e.g. [[3,10],[3,10,2],[5],[5,2],[5,3],[5,3,2],[5,3,10]], I want to go over each sublist and count how many of them sum to 15. In this case that would be 1, for the sublist [3,10,2].
I am aware of the predicate aggregate_all/3, but I'm having trouble writing a predicate to check each element of the list, what I have now is something like
fifteens([X|Xs]) :-
sum_list(X, 15),
fifteens(Xs).
and within another predicate I have:
aggregate_all(count, fifteens(Combinations), Value).
where Combinations is the list of lists of integers in question.
I know my fifteens predicate is flawed since it's saying that all elements of the nested list must sum to 15, but to fix this how do I take out each element of Combinations and check those individually? Do I even need to? Thanks.
First of all your fifteens/2 predicate has no because for empty list and thus it will always fails because due to the recursion eventually fifteens([]) will be called and fail.
Also you need to change completely the definition of fifteens, currently even if you add base case, it says check ALL elements-sublists to see if they sum to 15. That's Ok but I don't see how you could use it with aggregate.
To use aggregate/3 you need to express with fifteens/2, something like: for every part of my combinations list check separately each sublist i.e each member:
ifteens(L) :-
member(X,L),
sum_list(X, 15).
Now trying:
?- aggregate_all(count, ifteens([[3,10],[3,10,2],[5],[5,2],[5,3],[5,3,2],[5,3,10]]), Value).
Value = 1.
This is a job for ... foldl/4. Functional programming idioms in logic programming languages? Yes, we can!
First, summing the summable values of a list:
sum_them(List,Sum) :-
foldl(sum_goal,List,0,Sum).
sum_goal(Element,FromLeft,ToRight) :-
must_be(number,Element),
must_be(number,FromLeft),
ToRight is Element+FromLeft.
Then, counting the ones that sum to 15:
count_them(List,Count) :-
foldl(count_goal,List,0,Count).
count_goal(Element,FromLeft,ToRight) :-
must_be(list(number),Element),
must_be(number,FromLeft),
sum_them(Element,15) -> succ(FromLeft,ToRight) ; FromLeft = ToRight.
Does it work? Let's write some unit tests:
:- begin_tests(fifteen_with_foldl).
test("first test",true(R==1)) :-
count_them([[3,10],[3,10,2],[5],[5,2],[5,3],[5,3,2],[5,3,10]],R).
test("test on empty",true(R==0)) :-
count_them([],R).
test("test with 2 hist",true(R==2)) :-
count_them([[15],[],[1,1,1,1,1,10]],R).
:- end_tests(fifteen_with_foldl).
And so:
% PL-Unit: fifteen_with_foldl ... done
% All 3 tests passed
true.
I have a list containing lists and I want to reverse every second list in it. I tried something but if I have odd number of elements in the list the last list element is lost... So the best solution would be to put the odd lists first and the even lists second till every second list is reversed.
I can't use any libraries. I need to do it recursively or split them and append them again. The best thing I made so far was to reverse only the first even list and append the first odd and even list in a new list.
I tried to do this:
reverselist(List, [List]).
reverselist([X,Y|Rest], [SnakeList|Rest2]):-
append(X, [], Odd),
reverse(Y, EvenList),
append(Odd, EvenList, SnakeList),
reverselist(Rest, Rest2).
And this:
reverselist(List1, List2).
reverselist([H|Ts], [Odd|R]):-
not(0 is H mod 2),
append(H, [], Odd),
reverselist(Ts, R).
reverselist([H|Ts], [Even|R]):-
0 is H mod 2,
reverse(H, Even),
reverselist(Ts, R).
Sample query:
?- reverselist([[a,b,c],[d,a,b],[c,d,o],[b,c,d],[e,e,d]], List).
I want the result to be:
List = [ [a,b,c],[b,a,d],[c,d,o],[d,c,b],[e,e,d] ].
You can also write mutual recursion:
reverselist([],[]).
reverselist([H|T],[H|T1]):-reverselist2(T,T1).
reverselist2([],[]).
reverselist2([H|T],[H1|T1]):-reverse(H,H1), reverselist(T,T1).
You were pretty close with your first variant.
Instead of your
reverselist(List, [List]).
reverselist([X,Y|Rest], [SnakeList|Rest2]):-
append(X, [], Odd),
reverse(Y, EvenList),
append(Odd, EvenList, SnakeList),
reverselist(Rest, Rest2).
just tweak it as
reverselist([], []). % additional clause
reverselist([List], [List]).
reverselist([X,Y|Rest], [X,EvenList|Rest2]):-
reverse( Y, EvenList),
reverselist( Rest, Rest2).
All three clauses are mutually exclusive and together they are exhaustive, i.e. they cover every possibility.
I believe this definition to be the most immediate and close representation of your problem. In Prolog, to formulate the problem means to have the solution for it.
We need to create another predicate with one more argument to keep track of odd or even position:
reverselist(InList,OutList):- reverselist(InList,OutList, 0).
reverselist([],[],_). %base case
%case of even position
reverselist([H|T],[H|T1], 0):- reverselist(T,T1,1).
%case of odd position
reverselist([H|T],[H1|T1], 1):- reverse(H1,H), reverselist(T,T1,0).
Im very confused on how to filter out the element (1,1) from this list in the code below.
take 10 [ (i,j) | i <- [1,2],
j <- [1..] ]
yields
[(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),(1,8),(1,9),(1,10)]
My thoughts were to use something like filter but Im not too sure where to implement it.
My go was Filter ((i,j) /=0) "the list"
Thanks
Your attempt
Filter ((i,j) /=0) "the list"
has a few problems, which can be fixed.
First, the function is called filter. Second, its first argument must be a function: so you can use \(i,j) -> ... to take a list as input. Third, you want (i,j) /= (1,1) -- you can't compare a pair (i,j) to a single number 0.
You should now be able to correct your code.
As an alternative to using filter, you can also specify that you don't want (1,1) as an element within your list comprehension by adding a guard expression (i,j) /= (1,1):
take 10 [ (i,j) | i <- [1,2], j <- [1..], (i,j) /= (1,1) ]
This is similar to how you might write a set comprehension (which list comprehensions mimic):
This answer gives a nice example ([x | i <- [0..10], let x = i*i, x > 20]) of the three types of expression you can have in the tail end of a list comprehension:
Generators, eg. i <- [0..10] provide the sources of values.
Guards, eg. x > 20 are arbitrary predicates - for any given values from the generators, the value will only be included in the result if all the predicates hold.
Local declarations, eg. let x = i*i perform the same task as normal let/where statements.
Names for the different expressions taken from the syntax reference, expression qual.
I'm attempting to sort a list of colors, by a given preffered order. For example a list [r,z,z,w,g,g,r,z] sorted in this order [z,b,g,r,w], will give an end result of [z,z,z,g,g,r,r,w].
I tried using a basic bubblesort algorithme and adding a check to see which of first two terms would be 'higher' on the order list.
% take the to-sorted list, the order in which to sort the list, and the
% result.
%colourSort([r,z,z,w,g,g,r,z],[z,b,g,r,w],X). returns X = [z,z,z,g,g,r,r,w]
colourSort(List,Order,Sorted):-
swap(List,List1,Order),
!,
colourSort(List1,Order,Sorted).
colourSort(Sorted,_,Sorted).
% check if the either the first or second letter is first in the order
% list, if neither check the next letter in the order list.
check(A,_,[H|_],A):-
A == H.
check(_,B,[H|_],B):-
B == H.
check(A,B,[_|T],R):-
check(A,B,T,R).
check(_,_,[],_).
%swap incase a set of letters isn't ordered, continues otherwise.
swap([X,Y|Rest],[Y,X|Rest],Order):-
check(X,Y,Order,R),
X == R.
swap([Z|Rest],[Z|Rest1],Order) :-
swap(Rest,Rest1,Order).
When I run the code, it ends up crashing my swi-prolog, I'm assuming it's getting stuck in a loop or something, but haven't been able to figure out why or how. Any advice or tips would be appreciated.
Here's a solution to the stated problem, which does not, however, use a custom sorting algorithm. Instead, it uses the common pairs data-structure (using the (-)/2 operator to form a list of items Key-Value) and the keysort/2 for sorting. Edit: this answer has been reworked in accordance with #mat's tip in the comments, and to provide a more succinct explanation).
Solution:
item_with_rank(Ranking, Item, Rank-Item) :-
nth0(Rank, Ranking, Item).
sort_by_ranking(Ranking, ToSort, Sorted) :-
maplist(item_with_rank(Ranking), ToSort, Ranked),
keysort(Ranked, RankedSorted),
pairs_values(RankedSorted, Sorted).
Explanation:
We define a predicate item_with_rank(Ranking, Item, Rank-Item) that uses a list of arbitrarily ordered terms as a Ranking, and associates with the given Item a Rank which is equivalent to the 0-based index of the first term in Ranking that unifies with Item. We then define sort_by_ranking(Ranking, ToSort, Sorted). sort_by_ranking/3 uses maplist/3 to call item_with_rank/3, with the given Ranking, on each element of the list ToSort, obtaining a list of pairs, Ranked, assigning a rank to each item. We use keysort/2 to sort the Ranked so that they order of elements accords with the value of their "ranks" (keys) in RankedSorted. When we extract just the values from RankedSorted, we are left with the Sorted items, which is what we were after:
Example of usage:
?- sort_by_ranking([z,b,g,r,w], [r,z,z,w,g,g,r,z], S).
S = [z, z, z, g, g, r, r, w] ;
false.
So I'm new to Erlang and still on the learning curve, one question asked was to return all elements in a list followed by an equal element, to which I could to.
For example...
in_pair_lc([a,a,a,2,b,a,r,r,2,2,b,a]) -> [a,a,r,2]
I was then asked to do the same using a list comprehension, and I hit my mental block.
My unsuccessful attempt was this:
in_pair_lc([]) -> [];
in_pair_lc([H|T]) ->
[X || X ,_ [H|T], X=lists:nth(X+1, [H|T]).
Although with no look ahead in list comp it doesn't work.
Thanks for any help in advance.
One way to do this with a list comprehension is to create two lists from the input list:
one containing all elements except the very first element
one containing all elements except the very last element
By zipping these two lists together, we get a list of tuples where each tuple consists of adjacent elements from the input list. We can then use a list comprehension to take only those tuples whose elements match:
in_pair_lc([_|T]=L) ->
[_|T2] = lists:reverse(L),
[H || {H,H} <- lists:zip(lists:reverse(T2),T)].
EDIT: based on the discussion in the comments, with Erlang/OTP version 17.0 or newer, the two list reversals can be replaced with lists:droplast/1:
in_pair_lc([_|T]=L) ->
[H || {H,H} <- lists:zip(lists:droplast(L), T)].
The first example will work on both older and newer versions of Erlang/OTP.
I'm not convinced the problem is really about list comprehensions. The core of the problem is zipping lists and then using a trivial "filter" expression in the list comprehension.
If you want to stick to basic, long existing, erlang list functions (sublist, nthtail) you could go with the following:
X = [a,a,a,2,b,a,r,r,2,2,b,a].
[A || {A,A} <- lists:zip(lists:sublist(X, length(X)-1), lists:nthtail(1, X))].