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Counting viable sublist lengths from an array


Counting viable sublist lengths from an array

By : DownPayment
Date : November 19 2020, 07:01 PM
Does that help You need a recursive algorithm which saves it's results to some external data structure.
It should take subarray start and end indexes as parameters, check whether it's good or not.
code :


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Counting Sublist Elements in Prolog

Counting Sublist Elements in Prolog


By : Laura Christie
Date : March 29 2020, 07:55 AM
I wish did fix the issue. How can I count nested list elements in Prolog? I have the following predicates defined, which will count a nested list as one element:
code :
len([H|T],N) :-
    len(H, LH),
    len(T, LT),
    !,
    N is LH + LT.
len([], 0):-!.
len(_, 1):-!.
?- len([a,b,[c,d,e],f],Output).
Output = 6.
Python find list lengths in a sublist

Python find list lengths in a sublist


By : user1671062
Date : March 29 2020, 07:55 AM
hope this fix your issue I am trying to find out how to get the length of every list that is held within a particular list. For example: , [len(x) for x in a[0]] ?
code :
>>> a = []
>>> a.append([])
>>> a[0].append([1,2,3,4,5])
>>> a[0].append([1,2,3,4])
>>> a[0].append([1,2,3])
>>> [len(x) for x in a[0]]
[5, 4, 3]
Counting how many time a list is sublist of another one

Counting how many time a list is sublist of another one


By : user7371296
Date : March 29 2020, 07:55 AM
Any of those help I guess everyone has their own simplified version. Mine's only slightly different with a tail-recursion:
code :
% headsub(S, L) is true if S is a sublist at the head of L
headsub([Hs|Ts], [Hs|Tl]) :-
    headsub(Ts, Tl).
headsub([], _).

subcount(L, S, C) :-
    subcount(L, S, 0, C).
subcount([Hl|Tl], S, A, C) :-
    (   headsub(S, [Hl|Tl])
    ->  A1 is A + 1     % if S is a sublist in front, count it
    ;   A1 = A          % otherwise, don't count it
    ),
    subcount(Tl, S, A1, C).   % Count the rest, adding it into the accumulator
subcount([], _, C, C).  % We're done here. Accumulator becomes the answer.
countSublist([H|T], L2, R):- size(L2, SizeL2), %% we need L2's size
                             subList([H|T], SizeL2, SubL1), %% we get L1's sublist
                             countSublist(T, L2, R),
                             SubL1 = L2,
                             R1 is R+1. %% we do R+1 only if L2=Sublist of L1
countSublist([H|T], L2, R):- size(L2, SizeL2), %% we need L2's size
                             subList([H|T], SizeL2, SubL1), %% we get L1's sublist
                             countSublist(T, L2, R1),
                             (   SubL1 = L2
                             ->  R is R1+1  %% we do R+1 only if L2=Sublist of L1
                             ;   R = R1
                             ).
| ?- countSublist([1,2,3,4,5,2,3,2,3,4], [2,3,4], R).

R = 2 ? a

R = 2

R = 2
Counting elements in a sublist of strings

Counting elements in a sublist of strings


By : Shadaab Khan
Date : March 29 2020, 07:55 AM
it helps some times Use nested list comprehension. Then, take the sum of elements in each sublist to get the corresponding total number of characters. Furthermore, I am presenting a shorter simplified version without using range(len(...)). You can directly loop over the list elements
code :
mylist2 = [sum([len(j) for j in subl]) for subl in mylist] 
# [29, 75, 77, 33]
mylist2 = [[len(j) for j in subl] for subl in mylist]  
# [[16, 13], [21, 33, 21], [28, 16, 33], [33]]
Java. Find the lengths of strings in an array list for which those string lengths occur the most

Java. Find the lengths of strings in an array list for which those string lengths occur the most


By : user3574503
Date : March 29 2020, 07:55 AM
Hope this helps For example, say I have the following array after splitting it. , You can make use of Java Stream Library's groupingBy collector
code :
String[] arr = new String[]{
        "Hello",
        "world",
        "what",
        "a",
        "fine",
        "day",
        "it",
        "is",
        "the",
        "date",
        "is",
        "01/10/2020"
};

List<Integer> maxRepeatingSizes =
    Arrays.stream(arr)
            .collect(Collectors.groupingBy(
                String::length,
                Collectors.counting()
            )).entrySet()
            .stream()
            .collect(
                    Collectors.groupingBy(
                            Map.Entry::getValue,
                            Collectors.mapping(
                                Map.Entry::getKey,
                                Collectors.toList()
                            )
                    )
            ).entrySet()
            .stream()
            .max(
                    Map.Entry.comparingByKey()
            ).get()
            .getValue();
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