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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 : ## 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

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] ?
code :
``````>>> a = []
>>> a.append([])
>>> a.append([1,2,3,4,5])
>>> a.append([1,2,3,4])
>>> a.append([1,2,3])
>>> [len(x) for x in a]
[5, 4, 3]
`````` ## 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

subcount(L, S, C) :-
subcount(L, S, 0, C).
subcount([Hl|Tl], S, A, C) :-
->  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

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], ]
`````` ## 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();
`````` 