list,recursion,prolog,sublist,variable-length

Something similar to this is what you might be looking for mylen([H|Lc],N) :- X is mod(H,2), X == 0, mylen(Lc,M), N is M+1. mylen([],0). mylen([H|Lc],N) :- mylen(Lc, M), N = M. ...

Here is one solution. I am first converting the list of lists into a list of sets. If you have any control over the lists, make them sets. def matching_sublists(dict1, dict2): result = dict() for k in dict1: assert(k in dict2) result[k] = 0 A = [set(l) for l in...

Prior to above and my correct understanding this is a problem that is solved all the time. The pseudo code would go something like this.. fn sortObjects(List objects) { var sortedParentHashMap = new HashMap(); foreach(object in objects) { // If the HashMap entry for the current parentId does not exist...

java,arraylist,collections,iterator,sublist

There are several problems with your code: 1) you build the tempList on every item, so while the result are correct, you will see them for every item in myList. you need to "remember" each suffix you process so that you won't process it again. a HashMap will do the...

According to the javadocs for ArrayList.sublist: "The semantics of the list returned by this method become undefined if the backing list (i.e., this list) is structurally modified in any way other than via the returned list. (Structural modifications are those that change the size of this list, or otherwise perturb...

I think the most readable way to do that is to create a Counter instance for each of your sublists, and them check with the list "count" method if it matches the requirement for each argument of the sublist: from itertools import Counter def checksub(main, sublist): c = Counter(sublist) for...

python,list,sum,sublist,round-robin

Here's a simple implementation. First, sort the input list by weight, then enumerate, adding each item to the least full bucket. def EvenlyDistribute(lst, n): """Distribute items in lst (tuples of (val, weight)) into n buckets""" buckets = [[] for i in range(n)] weights = [[0, i] for i in range(n)]...

android,listview,arraylist,sublist

The problem is here: List subList = listItems.subList(1,3); if listItems, is empty or has less than four elements subList throws IndexOutBoundException. To be more precise it throws IndexOutBoundException when (fromIndex < 0 || toIndex > size || fromIndex > toIndex). Edit: I would like to put the title in the...

If I got it right, you need something like: split [] = [] split xs = case foo xs of (ys,r) -> r : split ys foo :: [a] -> ([a],r) foo = undefined In foo, the list should get partially consumed and returns the rest of the list and...

lst = [['a', 'b', 'a'], ['a', 'b', 'c'], ['a']] def count(lst): # declare dictionary that we are going to return foo = {} # iterate sublist for sublist in lst: # make sublist into unique element list sublist = list(set(sublist)) for element in sublist: # if element found in foo...

r,list,extract,dataframes,sublist

Not tested: setNames(data.frame(do.call(rbind,lapply(1:length(parsedData),function(i)cbind(parsedData[[i]][1],parsedData[[i]][2])))),c("Date","Volatility") OR: setNames(data.frame(do.call(rbind,lapply(1:length(parsedData),function(i)t(parsedData[[i]][1:2])))),c("Date","Volatility"))...

Straight-up data processing using your ad-hoc requirement above, I can come up with the following algorithm. First sweep: collect frequency information for every key (i.e. 'A', 'B', 'C'): def generate_frequency_table(lst): assoc = {} # e.g. 'A': {'id1': 3, 'id2': 2} for key, unused, val in list: freqs = assoc.get(key, None)...

Use a simple list comprehension, which iterates over the length of original list. Also, I have used variable name lst, since list is a python type. Here you go: >>> lst = [0,1,2,3] >>> [lst[i:] for i in range(len(lst))] [[0, 1, 2, 3], [1, 2, 3], [2, 3], [3]] ...

java,list,sorting,arraylist,sublist

The subList() method is actually explicitly documented to give you a view of the original: Returns a view of the portion of this list between the specified * {@code fromIndex}, inclusive, and {@code toIndex}, exclusive. (If * {@code fromIndex} and {@code toIndex} are equal, the returned list is * empty.)...

The modifyList function might be useful in this case. Just run lapply(m, modifyList, list(d=1:3)) And that will replace the values of the sublists in M with the values in the list you specify. If you had a vector of d values and only wanted to add one to each list,...

If I understood correctly what You want to achieve, this is the solution: sub_pos_list = [] last_found_pos = 0 for i in xrange(last_found_pos, len(pos_list)): difference = pos_list[i] - pos_list[last_found_pos] if difference >= 1000000: sub_pos_list.append(pos_list[last_found_pos:i + 1]) last_found_pos = i from: pos_list = [0, 5, 1000000, 1000005, 2000005] it will give...

So you want to keep only sublist. You can easily do that with remove-if: (defun remove-atoms (lst) (remove-if #'atom lst)) In your recursive code there is a problem with: (or (unterlisten_zurueckgeben (first lst)) (setq unterlisten (cons (first lst) (unterlisten_zurueckgeben (rest lst)))))) Here if the result of (unterlisten_zurueckgeben (first lst)) is...

python,comparison,append,sublist

You will have to catch the duplicate indexes but this should be a lot more efficient: gr = [] it = iter(mylst) prev = next(it) for ind, ele in enumerate(it): if ele[0] == prev[0] and abs(ele[1] - prev[1]) <= 2: if any(abs(ele[i] - prev[i]) < 10 for i in (2,...

Actually client side is what you have to do. You create a client script to go along with your suitelet. use form.setScript... to associate it. in the client script create an initLine function. That function can use jQuery (automatically included by Netsuite) to find and remove the Remove button. This...

tmp.subList() returns a new List instance that is different from the first element of dList. That's why the original List was unchanged. You need to set the first element of dList to refer to the sub-list you created : List<Double> tmp = dList.get(0); tmp = tmp.subList(1, 3); tmp.set(0, 4.5); dList.set...

Perhaps: lapply(s, function(i) originaldf[ i[ i %in% 1:length(originaldf) ] ] ) Unclear what you mean by "retaining the sublist number", since operations are rarely destructive and the numbers do not appear in your desired result. Also you show an output with varying numbers of elements which would generally need to...

python,list,count,nested-lists,sublist

Why not simply use set? Convert b to set a see if it is a subset of the items in the list using the set.issubset method: b = {11,14,15} #notice the {} braces or use `set([11,14,15])` print sum(b.issubset(x) for x in a) ...

Its what that itertools.groupby is for : >>> from operator import itemgetter >>> from itertools import groupby >>> data=[["apple",2],["cake",5],["cake",8],["chocolate",3],["chocolate",9],["chocolate",10],["grapes",6]] >>> [list(g) for _,g in groupby(sorted(data,key=itemgetter(0)),itemgetter(0))] [[['apple', 2]], [['cake', 5], ['cake', 8]], [['chocolate', 3], ['chocolate', 9], ['chocolate', 10]], [['grapes', 6]]] >>> You can use operator.iemgetter as the key of your sorted...

python,search,dictionary,sublist

So, your fixed program, based on the comments would look like this >>> dict2 = {} >>> for word in all_words: ... # Iterate over the dict1's items ... for key, sublist in dict1.items(): ... # If the word is found in the sublist ... if word in sublist: ......

python,list,dictionary,sublist

If you have small sublists for simplicity you can sort and return the last n sublists after sorting,this presumes biggest to be the sublists with largest sum if you mean has the largest subelement change sum to max: def create_champion_index(d, n): new_d = {} # iterate over dict items key/values...

Since you're looking for a non-contiguous sublist or ordered subset, and not wanting to include the empty list, then: sub_list([X], [X|_]). sub_list([X], [Y|T]) :- X \== Y, sub_list([X], T). sub_list([X,Y|T1], [X|T2]) :- sub_list([Y|T1], T2). sub_list([X,Y|T1], [Z|T2]) :- X \== Z, sub_list([X,Y|T1], T2). Some results: | ?- sub_list([1,4], [1,2,3,4]). true ?...

https://docs.python.org/2/library/itertools.html#itertools.chain is the place you want to start. to make your first pass easier for you to understand and tune-up i recommend declaring the sublists as their own variables and breaking out the slices you want to use ahead of the .chain() call. it will be much easier to understand...