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How can I compare two lists (containing character and number in it) in tcl?
I don't have VMD so I made proc from example output. This script finds indexes with the same attribute value. You need to set the attributes and use them in your get. Example: set attrs {name backbone} set values [$sel get $attrs] set answer [same_values $attrs $values] Test script #!/usr/bin/tclsh # http://www.ks.uiuc.edu/Research/vmd/vmd-1.7/ug/node181.html # $sel get {attr1 attr2} # is a

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Prolog list search returns false if list is predefined
someList is the name of a proposition that does not exist, not a list, so it'll always produce false/no. Prolog accepts queries and produces answers. What you seem to be asking for is the concept of variables in an imperative language, which doesn't exist here. You can, for example, ask Prolog for all of the lists that have 1 as their member with member(1, X), or for a particular list member(1, [

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Lists, and an unknown error
That's not a list, it's an object. Objects in Javascript coerce their keys to strings. What you probably want is an array, like so: function testFunction() { var test = ['test1', 'test2', 'test3']; Logger.log(test[0]); } Note that arrays are 0-based, and the use of brackets.

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Term for sublist of all elements except last?
Without context, tail is the end of a list. Head is the beginning of a list. Your clarification of "all except the first" must depend on the environment in which you are working. Tail can refer to any number of elements at the end of a list.

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Why doesn't this list check work?
Your issue is that you're reading newline characters from your data.txt file. Try this: for line in check: data.append(line.rstrip(' ')) When you read the lines. This will strip out trailing new line characters.

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How do I recursively calculate the number of elements in a list?
Your last value is returning 0 because this.next is null. So you are adding everything except the last element, hence you are returning lenght -1. I come from a Ruby background and haven't programmed Java in 10 years, but it would be more like this below, syntax may be off. Also, generally for recursive stuff your base case for returning is going to be the first line of the function. Public

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Abstract List Functions in Racket/Scheme - Num of element occurrences in list
It's a bit trickier than you imagine. As you've probably noticed, we must remove duplicate elements in the output list. For this, is better that we define a remove-duplicates helper function (also using abstract list functions) - in fact, this is so common that is a built-in function in Racket, but not available in your current language settings: (define (remove-duplicates lst) (foldr (lambda (

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Is an infinite list of ones sane?
You don't get an infinite number of ones, of course, but what's called a rational or cyclic term. Not all Prolog systems support cyclic terms, however. Systems that provide some support for rational terms include CxProlog, ECLiPSe, SICStus, SWI-Prolog, and YAP. But be aware that there are differences between them regarding the computations that you can perform with rational terms. A query such as

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Using foldl on 2dimentional lists in Haskell
Is this what you mean by bfs order? val (OrdTree a _) = a children (OrdTree _ ts) = ts bfs :: [OrdTree a] -> [a] bfs ts = (map val ts) ++ bfs (concatMap children ts) Note that I've written bfs to take a list of Ordtree a not just a single tree. The argument to bfs is the list of trees at the current level. map val ts are the values at the current level concatMap children ts are all of the

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Begin at head of a list after iterating through it
Remember what the beginning of the list was and you'll be able to go back to it. existing :: Eq a => [a] -> [(a, b)] -> [a] existing allxs allys = go allxs allys where go [] _ = [] go _ [] = [] go (x:xs) (y:ys) = if x == fst(y) then x:(go xs allys) else (go (x:xs) ys)

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Infinite list not being terminated after appropriate value is found.
The problem lies in how the list is generated. Lets hop in ghci and look how the infinite list looks: ghci> [(a, b, c) | a <- [2..], b <- [a+1..], c <- [b+1..]] We get something which looks like this: [(2,3,4),(2,3,5),(2,3,6),(2,3,7),(2,3,8),(2,3,9),(2,3,10),(2,3,11),(2,3,12),..] Because you have not specified when to stop iterating through c and to go to b, the above sequence wi

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sencha horizontal scroll with a list
First of all, you need to learn Sencha layouting. You can refere here. You were creating a List and then trying to put Dataview inside that. So if you need Dataview, directly extend Dataview like below: Ext.define('Styles.view.Beginning', { extend: 'Ext.dataview.DataView', xtype: 'beginninglist', config: { inline: { wrap: false },

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How to convert (int* int) list to (int: list) in F#
[(0,0); (1,2); (3,4)] |> List.map(fun (x,y)->[x; y]) |> List.concat ie. map tuple sequence to a list of lists, and then join the lists together or [(0,0); (1,2); (3,4)] |> Seq.map(fun (x,y)->seq { yield x; yield y}) |> Seq.concat |> Seq.toList I came up with the longer Seq version first, because the Seq module traditionally had more operators to work with. However, come

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print the elements of a list with brackets
JAVA for(int i:list) { System.out.print("["+i+"]"); } C# foreach(int i in list) { Console.Write("["+i+"]"); }

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Individual object map values in s:select in Struts2
The .{} in Clause.{clause} is called a projection in OGNL and it creates a list of properties of that object. That is why you get all values in option tag. Usually you should put just property name in listValue attribute but in your case you need to get value from map value then reference it like that value.property_of_object <s:select list="Clause" listValue="value.clause" />

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How to add a element to the next list everytime a new value is entered?
Please see the comments in the code # list is a function used by python to create lists, # do not use builtins' names for your variables my_list = [[['Jan'],1], [['Feb'],2],[['Mar'],3]] def month(): # print("Don't use globals! "*100) # you can pass arguments to a function, like in "def mont(my_list)" global my_list # cycling on a numerical index is complicated, use this syntax

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difference between [] and list() in python3
[a_dict.keys()] This one puts a single element in the list. Just as if you were to write [1]. In this case that one element is going to be a list. list(a_dict.keys()) The constructor accepts a sequence and will add all elements of the sequence to the container.

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Tcl pass args "as is" to proc
With $cmd [join $args] you are passing to command just one argument that is a string with all arguments. To pass to the command a list of arguments, you need to put in a string command and arguments and then evaluate it: eval "$cmd [join $args]" so, the resulting command is [catch [eval "$cmd [join $args]"] results] I think that so it can work. Let me know if there are problems. I don't

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Saving items in List Component using Shared Objects (flash AS3)
The problem is how you access SharedObject. you have: saveListObject.data.savedList = value; I would use: saveListObject.data["savedList"] = value; Hope that helps

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Pulling elements from dict embedded in a list in Python 3
{'a': [1,2], 'b':[3,4]}, and the LDAP data structure {'department': ['DepartmentName'], 'memberOf': ['CN=example,OU=of,DC=domain,DC=com', 'CN=example,OU=of,DC=domain,DC=com']} are key-value dicts. Values are retrieved by key. d = {'department': ['DepartmentName'], 'memberOf': ['CN=example,OU=of,DC=domain,DC=com', 'CN=example,OU=of,DC=domain,DC=com']} dept = d['department']

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Ocaml Pattern matching not doing as I expect
In fact, replacex function doesn't work correctly if you're trying to replace second element of the list. This leads to the call to firstx with second argument equal to 0. And your implementation of firstx returns a full list in that case. You should either fix indices in, or rethink what is first zero elements.

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Add each preceeding value of a list to the next
You can do it in a single list comprehension, just add the current item to the sum of the slice of all preceding items: mylist = [item + sum(mylist[:index]) for index, item in enumerate(mylist)]

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indexOf('?') for a list not working
Try lst.indexWhere(_.contains("?")) indexOf "Finds index of first occurrence of some value in this list." link Therefore it finds string that equals "?" in the list.

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jaxb unmarshlling object with list of objects returns the list with one element and all properties null
Use this XML <?xml version="1.0" encoding="UTF-8" standalone="yes"?> <DealLookupResponse xmlns="http://www.starstandards.org/webservices/2005/10/transport" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xsi:type="xsd:anyType"> <Deal> <CompanyNumber>CN7</CompanyNum

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Scheme - Testing if argument is a list (proper or improper)
You have right idea! But the main reason why your code is not working is that on fourth line you have wrong brackets. This is right transcription of your code: (define list? (lambda (ls) (if (or (eq? ls (quote())) (cons (car ls) (cdr ls))) #t #f))) But in my opinion this solution is bad for these reasons: If you supply improper list to this procedure it will end by

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Parse a file in various parts
This line: enumerate(all_file_contents, counter) Doesn't do what you think it does; it iterates over everything in the lists, but numbers them starting from counter instead of 0. A minimal fix would be: for i, line in enumerate(all_file_contents): if i >= counter: tail_list.append(line) However, much better would be to not iterate over the whole file three times. In particul

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Prolog - Replacing X number of elements of a list after element of order I
handling numeric indexing can be tedious in Prolog, I would write replace(L, From, To, X, R) :- findall(S, (nth0(I,L,E), (I>=From, I<To ->S = X ; S = E)), R). that yields ?- replace([0,1,2,3,4,5,6], 2,4,a, R). R = [0, 1, a, a, 4, 5, 6].

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Abstract List Functions in Racket/Scheme
Given the language restriction, we can do this using foldl: (define lst '(-4 -3 -2 -1 0 1 2 3 4 5)) (define (adder e acc) (if (> e 1) (add1 acc) acc)) (foldl adder 0 lst) => 4 If it weren't for the restriction, there's a simpler way in Racket: (count (lambda (x) (> x 1)) lst) => 4 In this case, the abstract list function in use is count, see the documentation for many more fu

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How to list subdirectories in Azure blob storage
If you would like to list all "subdirectories" in "CommonService1" directory you can use something like this: var directory = blobContainer.GetDirectoryReference(@"Common/Service1"); var folders = directory.ListBlobs().Where(b => b as CloudBlobDirectory != null).ToList(); foreach (var folder in folders) { Console.WriteLine(folder.Uri); } Full code sample: va

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What's the name of this algorithm?
Look at assignment problem and stable roomates problem. Addition: In regards to your question, search for the problem first and then choose an algorithm that fits your needs most. Normally, there are a number of algorithms to solve such problems you describe.

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Flex List Within a List?
You could do it that way and it would work, But its not a good idea if you have a lot data. You would have to worry about item renderers. I would probably use a tree and create custom item renderers if need be. May be some code might help.

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Scala getting a collection back when the compiler says that it has type Any
List is defined as List[+A]. So it has one type, not two types. The combination of Double and String is Any. In REPL, the type is shown as List[List[Any]] which is correct. scala> List(List(0.0,"hello world"), List(0.0, "huzzah!"), List(0.0,"herp a erp"), List(0.0, "Open up"), List(0.0, "WHY!! That FACE!")) res30: List[List[Any]] = List(List(0.0, hello world), List(0.0, huzzah!), List(0.0, her

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Multiply a list of lists by a list
import itertools L = [[2,3,6],[3,4,9],[4,8,13]] M = [2,3,4] [[i*j for i,j in zip(subl, subm)] for subl, subm in zip(L, itertools.cycle([M]))] Ouptut: Out[101]: [[4, 9, 24], [6, 12, 36], [8, 24, 52]] Without itertools: answer = [] for subl in L: temp = [] for i in range(len(subl)): temp.append(subl[i] * M[i]) answer.append(temp) Output: In [103]: answer Out[103]: [[4,

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f# sum list in other way
Here's how I would do this with a fold (with type annotations): let orig = [2; 3; 2; 6] let workingSum (origList:int list) : int list = let foldFunc (listSoFar: int list) (item:int) : int list = let nextValue = match listSoFar with | [] -> item | head::_ -> head + item nextValue::listSoFar origList |> List.fold foldFunc [

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Maple Factorials
Here are a few ways. restart: fact:=proc(n::posint) local i, a, b; a := 1; b[1] := 1; for i from 2 to n+1 do a := a*i; b[i] := b[i-1] + a; end do; return seq(b[i],i=1..n+1); end proc: for k from 1 to 10 do fact(k); end do; restart: fact2:=proc(n::posint) local i; seq(add(i!,i=1..k),k=1..n+1); end proc: for k from 1 to 10 do fact2(k); end do; restart: fact3:=pro

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filter a List according to multiple contains
Use filter method. list.filter( name => name.contains(pattern1) || name.contains(pattern2) ) If you have undefined amount of extentions: val extensions = List("jpg", "png") list.filter( p => extensions.exists(e => p.matches(s".*\.$e$$")))

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Implementing iter for list in F#
The type of iter is ('T -> unit) -> 'T list -> unit, i.e. it is used only for the side effects of the function. The main problem is that you're passing too many arguments to func let rec iter func list = match list with | [] -> [] | hd::tl -> func hd iter func tl;; ^ ^ ^ ^ 1 2 3 4 Four of them, but func onl

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Implementing collect for list in F#
The collect function is tricky to implement efficiently in the functional style, but you can quite easily implement it using the @ operator that concatenates lists: let rec collect f input = match input with | [] -> [] | x::xs -> (f x) @ (collect f xs)

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Using length of list X as an argument for a constructor of X in Coq
This is a nice problem indeed, I hadn't encountered it in this form before. Quite frankly, unfortunately, I don't see a way of getting Coq to accept your definition as is. Here are some options for mitigating this problem: Use a separate inductive predicate for term well-formedness. Manipulating terms will be much more convenient, because you won't have to worry about the length constraints whe

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