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List manipulation

E
elysha last edited by
Help me write a program that changes the elements of the list by the formula first.
```
a[i]=a[i]+a[i-1] для i>0
```
then removes all uneven elements that are less than the arithmetic mean.
Didn't find how to index the list to the maximum that it was possible to add or remove anything from the element or add or remove a specific element:
```
func (xs) = [ x+xs!!0 | x <- xs]
```
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Marcee last edited by
" Indicating on the list " is, in principle, possible, but I understand it is a very " unfunctional path " .
Changes the elements of the list by the formula a[i]=a[i]+a[i-1] for igrad
If the numbering begins at zero, for example:
```
f [] = []
f (x:xs) = x:(f1 x xs) where
  f1 _ [] = []
  f1 n (x:xs) = n+x : f1 (n+x) xs
```
Verification:
```
> f [1,2,3,4]
[1,3,6,10]
```
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How do you take UTCTIme from UTCTime and get the difference the other day?
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If I understand correctly, nominalDiffTimeToSeconds :: NominalDiffTime -> Pico gives the number of seconds. NominalDiffTimewhich can then be divided into 60*60*24 to get the number of days.UPD♪ Best option, I guess, just split it up. nominalDay :: NominalDiffTime And you have to do what you need. fromRational . toRational♪
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Software Programming • haskell • • elysha

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I take it you're trying to define a function, and then you should indicate the names of its parameters.GHCI for default does not support the introduction of multi-string teams, but it can be included with the help of :set +mPrelude> :set +m Prelude> let Prelude| sumSquareOrSquareSum x y = Prelude| let Prelude| sumSquare = x^2+y^2 Prelude| squareSum = (x+y)^2 Prelude| in Prelude| if squareSum > sumSquare Prelude| then squareSum Prelude| else sumSquare Either a multi-layer code in special brackets can be usedPrelude> :{ Prelude| let Prelude| sumSquareOrSquareSum x y = Prelude| let Prelude| sumSquare = x^2+y^2 Prelude| squareSum = (x+y)^2 Prelude| in Prelude| if squareSum > sumSquare Prelude| then squareSum Prelude| else sumSquare Prelude| :} Of course you can put everything in one line, but in that case, syntaxis will change a little.Prelude> let sumSquareOrSquareSum x y = let {sumSquare = x^2+y^2; squareSum = (x+y)^2} in if squareSum > sumSquare then squareSum else sumSquare Well, I guess it's worth noting that the function was overly detailed, which only hurts readability.Prelude> let sumSquareOrSquareSum x y = max (x^2+y^2) ((x+y)^2)
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How to add 3 list information to another list?
Software Programming • python • • Jolied

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A very practical way as suggested by @FjSevilla, is to use a https://docs.python.org/3.6/tutorial/datastructures.html#dictionaries . This structure is similar to a physical dictionary, where you have a clave and one valor. Some considerations of dictionaries:The claves cannot be repeatedAccess through the key is highly efficient, much better than looking for values on a listvalor can be any type of data, even another dictionaryValues are mutable the keys are notAdapting your code a bit:times=int(input("Cuantos nombres? ")) cumples = {} for i in range(0,times): nombre = input("Nombre? ") dia = input("dia de cumpleaños? ") mes = input("mes de cumpleaños? ") lista = cumples.get(mes, []) lista.append((nombre,dia)) cumples[mes] = lista print(cumples) Explanation:The idea is to save in every "month" a list of birthdays (name and day).First, we declare an empty dictionary with cumples = {}Already in the cycle, we search for the key mes the valor, doing: cumples.get(mes, []). This way of access has the particularity that: if the key already exists, it will return the associated value, otherwise it will return an empty list.With lista.append((nombre,dia)) what we do is add a tupla with the name and day entered by the user to the list associated with the key mes. Finally we update the value of the key with cumples[mes] = listaIf we run this code and charge two people with birthdays in the same month and another in another month, our dictionary will be something like this:{'enero': [('juan', '12'), ('pedro', '26')], 'junio': [('natalia','1')]} If you need to access the set of all birthdays and put together a flatter list, you could do this:for k, v in cumples.items(): for t in v: print("mes {0}: día: {1} nombre: {2}".format(k, t[1], t[0])) And you'd get a way out like this:mes enero: día: 12 nombre: juan mes enero: día: 26 nombre: pedro mes junio: día: 1 nombre: natalia
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How to convert lists into list of strings into Spark
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Just do a mapping. lambda on the list in question and operate with join on every inner list.>>> lista = lista.map(lambda x : ''.join(x)) >>> lista.take(2) ['Hola qué tal', 'muy bien']
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Compiling Archive .ly
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haskell font files notmally has the extension .hs (for normal Haskell files) or .lhs (for "literate" files, in which the default for each line is to be a comment and code lines are marked with >).This file .ly that you have actually a grammar description for the parsers generator https://www.haskell.org/happy/ . Following the tradition of yacc, a parsers generator for C that used the extension .y, Happy's incoming files get extensions .y or .ly (for the "literate" version).To compile your grammar file https://www.haskell.org/happy/doc/html/sec-invoking.html .happy Grammar.ly The output should be a "Grammar.hs" file that you must pass pro ghc. For more information, see https://www.haskell.org/happy/doc/html/index.html .
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Make notes for Haskellテラal literals
Software Programming • haskell • • fbio

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It seems to have been solved in the comment field a few minutes ago.What's it? This optimization is called "Remove common partial expression" (CSE: Common Subexpression。。), but it does not work at all for reasons that are unclear. GHC was originally not the most common CSE, but it was a simple CSE. If you do not make thunk correctly, Example: let x = trace "test" 1 in x + x + trace "test" 1 ⇒ let x = trace "test" 1 in x + x + xHome )But now, it seems that even CSE is easy. I don't know when to do this. – https://ja.stackoverflow.com/users/2860/darius-jahandarie January 6, 15 7 15
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Fixed an error in the typical reason for this is that there is no memory enough available (e.g. the OS killed a process using lots of memory)
Software Programming • mac languageos doc languageker haskell • • esther

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I don’t know how to increase memory to be assigned to the image, but there is a way to avoid this error more easily.It is a way to build only the affected package, while limiting the number of CPUs to use.In this case:# ライブラリーのinstallをするときは--libオプションを付けてください。 cabal install --lib -j1 pandoc-types HomeOnce you have successfully built it againcabal install pandoc pandoc-citeprocexecute.As a result, the entire Dockerfile will change:FROM haskell:8 RUN cabal update && cabal install --lib -j1 pandoc-types && cabal install pandoc pandoc-citeproc ENTRYPOINT ["pandoc"] If this method does not work well, it may be better to increase the memory value of your machine. I don’t know how to solve this problem in Docker settings.Production: https://haskell.e-bigmoon.com/posts/2017/12-31-travis-out-of-memory.html This is how to use stack....and then, after answering this far, but if the previous question and purpose have not changed, I would like to install a random package.The Dockerfile described here is to install the famous commands from Haskell called pandoc.So, if you want to use the random package, you don’t have to install pandoc. Sorry, this entry is only available in Japanese.The pandoc is very dependent and takes time to build.If you really want to use a pandoc image, it would be more efficient to create a built image. It is suspicious if there is something that is mented properly.It seems that the sample code is written as a typical application.
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Problem in implementing a chained list - list insert
Software Programming • c language • • zymir

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Let's go for parts.First, in print_names the use of for In this case it is a bad programming practice as you do not know the number of elements that will have the chain list. For this case, use while in this way:void imprime_nomes(nodo* lista) // função que imprime os valores { nodo* p = lista; while (p) { printf("Nome eh: %s\n", p->dado->nome); p = p->prox; } } Second, the problem is in the structure statement nodo_st. You must point the member data as a pointer, and not as a given by itself. It is for this reason that was giving the problem with the pointer arrow (-), because the structure was not indicated as a pointer in the way it should. This will look like this:typedef struct nodo_st { registro * dado; } Third, the parameter used in Creature must be a pointer, and therefore must have asterisk before the variable name:nodo* CriarNodo(registro * p) { nodo* n; n = (nodo*)malloc(sizeof(nodo)); n->dado = p; n->prox= NULL; return n; } The same occurs in the function insert_ini. Correcting, stay like this:void insere_ini(lista *l, registro * dado) { nodo* novo = (nodo*)malloc(sizeof(nodo)); if(novo=NULL){ return 0; //falta de espaço }; novo->dado = dado; novo->prox = l->cauda; //antigo primeiro aponta para o próximo l->cauda = novo; // novo nodo recebe ponteiro para começo return novo; } Finally, it is also necessary to adjust this operation for the modifications made, indicating the arguments as pointers:insere_ini(&p2_main, &p1_main); This will still generate multiple Warnings that you will have to correct gradually. I recommend an in-depth study of pointers as it is one of the most essential knowledge to program well in C.
Cyclical rotation of the Haskell list
Software Programming • list lists haskell • • emmalee

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rotate :: Int -> [a] -> [a] rotate _ [] = [] rotate _ xs@[x] = xs rotate 0 xs = xs rotate n xs'@(x:xs) | n > 0 = rotate (n-1) (xs ++ [x]) | n < 0 = rotate (n+1) (last xs' : init xs') With regard to endless lists: there is no possibility of checking the list indefinitely, so the endless or no list should be determined by the programmer itself, the function of the endless list rotation is therefore limited to the removal of the left element and the transition to the following, i.e., no negative rotation, and the task of rotation of the endless list is not correct, in my view.
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As to add elements to the List<list<string>?</list<string>
Software Programming • c sharp • • Jolied

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As an option:var sentenseList = inputText .Split(new[] { "." }, StringSplitOptions.RemoveEmptyEntries) .Select(x => x.Split(new[] { "-", " ", "," }, StringSplitOptions.RemoveEmptyEntries).ToList()) .ToList(); Where? inputText This is a string variable containing the input text for the sampling.
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Preparing for functional programming and Haskell. Books, articles, councils
Software Programming • haskell • • Saumya

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No higher mathematics for Haskell programming and functional languages should be known at all. It's a stereotype from a series that C++ ancient, Python is a horrific brake and hygigabytes' memory, Perl scruples are totally unreadable, PHP only writes 8th and 9th grades and so on. In reality, functional languages are more common with procedural and objective guidance than differences. You'll understand it quickly when you start to figure it out.About the materials. http://learnyouahaskell.com/ Good textbook. If you need a Russian, please draw your attention. http://ru.wikibooks.org/wiki/%D0%AF%D0%B7%D1%8B%D0%BA_Haskell:_%D0%9E_%D0%BF%D0%BE%D0%BB%D1%8C%D0%B7%D0%B5_%D0%B8_%D0%B2%D1%80%D0%B5%D0%B4%D0%B5_%D0%BB%D0%B5%D0%BD%D0%B8 And some references I've been collecting in this note: http://eax.me/why-haskell/ ♪ Pay attention to the blog. http://users.livejournal.com/_darkus_/ - there's a functional software contest every month that will help you outmove Haskele's possession. I can suggest http://it-talk.org/forum49.html ♪
C

List within JS list
Software Programming • javascript jquery something • • courtlanda

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People, I thank you all for your time, but I managed to solve by breaking my head a little bit more.The solution was as follows:var palavraChave = $("#palavra-chave").val(); // Alterar para o ID do campo var conteudo = editor.getData(); // Pega os valor do textarea CKEDITOR var arrayExterno = conteudo.split(/<p.*?>(.*?)<\/p.*?>/g); // Coloca cada paragrafo dentro de um array // Pega cada palavra dentro do paragrafo e coloca em um array interno ao array anterior var arrayInterno = []; for(i=0; i < arrayExterno.length; i++){ arrayInterno.push(arrayExterno[i].split(" ")); } // Quantidade de vezes que a palavra chave foi encontrada contPalavrasChaves = 0; for(i=0; i<arrayExterno.length; i++){ for(j=0; j<arrayInterno[i].length; j++){ // SOLUÇÃO -> Add o index [i] ao arrayInterno e assim me retornou o valor esperado if(arrayInterno[i] == palavraChave){ console.log(arrayInterno[i][j]); } } } SOLUTION(Ad to code comment line) -> Add index [i] to arrayInterno and thus returned the expected value
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Guards vs roles with pattern matching
Software Programming • haskell pattern matc languagehing • • Amritpal

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First, its function is that it uses guards (guardians), while that of the other participant uses wedding patterns (patter matching). In addition, both will always have the same results and are semantically equal.Choosing to write one way or another is a matter of style or to facilitate reading. In this case, as the comparison of each guard is only equality in all cases (nucleotide == ...), the most common is to opt for marriage of patterns, even knowing that it doesn't make a difference.Maybe you want to read a little about marriage patterns: https://pt.wikibooks.org/wiki/Haskell/Casamento_de_padr%C3%B5es . http://haskell.tailorfontela.com.br/syntax-in-functions#pattern-matching .
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How can I flatten a list of lists and get a list with all the sublist elements?
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With https://docs.python.org/3.6/library/itertools.html#itertools.chain you can use the expansion of elements:list(itertools.chain(*A)) If not, use a double understanding of lists:[item for lista in A for item in lista] Done A=[[1,2,3],[4,5]] in both cases we get [1, 2, 3, 4, 5].
List C++ does not remove the list
Software Programming • c++ stl iterators std list • • briley

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It doesn't turn out because in the second cycle. v==v1 And he just doesn't do it. We need to do this.for ( list<int>::const_iterator it = myList.begin(); it != myList.end(); it++) { cout << *it << " "; }

List manipulation

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