You can use two things to change the coma the first would be to tell the separatorsDecimalFormatSymbols otherSymbols = new DecimalFormatSymbols(currentLocale); otherSymbols.setDecimalSeparator(','); otherSymbols.setGroupingSeparator('.'); DecimalFormat df = new DecimalFormat(formatString, otherSymbols); And the other option is to tell you a language that has coma as decimal is in every europe with what any European country serves you.NumberFormat nf = NumberFormat.getNumberInstance(Locale.GERMAN); DecimalFormat df = (DecimalFormat)nf;

Allocations: df["Year"] = a df["Week"] = s replaces the entire column Year with value aand the entire column Week with value s.If you want to replace only one cell, you have to indicate the row number, so df["Year"][fila] = a df["Week"][fila] = s Without knowing the structure of the dataframe or the logic of the problem, you cannot say what is the right value for fila.

This code prints up to the last full value and then the value with decimalsvalor = "12.5" for x in [*range(1, int(float(valor)) + 1), valor]: print(x) produces:1 2 ... 12 12.5 And with valor = "12.0005" produces1 2 ... 11 12 12.0005

when it's time to print the results on screen, it only shows me "0.00".You printf(), print the memory direction and not variable value. A memory address occupies 8 bytes in memory, while the type float 4 bytes. By formatting an element with a type that has less bytes capacity, it reduces to 0.char var = 'a'; printf("%d\n",sizeof(float)); //4 printf("%d\n",sizeof(&var)); //8 Do not print the memory address of the variable, print the variable:printf(" %.2f:",var); I'm supposed to occupy a block forWhen making a data request for each variable, it has no automation.You can.Define an array that stores all the darts of a necessary type.Collect those and store them in the array, through a cycle for.Access each element through the indexes of the array.char datos[4][10]; //... for (int i = 0;i < 4;i++) { printf("Dato %d\n",i); scanf("%s",datos[i]); } //... Obviously this example is not specific to your program, the idea is that you find how to implement it to your needs.Greetings.

I think you haven't noticed, but what you're actually doing is running. https://www.postgresql.org/docs/9.2/ecpg-dynamic.html , invoking execute.This is called so, because what you do is build valid SQL sentences within character strings and then ask the engine to execute them.For example, the sentence:execute 'insert into ruta' ||i|| '(id, origen, destino, aerolinea, precio_base, distancia, orig_geom, dest_geom, ruta_geom, source, target, seq) select id, origen, destino, aerolinea, precio_base, distancia, orig_geom, dest_geom, ruta_geom, source, target, seq from aerolineas v natural join (select edge id, * from pasos_ruta'||i||') r where r.id=v.id'; It valued your control variable. ito generate sentences insert to different tables, when i Okay. 1, make a insert into ruta1When i 752, make a insert into ruta752etc.If you compare that to your sentence if, you will notice that you are actually comparing two strings of characters (which contain SQL sentences, yes, but they are never executed).if ('select aerolinea from ruta'||i||' where seq=1')<>('select aerolinea from ruta'||i||' where seq=3') then It would be more or less equivalent to doing this (I've removed everything that makes it look like there's SQL through and I've left what makes the strings different:if ('ruta'||i||'1')<>('ruta'||i||'3') then And chains are always different, so it will always come in to execute what you have within the if.Having said this, I have to say that I am not a regular postgresql user and do not have it on hand to test, so from here on I will talk based on my expectations and I will give you alternatives.If the dynamic SQL postgresql supports execution of sentences if (thing I didn't find in a search for documentation), you could generate a string that contains the whole if complete and send to execute it:That, taking into account your example of what if it works, would be something like:execute 'if (select aerolinea from ruta'||i||' where seq=1)<>(select aerolinea from ruta'||i||' where seq=3) then update ruta'||i||' set precio_base=((150*.10)+150) where seq=3; end if;' It is likely that for this to be executed correctly you have to turn it into a complete block (including the sentence do inside the chain, I'll let you do the test if it doesn't work for the first one.Since it is possible that the engine does not support running an if or within a block do, another option is to completely replace the if with a single sentence update make the update only under the right conditions.For me, this sentence would be something likeupdate ruta1742 set precio_base = 165 where seq=3 and aerolinea <> (select b.aerolinea from ruta1742 b where b.seq = 1) I have removed the formula to calculate the new price_base, as since it is all the constant values, it seems to me that it is not necessary, but you can return it.The update will only happen if the airline seq = 1 is different from the airline seq = 3That's what you're looking for.I'll leave you the exercise of putting this in a chain, concatenating i where it takes and throw it with execute.

You can declare a static variable:public static int contadorDePruebaServicio = 0; To that variable you add 1 within the method ejecutarPruebaServicio// Esta línea va dentro del método ejecutarPruebaServicio // Al inicio o al final no hay diferencia, solo asegúrate // de que no quede dentro de un if. contadorDePruebaServicio++; Note the following:If the method is executed from multiple processes all executions are added to the same variable.At some point it will be necessary to restart the counter, by the content of the question I suppose it is after the comparison in the last if.

You're focusing the wrong problem.Right now your algorithm intends to create five random numbers between the one and the five without repeating any of them; for this:You generate a number.You see if the number was already generated.If so, you go back to the first point, otherwise you keep it.This algorithm can potentially be infinite and the more numbers you have saved the more likely to generate a repeat number. https://tio.run/##bU/bSgMxEH3efMWAYDcQwVqLdHfxR1RKTGbTQDZTJokvpd@@xm5BxL7NuXDmHHM8Ppigo5vnOx9NKBZh8JQyo55exS/HOlr6w@jgiH0@VFL4mGHSPrZSnETzgzQzukB7HVDn6qP09gGntYInBRsFzwq2514IaFK2Xbek7y1@eYPAthcLP@X1brd5AdeybaXsBVyFdCjjGLC9gE909fX/j1LBRcdob6tuSRyJoTUUU61dMsF9LBMyQXdjhRTN0sBQyTAMcPXWa/UeV3VSw5gLR3jsxXmevwE :Generates numbers from one to five.Disorder them.int arreglo_aleatorios[] {1, 2, 3, 4, 5}; std::random_device rd; std::mt19937 g(rd()); std::shuffle(std::begin(arreglo_aleatorios), std::end(arreglo_aleatorios), g); for (const auto &numero : arreglo_aleatorios) std::cout << numero << '\n'; You can't have repeated numbers if parts of an unrepeated arrangement.

Based on the solution already proposed we can:Save the path (at least visibly) using filterNor do we need an empty vector to fill because filter will return a vector with the elements that meet the conditionThe conditions for the filter I share them entirely with who responded, I invite you to read https://developer.mozilla.org/es/docs/Web/JavaScript/Referencia/Objetos_globales/Number/isInteger Proposallet numeros = [true, 1, 6.8]; let filtrados = numeros.filter(numero => Number.isInteger(numero)); console.log(filtrados); Observation of Mauritius ContrerasThe use of typeof is not technically necessary. You can use only isInteger() as any analyzed value that is not integer (number or not) will return false

When you make a loop, it is never advisable to modify the object you are traveling for. The best thing is to think that what you need is transformation, starting from an iterable to reach another different.Let's assume what you want is to extract data from a field in the dataframe. This operation can be declared as follows:def get_publish_date(url): article= Article(url) article.download() article.parse() nltk.download("punkt") article.nlp() return article.publish_date In pandas, you can transform a dataframe into another by https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.transform.html and assign it to a column with https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.assign.html :medios_with_publish_date = medios.assign(publish_date=medios['link'].transform(get_publish_date))

The shared code is not right (lack quotes, the number of items in each column are different, etc.), but something like that should do the job more or less.library(data.table) df1 <- data.table(df1) df2 <- data.table(df2) df3 <- merge.data.table(df1, df2, by = "ID") for (cID in unique(df3[, ID])) { write.table(df1[ID == cID], col.names = F, row.names = F, quote = F) if (df1[ID == cID, ntrip] != 0) { write.table(cbind(cols = cols[1], df2[ID == cID, c(2, 3)]), col.names = F, row.names = F, quote = F) write.table(cbind(cols = cols[2], df2[ID == cID, c(2, 3)]), col.names = F, row.names = F, quote = F) } }

One solution is to add the beat hash in the requested class,def __hash__(self): return hash((self.srcIP, self.dstIP, self.prtcl)) By defining it you already evaluate the condition correctly.

I understand that the operator .= adds the value to the initial value, does not replace it, that is why you get doubled. The operator outside the for that is duplicating its same value. This: $friends .= substr($friends,0,-2)

You can add content to the current text by adding to the existing current text. I included a line prior to the for to clean up the area. If you wish not to lose the current data, that line would not be necessary.txtbinario.setText(""); //limpiar txtbinario for(i=0;i<pro;i++){ System.out.println(Elementos[i]); txtbinario.setText(txtbinario.getText()+"S"+(i+1)+" = "+Elementos[i]+"\n"); }

Why is this happening?The problem is with remove()the list in question is being modified. Therefore the values will not be the same in each cycle. The time variable that of the cycle will take the values of each element in the iterable, means that iteration will follow by the following element and will leave certain values behind.for i in lista: print(i) if i % 2 != 0: lista.remove(i) With this cycle you can visualize that values take i, for each iteration. On screen you get29 23 16 12 5 -12 As you can, values are never evaluated 24, 27 y -5, they stay on the list no matter whether the condition is met or not.How do I solve this?The most practical thing is to clone the list to an auxiliary list, and iterate on this auxiliary list. To avoid affecting the values that the time variable takes:lista = [29, 24, 23, 17, 16, 12, 5, -5, -12] lista_aux = lista[:] for elemento in lista_aux: if elemento % 2 != 0: lista.remove(elemento) Here's what's going on. lista_aux, all list values are assigned. The slice [:] refers to all elements of principle at the end of the iterable.This piece of code would show with print(lista):[24, 16, 12, -12] If you're upset about having created another whole list just for this, after the end of the cycle you can erase to lista_aux:del lista_aux Why should you copy a list with a list slice?Let's see what happens when a sentence is madelista = [8,3,45,6] lista_copia = lista He's being assigned. value of lista a lista_copia. The first thing that can alarm us, is when it comes to verify if they are the same element:print(lista is lista_copia) sampleTrue If we try to modify the values of one, the other will also be modified. Because they are the same objectlista.pop() print(lista_copia) sample[8, 3, 45] Here you can see that the last element of listaand this also affects lista_copia.And... Why is this happening?The trick is on the operator =This is an assignment operator. The value of the right is assigned to the variable on the left. So you don't have to assign the value of lista a lista_copia, you have to assign him to lista_copia each of the listalista = [8,3,45,6] lista_copia = lista[:] This time print(lista is lista_copia) return False.I hope it served you something.

You actually have problems with the logic that it is applying, however, is a basic problem of accumulators and counters, with the accumulators sums the matches you want to add is usually of the acum type = acum + num, and then since you have to draw averages you must be able to count each match separately, the counters are of the type, cont = cont + 1. And finally, nothing prevents you from using another one within a decision to segregate 0 of the elders to 0. Greetings!

First we built a generator function. This is useful when you want to have two processes, where one generates data that should be processed by the other as they are being generated.The function receives the name of a file with tweets (one per line). The function runs through the file, producing a text at the same time:def sig_texto(nombre_archivo): """ Función generadora. En cada llamada retorna el texto de otro tweet. @param nombre_archivo El nombre del archivo de tweets. @yield El texto del tweet """ with open (nombre_archivo) as tweets: for line in tweets: contenido=json.loads(line) if 'text' in contenido: yield contenido['text'] We also need to build a dictionary with valued feelings:with open("sentimientos.txt") as sentimientos: valores = {} for linea in sentimientos: termino, valor = linea.split("\t") valores[termino.lower()] = int(valor) We take the precaution of minisculating the text of feelings.We now have to go through the tweets, separating each text in words, discarding all the special characters (.,?-, etc) that they can bring glued. We do that with a regular expression and findall().import re palabra = re.compile("[\w]+") Now we go through the tweets by valuing every word. If the tweet has a value other than zero, we show itfor text in sig_texto('Tweets.json'): # Por cada tweet. valor_tweet = 0 # Recorrer las palabras dentro del tweet for termino in palabra.findall(text): term = termino.lower() if term in valores: valor_tweet += valores[term] if valor_tweet &gt; 0: print(valor_tweet, text) TestingNow, with everything together, let's tryimport json import re palabra = re.compile("[\w]+") def sig_texto(nombre_archivo): """ Función generadora. En cada llamada retorna el texto de otro tweet. @param nombre_archivo El nombre del archivo de tweets. @yield El texto del tweet """ with open (nombre_archivo) as tweets: for line in tweets: contenido=json.loads(line) if 'text' in contenido: yield contenido['text'] Cargar el diccionario de sentimientos. with open("sentimientos.txt") as sentimientos: valores = {} for linea in sentimientos: termino, valor = linea.split("\t") valores[termino.lower()] = int(valor) for text in sig_texto('Tweets.json'): # Por cada tweet. valor_tweet = 0 # Recorrer las palabras dentro del tweet for termino in palabra.findall(text): term = termino.lower() if term in valores: valor_tweet += valores[term] if valor_tweet &gt; 0: print(valor_tweet, text) with this file of feelings:love 1 amor 1 mariage 1 ugly -1 happy 1 sincero 1 produces:1 RT @byunghns: I LOVE #틴탑 SO MUCH #쉽지않아 IS GOING TO BE SO GOOD 2 RT @Positivamos: tudo o que eu quero é um amor sincero. 1 RT @GossipRoomOff: Si Nathalie sait pertinemment que son couple ne va pas durer, alors pourquoi accepter la demande en mariage de Vivian ? … 2 RT @Positivamos: tudo o que eu quero é um amor sincero. 2 RT @Positivamos: tudo o que eu quero é um amor sincero. 2 RT @Positivamos: tudo o que eu quero é um amor sincero. 2 RT @Positivamos: tudo o que eu quero é um amor sincero. 2 RT @Positivamos: tudo o que eu quero é um amor sincero. 1 RT @Nashgrier: Happy Birthday to one of my best friends @jackgilinsky Can't wait till you're out in LA! Have a good one! 2 RT @Positivamos: tudo o que eu quero é um amor sincero. 2 RT @Positivamos: tudo o que eu quero é um amor sincero. 2 RT @Positivamos: tudo o que eu quero é um amor sincero. 2 RT @Positivamos: tudo o que eu quero é um amor sincero. 2 RT @Positivamos: tudo o que eu quero é um amor sincero. 2 RT @Positivamos: tudo o que eu quero é um amor sincero. 2 RT @Positivamos: tudo o que eu quero é um amor sincero. 2 RT @Positivamos: tudo o que eu quero é um amor sincero. 1 I thought I would be happy by now 1 RT @JeremiahTaylorr: happy birthday @MeghanNally ❤️❤️ imysm i found this in my dropbox http://t.co/NEFZJVWs6I 1 Happy Birthday to Robbie , @arsenalfantv you offer excellent insight &amp; coverage on The Mighty Arsenal hope you continue to expand &amp; Innovate 2 RT @Nashgrier: Happy birthday @tfionash I love you soooo much! Thank you for everything, you have no idea how much you mean to me. Hope it'…

The most basic sort algorithm is bubble sweep, or bubble draw https://es.wikipedia.org/wiki/Ordenamiento_de_burbuja The idea is to go through the list by examining each element against which one is ahead. If there is an element that is larger, you exchange it and keep looking until you reach the end of the list. Thus, each element will be greater than all who come later.The problem with your code is that you only compare each element to the next. You must compare it against the rest of the list ahead. That means you need two. for nest.Here is an implementation:lista_empleados=["Juan", "Pedro", "Agustin", "Maria", "Isabel", "Ana"] ingreso_acumulado=[100, 110, 90, 120, 70, 60] def bubble_sort(lista1, lista2): """ Bubble sort de dos listas paralelas. Ordena la primera lista de mayor a menor. Ordena la segunda lista para mantener la informaciòn consistente. """ num = len(lista1) for x in range(num - 1): for y in range(x+1, num): < lista1[y]: lista1[x], lista1[y] = lista1[y], lista1[x] lista2[x], lista2[y] = lista2[y], lista2[x] def print_lista(titulo, lista1, lista2): print(titulo) for i in range(len(lista1)): print("%4d %s" % (lista1[i], lista2[i])) print_lista("Lista original", ingreso_acumulado, lista_empleados) bubble_sort(ingreso_acumulado, lista_empleados) print_lista("Lista ordenada", ingreso_acumulado, lista_empleados) produces:Lista original 100 Juan 110 Pedro 90 Agustin 120 Maria 70 Isabel 60 Ana Lista ordenada 120 Maria 110 Pedro 100 Juan 90 Agustin 70 Isabel 60 Ana CommentsThe bubble draw is inefficient. Running out in situations with a minimo data game (about 10-20 items) when the environment does not offer an alternative draw function.

Note: read the comments in the code.Compare valuesYou just have to compare the values that are found Array.For example:// Array de números: const numeros = [1, 2, 3, 30, 5, 6]; // Nuestra variable auxiliar para almacenar en ella // el valor más grande del array: let max = 0; // Recorres el valor del array «numeros»: for ( let numero of numeros ) { // Evalúa si «max» es menor que «numero» para almacenar // en él el número más grande hasta el momento: if (max < numero) max = numero; } // Muestra en la consola el valor numérico más grande del Array: console.log(max);You just need to create an auxiliary variable (maxin this case) to store in it the largest value of Array.

What happens is that inside for You are no longer evaluating if the number is negative, add this evaluation also, so you can get out of the loop:public class bucles4{ public static void main(String[]args){ int numero, contador; contador = 0; numero = Integer.parseInt(JOptionPane.showInputDialog("Digite un numero")); boolean condicion = numero >= 0; for(int i= 0; condicion; i++){ contador++; numero = Integer.parseInt(JOptionPane.showInputDialog("Digite otro numero")); //*Evalua si es negativo. condicion = numero &gt;= 0; } JOptionPane.showMessageDialog(null, "Has tecleado: "+contador+" cantidad de numeros"); } } }

The proposed solution consists of three parts:First we need a function valor_celdawhich recovers the value of a given cell (fila, column). If the cell doesn't exist, it returns zero. This function is responsible for the problem of seeing whether or not a cell exists.Second. With that function we can build the function cuenta_minasthat counts the mines around a row, column.Third, we went through the board counting the mines around each cell.The beginning is called "dividing to reign"The code is:tablero = [["A",0,0,0,0,8,0,0,0,0,0,8,0,0,0,0], ["B",0,0,0,0,0,0,0,0,0,0,8,0,0,0,0], ["C",8,0,0,0,0,0,0,0,0,0,0,0,0,0,0], ["D",0,0,0,0,0,0,0,0,0,0,8,0,0,0,0], ["E",0,0,0,0,0,8,0,0,0,0,0,0,0,0,0], ["F",0,0,8,0,0,0,0,0,0,0,0,0,0,0,0], ["G",0,0,0,0,0,0,0,0,0,8,0,0,0,0,0], ["H",0,0,0,0,0,8,0,0,0,0,0,0,0,0,0], ["I",0,8,0,8,0,0,0,0,0,0,8,0,0,0,0], ["J",0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], ["K",0,0,0,0,0,0,0,0,0,8,0,8,0,0,0], ["L",0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], ["M",0,0,0,8,0,0,0,0,0,0,0,0,0,0,0], ["N",0,0,0,0,0,0,0,0,0,8,0,0,0,0,0], ["O",0,0,0,0,0,8,0,0,0,0,0,0,0,0,0]] n_filas = len(tablero) n_columnas = len(tablero[0]) def valor_celda(fila, columna): """ Recupera el valor de la celda en fila, columna. Si la celda no existe, retorna cero. """ valor = 0 if fila >= 0 and fila < n_filas: if columna > 0 and columna < n_columnas: valor = tablero[fila][columna] return valor def cuenta_minas(fila, columna): """ Cuenta las minas alrededor de fila, columna. Actualiza la cuenta directamente en el tablero. """ if tablero[fila][columna] == 0: for row in range(fila - 1, fila + 2): for col in range(columna - 1, columna + 2): if valor_celda(row, col) == 8: tablero[fila][columna] += 1 Recorrer el tablero contando las minas. for fila in range(n_filas): for columna in range(1, n_columnas): cuenta_minas(fila, columna) for fila in tablero: print(fila) It produces:['A', 0, 0, 0, 1, 8, 1, 0, 0, 0, 2, 8, 2, 0, 0, 0] ['B', 1, 1, 0, 1, 1, 1, 0, 0, 0, 2, 8, 2, 0, 0, 0] ['C', 8, 1, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 0, 0, 0] ['D', 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 8, 1, 0, 0, 0] ['E', 0, 1, 1, 1, 1, 8, 1, 0, 0, 1, 1, 1, 0, 0, 0] ['F', 0, 1, 8, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0] ['G', 0, 1, 1, 1, 1, 1, 1, 0, 1, 8, 1, 0, 0, 0, 0] ['H', 1, 1, 2, 1, 2, 8, 1, 0, 1, 2, 2, 1, 0, 0, 0] ['I', 1, 8, 2, 8, 2, 1, 1, 0, 0, 1, 8, 1, 0, 0, 0] ['J', 1, 1, 2, 1, 1, 0, 0, 0, 1, 2, 3, 2, 1, 0, 0] ['K', 0, 0, 0, 0, 0, 0, 0, 0, 1, 8, 2, 8, 1, 0, 0] ['L', 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 2, 1, 1, 0, 0] ['M', 0, 0, 1, 8, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0] ['N', 0, 0, 1, 1, 2, 1, 1, 0, 1, 8, 1, 0, 0, 0, 0] ['O', 0, 0, 0, 0, 1, 8, 1, 0, 1, 1, 1, 0, 0, 0, 0]