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to remove words in order to reduce their length

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emerson last edited by

Speak in order to lose their length, from the longest to the shortest.
Initiate in the programme code:
So she was considering in her own mind, as well as she could, for the hot day made her feel very sleepy and stupid, whether the pleasure of making a daisy-chain would be worth the trouble of getting up and picking the daisies, when suddenly a White Rabbit with pink eyes ran close by her.
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carriann

src = 'So she was considering in her own mind, as well as she could, for the hot day made her feel very sleepy and stupid, whether the pleasure of making a daisy-chain would be worth the trouble of getting up and picking the daisies, when suddenly a White Rabbit with pink eyes ran close by her.'
print '\n'.join(sorted(set([x.strip('.,').lower() for x in src.split(' ')]), key=lambda x: len(x), reverse=True))

python 57244

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The schedule needs to be painted once, and we need to transmit the list.import matplotlib.pyplot as plt import matplotlib from math import sin, pi from pylab import * import os xmax = 10 #интервал в секундах x = 0 freq = 4 #частота дискретизации в Гц ff = 0.6 xstep = 1 / freq bits = 6 mult = 1 << (bits - 1) result = [] while x < xmax: y = math.sin(x*ff) dig = int(y * mult) result.append(dig) x += xstep plt.plot(result) plt.grid(True) plt.show()
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We've got to stop the skirts in the same line and not forget the end of the line:[^"]*(?="$)
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Removal of words on the list
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Let's set up a method. RemoveWords(str, word)♪In the body of the method, the index is obtained by the method IndexOf(str, word) and Remove(startIndex, length) remove all the symbols.Read it. StringBuilder You can. https://docs.microsoft.com/ru-ru/dotnet/api/system.text.stringbuilder?view=net-5.0 Read it. IndexOf() You can. https://docs.microsoft.com/ru-ru/dotnet/api/system.string.indexof?view=net-5.0 Read it. Remove() You can. https://docs.microsoft.com/ru-ru/dotnet/api/system.string.remove?view=net-5.0 static void Main(string[] args) { string str = "84ыва846ыв88сСлово545выщв"; string removedWordString = RemoveWord(str, "Слово"); Console.WriteLine(removedWordString); Console.ReadKey(); } static string RemoveWords(string str, string word) { StringBuilder builder = new StringBuilder(str); int index = str.IndexOf(word); builder.Remove(index, str.Length - index); return builder.ToString(); }
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Basically there are ":" at the end of the line if and elseprint("Agora eu vou dizer algo para você.") idade = int(input("Me diga sua idade.")) if idade < 18: print("Me desculpe, você não pode dirigir.") else: print("Que bom, você pode dirigir!") wait = input("Aperte qualquer tecla para sair do CMD.") exit Also lacked indentation (recuo) on the lines of print, but this may have been a problem when you posted the question, I imagine that in your original code there are the indents (if not, add).See the code working perfectly on https://ideone.com/pFrKa2 .(with simulated input 19 for the first input, and x for the second)
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The module https://pypi.python.org/pypi/pyttsx/1.1 the latest version in pypi is 1.1 in 2012 (although there seems to be a more recent version according to documentation, 1.2 in 2015) and is written in Python 2.Possibly fixing the imports in the source code can make it work. You can also use Python 2, but it is preferable to be carrying the entire Python 2 code to Python 3 given the more than next death of branch 2 (if nothing changes to PyCon 2020 you will stop supporting Python 2). However, the module is still quite outdated.There is a recent and maintained for Python 3 / Python 2 that you should consider using called pyttsx3: https://pypi.python.org/pypi/pyttsx3/2.6 Its handling is supposed to be identical to pytsx (although I have never used this last one). I've used pytsx3 before and it works perfectly at Python 3.6 and 3.5 which is where I've tried it. Your code doesn't change anything, just the name of the module:import pyttsx3 engine = pyttsx3.init() engine.say('Sally sells seashells by the seashore.') engine.say('The quick brown fox jumped over the lazy dog.') engine.runAndWait()
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Remove word from char array that starts and ends by 'a' in java
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I recommend using you split so as to separate the word. and then check if the first and last character is a. Make sure you have the letters in lowercases or uppercases (I chose lowercases), then create a ArrayList to save all the words that begin and end with a In addition, add that if the word comes with accents, look for them. He'll find matches with Amiga (without accent) but show the result with AmigáString string = "la Amigá es mía azula"; String palabraLimpia = new String(string); palabraLimpia = Normalizer.normalize(palabraLimpia, Normalizer.Form.NFD); palabraLimpia = palabraLimpia.replaceAll("[^\\p{ASCII}]", ""); System.out.println(palabraLimpia); ArrayList<String> lista = new ArrayList<String>(); String[] parts = string.split(" "); String[] parts2 = palabraLimpia.split(" "); for (int i = 0; i < parts2.length; i++) { char[]caracteres = parts2[i].toCharArray(); int largo = caracteres.length; if (Character.toUpperCase(caracteres[0]) == 'A' && Character.toUpperCase(caracteres[largo-1]) == 'A') { lista.add(parts[i]); } } for (String palabra: lista) { System.out.println(palabra); }
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to put the word in parts in order of increase. Python
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We shall follow each symbol in the line:s = input("word: ") for i in range(len(s)): print(s[:i + 1])
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Code:public class MyClass { public static void main(String args[]) { // Аргументов может быть только 2 if (args.length != 2) { System.out.println("Run with two arguments!"); return; } String word = args[0]; String posStr = args[1]; int pos; // Проверка на корректное число try { pos = Integer.parseInt(posStr); } catch (java.lang.NumberFormatException e) { System.out.println("Invalid number: " + posStr); return; } char symbol = word.charAt(pos - 1); String text = String.format("В слове %s %s-й символ %s", word, posStr, symbol); System.out.println(text); } }
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Read this: https://stackoverflow.com/questions/51839937/beautiful-soup-returns-empty-list The reason for the failure is that when you open that website reviews are not loaded automatically. The reviews are loaded only when you scroll down to review section. So you have to inspect the netwrok and find the correct url that loads review ALso try: Try review = soup.findAll("p",class_="css-oals0c-unf-heading e1qvo2ff8") or review = soup.findAll("p",attrs={"data-testid"="txtDateGivenReviewFilter0"}) data-testid , is a custom attribute and not a id attribute. as per the documentation custom attribute should be provided as dictionary https://www.crummy.com/software/BeautifulSoup/bs4/doc/#find-all Some attributes, like the data-* attributes in HTML 5, have names that can’t be used as the names of keyword arguments: data_soup = BeautifulSoup('foo!') data_soup.find_all(data-foo="value") SyntaxError: keyword can't be an expression You can use these attributes in searches by putting them into a dictionary and passing the dictionary into find_all() as the attrs argument: data_soup.find_all(attrs={"data-foo": "value"}) [foo!] You can’t use a keyword argument to search for HTML’s ‘name’ element, because Beautiful Soup uses the name argument to contain the name of the tag itself. Instead, you can give a value to ‘name’ in the attrs argument: name_soup = BeautifulSoup('') name_soup.find_all(name="email") [] name_soup.find_all(attrs={"name": "email"}) []
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Each comedian contains (as an integral part) a reference to the " parent " committee (the merging units contain more than one reference to " breeders " ). The " parent " Committ, in turn, contains a reference to its " parent " , and so forth. Link History of change.If a comedian is removed, it will be disrupted by " time interconnection " , since his " exodus " does not contain references to " deda " .It's impossible to re-establish communication without writing history, and if you're thinking about rewriting, you'd be ready for it. https://ru.stackoverflow.com/a/572907/178576 ♪
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Streamlining of the line, removal of single astrophic symbols from the list of words containing aspostrophe
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words = ["we're", "'", "'''", "he's", "'abc", "ghe'", 'hello', "'h", "h'"] res = ([x for x in words if "".join(set(x))!="'"]) print(res) [ "we're," "he's," "'abc," "ghe'," "hello""
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I think you should. https://scikit-learn.org/stable/modules/generated/sklearn.feature_extraction.text.CountVectorizer.html ♪Example:import pandas as pd from nltk import sent_tokenize, word_tokenize from sklearn.feature_extraction.text import CountVectorizer text = """Всем доброго времени суток! У меня есть два списка: в первом предложения(их 22), во втором слова(их 254,все уникальные). Нужно получить результирующую матрицу, в которой элемент [i][j] представляет собой количество вхождений j-ого слова в i-ое предложение. Я пробовал сделать через .count(), но получается что-то неадекватное(например, 7 вхождений одного слова в одно предложение)""" vocabulary = ['слова', 'списка', 'предложения', 'вхождений'] cv = CountVectorizer(vocabulary=vocabulary) r = pd.SparseDataFrame(cv.fit_transform(sent_tokenize(text)), columns=cv.get_feature_names(), default_fill_value=0) result:In [75]: r Out[75]: слова списка предложения вхождений 0 0 0 0 0 1 1 1 1 0 2 1 0 0 1 3 1 0 0 1
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Look I’m creating an app and work on the local network with shampoo I think it could be a solution. And I would use HttpURLConnection for the connection between apps. But that's just an idea. There may be a class to work directly with the network. Maybe I'm talking bullshit too. It's just a blz idea.

to remove words in order to reduce their length

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