Ok, continuing from the comment, and just as laid out in the comment, the first thing you need to do with any data handling problem is to read all of your data into a form that allows you to work with it as needed. Here you have varying types of data that all make up … Read more
you just need to change you code likewise, if(str.substring(1,3).equals(a)){…} what you did wrong is , you have been used ‘==’ assignment operator, it is not comparing content(here, string) instead of it, it compare’s memory location between two compared string. it’s obvious to get false, because how is it possible ? that two different string refer … Read more
The Best answer is from Yousaf . x = [“apple”, “banana”, “carrot”, “cheese”, “pear”] print(“\””+’s ‘.join(x[:-1])+”s”+” and “+(x[4])+”s”+”\””) solved How removing commas in a list with strings, write in one line of code [duplicate]
Check if there’s a whitespace (\s) before or if it’s the start of the string (^). (?:\s|^)@.* Live Demo If you only want to match word characters, then instead of using . you could use \w. Also note that * would still match just “@”, so you might want to use + instead to match … Read more
(=[A-Z])(\w|\%20|\.)+ This will capture only the team names and the space characters between them, with leading “=”. Then you can replace “%20″ with ” ” and split the resultant string along “=” so you’ll have a list of strings, each separately a team name. At least, that’s what I’d do in Python. Not sure what … Read more
User Himal say use typeof.. so I can found the answer var myvar=5 alert(typeof myvar); //alerts “number” so to know kind of varible use typeof source : http://www.javascriptkit.com/javatutors/determinevar2.shtml solved How can I identify kind of variable I’m trying to compare in JavaScript?
I think using Regular Expression works better. private static String setSize(String htmlString) { String reg = “size=”[0-9]+””; Pattern pattern = Pattern.compile(reg); Matcher matcher = pattern.matcher(htmlString); while (matcher.find()) { String sizeString = matcher.group(); pattern = Pattern.compile(“[0-9]+”); Matcher numMatcher = pattern.matcher(sizeString); if (numMatcher.find()) { String size = numMatcher.group(); int realSize = Integer.parseInt(size); int resultSize = realSize / … Read more
If you want to check for any word to be used twice, use the Split function to make a string into words, and then Group to get counts: string input = “MyString MyString”; var words = input.Split().GroupBy(s => s).ToDictionary( g => g.Key, g => g.Count() ); The dictionary words will be a set of key … Read more
public void checkout(String answer,int missedcallcount) Declaration of this method shows, that you need to pass two parameters to call it properly – in this example, if you want to call this method you need to pass to it String as a first argument, and int as a second argument. Here is an example of correct … Read more
Most Unicode characters have multiple versions that can look very alike. For example: 1????１①⑴ var s = “æӕ”.Replace(“æ”, “ae”); // s = “aeæ” var v = “æӕ”.Select(c => (int)c).ToArray(); // { 230, 1237 } I consider it a good practice to expect the unexpected (especially when it comes to user input) var s = “æӕ”; … Read more
It’s pretty simple why it’s crashing when you change that line. In the version that crashes you check whether words[j].find(new_word) is equal to std::string::npos and then call s.erase(s.find(new_word), …). Of course, s.find(new_word) could still be std::string::npos, and string::erase throws an exception when the first argument is out of range. In the version that you say … Read more
x = [‘cat’, ‘dog’, ‘chicken’] y = [word[0] + ‘_’*(len(word)-1) for word in x] print(y) # [‘c__’, ‘d__’, ‘c______’] solved how to show only the first letter of an element of a string [closed]
From the standard N4687: 20.5.4.3.2 Macro names 1 A translation unit that includes a standard library header shall not #define or #undef names declared in any standard library header. 2 A translation unit shall not #define or #undef names lexically identical to keywords, to the identifiers listed in Table 4, or to the attribute-tokens described … Read more
A simple way to do it is to iterate over the string, that will give you each letter separately: alphabet = “abcdefghijklmnopqrstuvwxyz” for letter in alphabet: print(“a” + letter) solved Adding first character of a string to each of the other characters [closed]
Another option is to parse the XML, and use the OutputKeys.INDENT option of the Transformer class to output the XML formatted. The following example Source source = new StreamSource(new StringReader(s)); TransformerFactory transformerFactory = TransformerFactory.newInstance(); transformerFactory.setAttribute(“indent-number”, 4); Transformer transformer = transformerFactory.newTransformer(); transformer.setOutputProperty(OutputKeys.INDENT, “yes”); StreamResult result = new StreamResult(new StringWriter()); transformer.transform(source, result); String xmlOutput = result.getWriter().toString(); System.out.println(xmlOutput); … Read more