Java works a little bit different than many other languages. What o is in the first example is simply a reference to the object. When you say o = new MyObject(), it creates a new Object of type MyObject and references o to that object, whereas before o referenced objects[index]. That is, objects[index] itself is … Read more
a = next(i for i in userInput if i in wordsTask) To break it down somewhat: [i for i in userInput if i in wordsTask] Will produce a list. What you want is the first item in the list. One way to do this is with the next function: next([i for i in userInput if … Read more
Use a dictionary instead. E.g: doubles = dict() for x in range(1, 13): doubles[x] = x * 2 Or if you absolutely must do this AND ONLY IF YOU FULLY UNDERSTAND WHAT YOU ARE DOING, you can assign to locals() as to a dictionary: >>> for x in range(1, 13): … locals()[‘double_{0}’.format(x)] = x * … Read more
(update) Actually – there is one scenario where the for construct is more efficient; looping on an array. The compiler/JIT has optimisations for this scenario as long as you use arr.Length in the condition: for(int i = 0 ; i < arr.Length ; i++) { Console.WriteLine(arr[i]); // skips bounds check } In this very specific … Read more
You may want to re-read The Rust Programming Language, specifically the sections on: mutability mutable references vectors We can also iterate over mutable references to each element in a mutable vector in order to make changes to all the elements. The for loop in Listing 8-9 will add 50 to each element. let mut v … Read more
It’s never a good idea to change the object you’re iterating over. Normally dict even throws an exception when you attempt it: name_dict = {1: 1, 2: 2, 3: 3, 4: 4, 5: 5, 6: 6} for k, v in name_dict.items(): name_dict.pop(k) RuntimeError: dictionary changed size during iteration However in your case you add one … Read more
You concrete problem is caused because you’re mixing discouraged and old school scriptlets <% %> with its successor EL ${}. They do not share the same variable scope. The allFestivals is not available in scriptlet scope and the i is not available in EL scope. You should install JSTL (<– click the link for instructions) … Read more
You could use the Array.reduce method: const myData = [ [‘2013-01-22’, 0], [‘2013-01-29’, 0], [‘2013-02-05’, 0], [‘2013-02-12’, 0], [‘2013-02-19’, 0], [‘2013-02-26’, 0], [‘2013-03-05’, 0], [‘2013-03-12’, 0], [‘2013-03-19’, 0], [‘2013-03-26’, 0], [‘2013-04-02’, 21], [‘2013-04-09’, 2] ]; const sum = myData .map( v => v[1] ) .reduce( (sum, current) => sum + current, 0 ); console.log(sum); See … Read more
The list.append function does not return any value(but None), it just adds the value to the list you are using to call that method. In the first loop round you will assign None (because the no-return of append) to a, then in the second round it will try to call a.append, as a is None … Read more
++i is slightly more efficient due to its semantics: ++i; // Fetch i, increment it, and return it i++; // Fetch i, copy it, increment i, return copy For int-like indices, the efficiency gain is minimal (if any). For iterators and other heavier-weight objects, avoiding that copy can be a real win (particularly if the … Read more