counter

问题 Is it possible to write a list comprehension for the following loop? m = [] counter = 0 for i, x in enumerate(l): if x.field == 'something': counter += 1 m.append(counter / i) I do not know how to increment the counter inside the list comprehension. 回答1: You could use an itertools.count: import itertools as IT counter = IT.count(1) [next(counter)/i for i, x in enumerate(l) if x.field == 'something'] To avoid the possible ZeroDivisionError pointed out by tobias_k, you could make enumerate

问题 This is my list a=[ ['a','b','a','a'], ['c','c','c','d','d','d']] I wanna find most common elemments. I have tried this from collections import Counter words = ['hello', 'hell', 'owl', 'hello', 'world', 'war', 'hello', 'war','aa','aa','aa','aa'] counter_obj = Counter(words) counter_obj.most_common() but it works just for simple list. my output should be like this [('a', 3), ('c', 3), ('d', 3), ('b', 1)] 回答1: Apply Counter().update() option on the elements of your list, Based on suggestion

问题 I want to get the elements for a given mixture. For examples, for a mixsture of Air (O2 and N2) and Hexane (C6H14) given by the dict with their respectives mole numbers mix = {'O2': 1, 'N2': 3.76, 'C6H14': 0.01} I want to get the following: {O: 2, N: 7.52, C:0.06, H: 0.14} Another example: mix = {'C6H14': 1, 'C9H20': 1} must yields {H: 34, C: 15} enter code here The sequence of the dict it's not important. I was trying with the re.split, but I don't get any progress. If anyone can help me I

问题 Im having a problem incrementing a counter in one of my while loops basically i just want to alternate between two image links that were fetched in my database but my counter wont increase and im not sure why can anyone help? while($row = $stmt->fetch(PDO::FETCH_ASSOC)) { $img_link = $row['Image']; $img_link_alt = $row['Image_alt']; $i = 0; echo '<div class="col-xs-6 col-sm-3 placeholder">'; $img = ( $i % 2 == 0 ) ? $img_link : $img_link_alt; echo $i; //'?' . date("h:i:sa").' echo '<img style

问题 So for sets you can do a symmetric difference (^) which is the equivalent of union minus intersection. Why is ^ an unsupported operand for Counter objects while union and intersection still work? 回答1: Expanding on my comment, turns out it was discussed at time, and rejected. Click the link for the full message (and its thread); I'll just quote the "high order bits" of Raymond Hettinger's reply: It's unlikely that I will add this [symmetric difference] method to the Counter API because the

问题 function parse($string){ $counter = 0; $string = preg_replace("_\[b\](.*?)\[/b\]_si", '<span class="b">'. $counter .'. $1</span>', $string, -1, $counter); return $string; } I'm trying to make a ubb parser, that parses tags and put the counting in front of it: [b]Hey[/b] [b]Hello[/b] Should return this: <span class="b">1. Hey</span> <span class="b">2. Hello</span> But is returning this: <span class="b">1. Hey</span> <span class="b">1. Hello</span> So beside the function above, I've tried this:

问题 I'm trying to count the number of objects created but it always returns 1. public class Drivertwo { public static void main(String[] args) { Employee newEmp = new Employee(); Employee newEmp2 = new Employee(); Calculate newcal = new Calculate(); Clerk newclerk = new Clerk(); float x; int y; newEmp.setEmp_no(2300); newEmp.setEmp_name("W.Shane"); newEmp.setSalary(30000); newEmp.counter(); newEmp2.setEmp_no(1300); newEmp2.setEmp_name("W.Shane"); newEmp2.setSalary(50000); newEmp2.counter();

问题 I'm doing a counter where I'm showing how many seconds the user has to make the input. The problem is that when I use scanf(), the program will stop and will wait for the answer, but I don't want that. I'd like to keep running the counter even thought the user doesn't put anything. Example: for(int i=10;i>=0;i--){ i==0? printf("time over\n"):printf("%d seconds left\n",i); scanf("%s", decision); sleep(1); } What can i do to solve this? 回答1: Like mentioned in the comments one possibility would

问题 Trying to create an infinite counter that starts at 10.41 and increases by 10.41 every second. Have looked at various tutorials but I cannot find one that will account for the increase with a decimal number unit (10.41) every second. This is the code I am using: setTimeout(start, 0); var i = 100; var num = document.getElementById('number'); function start() { setInterval(increase, 100); } function increase() { if (i < 100000) { i++; num.innerText = i; } } 回答1: If I understand correctly, all

问题 I have a private bot on discord, i've been trying to make him count a command and adding '+1' everytime I write that one command but it stays at 1 and can't go further : like this I think what I want to do is to make it save the number of time the command has been writen and add +1 to this number ; Should I do a loop or something ? Basically what I want is something like this in python for a discord bot : https://docs.nightbot.tv/commands/variables/count 回答1: You're resetting your counter