counter

问题 I've a dictionary such as this: my_dict=collections.OrderedDict([((123, 1), 'qwe'), ((232, 1), 'asd'), ((234, 2), 'zxc'), ((6745, 2), 'aaa'), ((456, 3), 'bbb')]) The combination of the tuple is always unique and I would like to maintain the order of insertion, and hence OrderedDict. I've a well over ~10K items in the dict. How can I efficiently maintain a counter that gives the count of the second element in the tuple? Basically, I need to know the count whenever I would like to add/delete an

问题 http://flipclockjs.com/ I need a counter which counts down to zero from a certain time (for example 2 hours). After those two hours the timer needs to reset and starts again. How can I do this? I can't find anything about it in their docs. What I got now: var clock = $('.your-clock').FlipClock({ countdown : true, }); 回答1: Try this one: var clock = $('.your-clock').FlipClock({ countdown : true, }); clock.setTime(10); clock.start(); setTimeout(function(){ checktime(); }, 1000); function

问题 http://flipclockjs.com/ I need a counter which counts down to zero from a certain time (for example 2 hours). After those two hours the timer needs to reset and starts again. How can I do this? I can't find anything about it in their docs. What I got now: var clock = $('.your-clock').FlipClock({ countdown : true, }); 回答1: Try this one: var clock = $('.your-clock').FlipClock({ countdown : true, }); clock.setTime(10); clock.start(); setTimeout(function(){ checktime(); }, 1000); function

问题 I have a 3 D numpy array which contains elements with repetition. counterTraj.shape (13530, 1, 1 For example counterTraj contains such elements: I have shown few elements only: array([[[136.]], [[129.]], [[130.]], ..., [[103.]], [[102.]], [[101.]]]) ``` I need to find frequency of different element: Example: 136 count 5 (say), 101 count 12 (say). The array elements are not fixed and changes with input data. I try following: from collections import Counter Counter(counterTraj) Following error

问题 Here's my method: public char[] ReturnAllVowels(String word) { for (int i = 0; i < word.length(); i++) { if (word.contains("a" || "e" || "i" || "o" || "u")) { } } } It says that || cannot be applied to String class. How can I do this then? 回答1: char ch = word.charAt (i); if (ch == 'a' || ch=='e') { } 回答2: Using regular expressions you can try. int count = word.replaceAll("[^aeiouAEIOU]","").length(); 回答3: String regex = "[aeiou]"; Pattern p = Pattern.compile(regex,Pattern.CASE_INSENSITIVE);

问题 I need to store word co-occurrence counts in several 14000x10000 matrices. Since I know the matrices will be sparse and I do not have enough RAM to store all of them as dense matrices, I am storing them as scipy.sparse matrices. I have found the most efficient way to gather the counts to be using Counter objects. Now I need to transfer the counts from the Counter objects to the sparse matrices, but this takes too long. It currently takes on the order of 18 hours to populate the matrices. The

问题 I would like to count how many times my signal goes to zero. For example having an impulse signal as input I want a variable which counts how many times the impulse goes to zero. I´m becoming crazy thinking of something....can anybody help me? Thanks 回答1: figure 1 is a pulse counter model and figure 2 is scope output which shows for 5 input pulses we get counter output as 5 回答2: There is a similar question here that might help. There are loads of ways to implement a counter, but a good way I

问题 I have created a custom class in xcode: PaperPack and defined 2 instant variables: title and author - Then I alloc 2 instances of the class as below: PaperPack *pack1 = [[PaperPack alloc] init]; pack1.title = @"Title 1"; pack1.author = @"Author"; PaperPack *pack2 = [[PaperPack alloc] init]; pack1.title = @"Title 2"; pack1.author = @"Author"; Then how do I count and return number of instances I have created with that class? 回答1: No you can not get directly. Whenever you create in instance add

问题 I have a csv file of data that has the columns ‘number’ , ’colour’ , ’number2’ , ’foo’ , ’bar’ , which looks like: 12, red, 124, a, 15p 14, blue, 353, c, 7g 12, blue, 125, d, 65h 12, red, 124, c, 12d I want to count the number of times number, colour and number2 occur together, so for example, the output from the above list would be: ’12, red, 124 :2’,’14, blue, 353: 1’, ’12, blue, 125: 1’ . I’ve done this by using: import csv datafile=open('myfile.csv','r') usefuldata=[] for line in datafile

问题 Given the big.txt from norvig.com/big.txt, the goal is to count the bigrams really fast (Imagine that I have to repeat this counting 100,000 times). According to Fast/Optimize N-gram implementations in python, extracting bigrams like this would be the most optimal: _bigrams = zip(*[text[i:] for i in range(2)]) And if I'm using Python3 , the generator won't be evaluated until i materialize it with list(_bigrams) or some other functions that will do the same. import io from collections import