benchmarking | 易学教程

Why play framework 2.4 is so slow when using @routes.Assets.versioned in twirl?

阅读更多关于 Why play framework 2.4 is so slow when using @routes.Assets.versioned in twirl?

问题 I have a controller, when i render a twirl template and return it in my action. the play performance decrease about 5 time (15000 Requests per second in apache benchmark). but when i return json in this action play performance increase about 5 time (79000 Requests per second in apache benchmark). I have a very simple twirl template: @(message: String) @main("Welcome to Play") { @message } Update : I find that when i use @routes.Assets.versioned instead of @routes.Assets.at to address assets

Python timing in order to trigger specific events

阅读更多关于 Python timing in order to trigger specific events

问题 In Python I would like to run a function calling a default action for 20 seconds. However, there are 5 specific timings within these 20 seconds when another function should be triggered. In order to simplify my code, I have replaced the "action" functions with simple printing commands. Here is what I have so far - The output seems ok, in that it lasts for 10 seconds and prints the time and the default state/action. But the triggered action is missing! Is there a better/correct way to do this?

Why is List<T>.ForEach faster than standard foreach?

阅读更多关于 Why is List.ForEach faster than standard foreach?

问题 Consider this: Requisite: //The alphabet from a-z List<char> letterRange = Enumerable.Range('a', 'z' - 'a' + 1) .Select(i => (Char)i).ToList(); //97 - 122 + 1 = 26 letters/iterations Standard foreach: foreach (var range in letterRange) { Console.Write(range + ","); } Console.Write("\n"); Inbuilt foreach: letterRange.ForEach(range => Console.Write(range + ",")); //delegate(char range) works as well Console.Write("\n"); I have tried timing them against each other and the inbuilt foreach is up

Any ASP.Net benchmark tool? [closed]

阅读更多关于 Any ASP.Net benchmark tool? [closed]

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 4 years ago . I would like to measure the performance of the .Net application, especially web application in ASP.Net when running in the server. I need to know any benchmark tool enable for me to know my source code can be optimize. As long as the benchmark tool can be helping me optimize the performance of the website. Any

Benchmarking data.frame (base), data.frame(package dataframe) and data.table

阅读更多关于 Benchmarking data.frame (base), data.frame(package dataframe) and data.table

问题 With the recent introduction of the package dataframe , I thought it was time to properly benchmark the various data structures and to highlight what each is best at. I'm no expert at the different strengths of each, so my question is, how should we go about benchmarking them. Some (rather crude) things I have tried: library(microbenchmark) library(data.table) mat <- matrix(rnorm(10000), nrow = 100) mat2df.base <- data.frame(mat) library(dataframe) mat2df.dataframe <- data.frame(mat) mat2dt <

How to benchmark Linux threaded programs?

阅读更多关于 How to benchmark Linux threaded programs?

问题 I'm trying to compare the performance of threaded programs (on Linux). Since the programs use different thread synchronization methods and different lock granularity, running the programs on a shared server or desktop would not be good, since the other tasks may interfere with the scheduling of my programs. I don't have dedicated hosts, so I thought that using qemu would be a good option. What I want to know is: Are there any alternatives for this task? I suppose that there is no way to

Speed of if compared to conditional

阅读更多关于 Speed of if compared to conditional

问题 I had the idea I would turn some of my if blocks into single lines, using the conditional operator. However, I was wondering if there would be a speed discrepancy. I ran the following test: static long startTime; static long elapsedTime; static String s; public static void main(String[] args) { startTime = System.nanoTime(); s = ""; for (int i= 0; i < 1000000000; i++) { if (s.equals("")) { s = ""; } } elapsedTime = System.nanoTime() - startTime; System.out.println("Type 1 took this long: " +

Why does IntStream.range(0, 100000).parallel.foreach take longer then normal for loop

阅读更多关于 Why does IntStream.range(0, 100000).parallel.foreach take longer then normal for loop

I am just starting to learn about the Streams and parallel in Java and I was wondering why a normal for loop takes less time than IntStream paralleled at adding items to an array. package parallel; import java.util.stream.IntStream; public class Parallel { public static void main(String[] args) { final int[] intArray = new int[100000]; long startTime = System.currentTimeMillis(); IntStream.range(0, 100000).parallel().forEach(i -> intArray[i]=i); long endTime = System.currentTimeMillis(); System.out.println("Parallel time: " + (endTime-startTime)); final int[] intArray2 = new int[100000]; try {

How does lmbench measure L1 and L2 cache latencies using C? (cannot understand explanation in manual)

阅读更多关于 How does lmbench measure L1 and L2 cache latencies using C? (cannot understand explanation in manual)

问题 I am trying to understand how lmbench measures latency for L1, L2 and main memory. The man page for lat_mem_rd mentions the method, but it's not clear to me: The benchmark runs as two nested loops. The outer loop is the stride size. The inner loop is the array size. For each array size, the benchmark creates a ring of pointers that point forward one stride. Traversing the array is done by p = (char **)*p; in a for loop (the over head of the for loop is not significant; the loop is an unrolled

Why is the first run always much slower?

阅读更多关于 Why is the first run always much slower?

问题 I wrote a macro that reports the time required to run a given operation. It runs it a number of times and prints out the time for each run in nanoseconds. The first run always takes significantly more time than subsequent ones. Why is that so? Here are the results of 10 x 10 runs, timing Thread.yield() : > (dotimes [x 10] (prn (times 10 (Thread/yield)))) [55395 1659 622 561 591 702 795 719 742 624] [3255 772 884 677 787 634 605 664 629 657] [3431 789 965 671 774 767 627 627 521 717] [2653 780