intervals

Generating random numbers on open-open interval (0,1) efficiently

时光毁灭记忆、已成空白 提交于 2019-12-04 08:46:36
I'm looking for an efficient way to generate random floating-point numbers on the open-open interval (0,1). I currently have an RNG that generates random integers on the closed-closed interval of [0, (2^32)-1]. I've already created a half-open floating point RNG on the interval [0,1) by simply multiplying my result from the integer RNG by 1/((2^32)-1) rather than dividing by (2^32)-1 since it's inefficient. The way I'm currently going about generating numbers on the interval (0,1) is with a conditional statement like the one below: float open_open_flt = (closed_open_flt==0) ? closed_open_flt :

python time interval algorithm sum

 ̄綄美尐妖づ 提交于 2019-12-04 08:16:42
Assume I have 2 time intervals,such as 16:30 - 20:00 AND 15:00 - 19:00, I need to find the total time between these two intervals so the result is 5 hours (I add both intervals and subtract the intersecting interval), how can I write a generic function which also deals with all cases such as one interval inside other(so the result is the interval of the bigger one), no intersection (so the result is the sum of both intervals). My incoming data structure is primitive, simply string like "15:30" so a conversion may be needed. Thanks from datetime import datetime, timedelta START, END = xrange(2)

Java check if number in interval [duplicate]

五迷三道 提交于 2019-12-04 06:59:07
Possible Duplicate: Does an open-ended interval implementation exist for Java? i have an int variable and i'd like to check if it's value is in an interval [a,b]. I know it's a simple matter of using x>=a and x<=b or implementing a simple method which can do this, but i'd like to know if there is something already done. Searched in Math class, but i wasn't able to find one. It's not that important and not that big of an issue, but i'm curious if there is something like this, so i can use it instead of implementing my own :) In all my coding i haven't come across a method like this. Maybe one

Time Interval overlap in python

元气小坏坏 提交于 2019-12-04 05:20:01
问题 Suppose I have list of time interval like a = [datetime.time(0,0),datetime.time(8,0)] Now I Have lacs of intervals in list like given below. b = [[datetime.time(0,0),datetime.time(8,0)], [datetime.time(0,0),datetime.time(10,0)], [datetime.time(0,0),datetime.time(23,59,59)], [datetime.time(15,0),datetime.time(9,0)], [datetime.time(9,0),datetime.time(15,0)]] We have to filter list b with intervals containing interval a. like in example result will be. b = [[datetime.time(0,0),datetime.time(8,0)

Finding “maximum” overlapping interval pair in O(nlog(n))

烈酒焚心 提交于 2019-12-04 01:37:00
Problem Statement Input set of n intervals; {[s_1,t_1], [s_2,t_2], ... ,[s_n,t_n]}. Output pair of intervals; {[s_i,t_i],[s_j,t_j]}, with the maximum overlap among all the interval pairs. Example input intervals : {[1, 10], [2, 6], [3,15], [5, 9]} -> There are possible 6 interval pairs. Among those pairs, [1,10] & [3,15] has the largest possible overlap of 7. output : {[1,10],[3,15]} A naive algorithm will be a brute force method where all n intervals get compared to each other, while the current maximum overlap value being tracked. The time complexity would be O(n^2) for this case. I was able

php - How To Convert 1 day To Seconds

旧巷老猫 提交于 2019-12-04 01:27:00
问题 I am creating a form to set publish interval and want to sum all of the submitted values into seconds. The form consist of fields to determine seconds, minutes, hours, day and so on. I am collecting the submitted values to string by giving date attributes, like second, minute etc. For example: $second second $minute minute $hour hour So if user input 1 in input field to determine the interval in day and 30 in minute , then my code will convert it to string 1 day and 30 minute . My questions:

How to stop $interval on leaving ui-state?

十年热恋 提交于 2019-12-04 00:22:02
Angular, UI-router. Using $interval in a controller of a state like so: $scope.Timer = null; $scope.startTimer = function () { $scope.Timer = $interval($scope.Foo, 30000); }; $scope.stopTimer = function () { if (angular.isDefined($scope.Timer)) { $interval.cancel($scope.Timer); } }; The problem? The timer persists upon leaving the state. My understanding was that the $scope and the controller are essentially "destroyed" when a state is left. So, based on that, the timer should stop (Within the controller, I am cancelling the timer when moving around, that works - but it persists if I navigate

How to spread the average between two intervals in oracle

*爱你&永不变心* 提交于 2019-12-03 21:48:06
If given the Average for 24 hours for each date in a year. I want to spread this hourly average to average at each minute. e.g. given Date Time Average 01-Jan-15 23:00 20 02-Jan-15 00:00 50 02-Jan-15 01:00 30 I want the output to be calculated something as below .... DateTime AVG_VALUE 01/01/2015 23:00:00 20 01/01/2015 23:01:00 20.5 01/01/2015 23:02:00 21 01/01/2015 23:03:00 21.5 01/01/2015 23:04:00 22 01/01/2015 23:05:00 22.5 01/01/2015 23:06:00 23 01/01/2015 23:07:00 23.5 01/01/2015 23:08:00 24 01/01/2015 23:09:00 24.5 01/01/2015 23:10:00 25 01/01/2015 23:11:00 25.5 01/01/2015 23:12:00 26 01

Angular2 Observable with interval

我的未来我决定 提交于 2019-12-03 20:42:41
I have a function that needs to be called about every 500ms. The way I am looking at doing it with angular2 is using intervals and observables. I have tried this function to create the observable: counter() { return Observable.create(observer => { setInterval(() => { return this.media.getCurrentPosition(); }, 500) }) } With this code for the subscriber: test() { this.playerService.initUrl(xxxx) // This works this.playerService.counter().subscribe(data => { res => { console.log(data); } }) } I am very new to observables and angular2 so I might be taking the wrong approach completely. Any help

Run a function periodically in Scala

我的梦境 提交于 2019-12-03 18:32:33
问题 I want to call an arbitrary function every n seconds. Basically I want something identical to SetInterval from Javascript. How can I achieve this in Scala? 回答1: You could use standard stuff from java.util.concurrent : import java.util.concurrent._ val ex = new ScheduledThreadPoolExecutor(1) val task = new Runnable { def run() = println("Beep!") } val f = ex.scheduleAtFixedRate(task, 1, 1, TimeUnit.SECONDS) f.cancel(false) Or java.util.Timer : val t = new java.util.Timer() val task = new java