tuesday

iota 1 //iota作用会自动递增,引用自己定义的表达式。 2 3 4 //定义所有星期,从0到6 5 package main 6 7 import "fmt" 8 9 const ( 10 Sunday = iota //重点 11 Monday 12 Tuesday 13 Wednesday 14 Thursday 15 Friday 16 Saturday 17 ) 18 19 func main() { 20 21 //从第二个开始,自动引用排头兵的表达式,但iota逐一递增. 22 fmt.Println(Sunday, Monday, Tuesday) //0 1 2 23 24 } 1 ////定义所有星期,从1到7 2 package main 3 4 import "fmt" 5 6 const ( 7 Monday = iota +1 //0 +1 = 1 //重点 8 Tuesday 9 Wednesday 10 Thursday 11 Friday 12 Saturday 13 Sunday 14 ) 15 16 func main() { 17 18 //从第二个开始,自动引用排头兵的表达式,但iota逐一递增. 19 fmt.Println(Monday, Tuesday) //1 2 20 21 } 1 //定义5大常任理事国编号100-500

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: How to set the first day of current week to Tuesday in Swift 3? The schedule will update every Tuesday so the first day have to start from Tuesday. and then display on label : thisWeekDate.text! = "\(first day of current week) - \(last day of current week)" Full code. let today = NSDate() let nextTue = Calendar.current.date(byAdding: .day, value: 6, to: today as Date) let formatter = DateFormatter() formatter.dateFormat = "yyyy-MM-dd" let todayString = formatter.string(from: today as Date) let nextString = formatter.string(from: nextTue!)

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I'm trying to replicate the graph similar to the following (originally found HERE ) It's conceptually simple, but I'm a bit stumped as to how to do it in R. To summarize: I want to generate histograms of behavioral frequency over the 24 hours of the day (24-level factor variable) by each day of the week. Then, I want to stack these histograms on top of each other so that the distribution of behavior over the hour of day can easily be compared (again, see example). For example, my data might look like this: weekday hour count Tuesday 15 553

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have start date and end date. I need to find out the day that is Sunday or Monday etc dependent upon user click on check box. How can I find/calculate that in PHP? 回答1: You could create a function that uses strtotime() recursively to count the number of days. Since strtotime("next monday"); works just fine. function daycount($day, $startdate, $counter) { if($startdate >= time()) { return $counter; } else { return daycount($day, strtotime("next ".$day, $startdate), ++$counter); } } echo daycount("monday", strtotime("01.01.2009"), 0);

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I am not understanding how the TDD FIRST principle isn't being adhered to in the following code. These are my notes about the FIRST principle: Fast : run (subset of) tests quickly (since you'll be running them all the time) Independent : no tests depend on others, so can run any subset in any order Repeatable : run N times, get same result (to help isolate bugs and enable automation) Self-checking : test can automatically detect if passed (no human checking of output) Timely : written about the same time as code under test (with TDD, written

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: What I have: I'm generating a JSON schema from a pojo. My code to generate the schema looks like so: ObjectMapper mapper = new ObjectMapper (); TitleSchemaFactoryWrapper visitor = new TitleSchemaFactoryWrapper (); mapper . acceptJsonFormatVisitor ( clazz , visitor ); JsonSchema schema = visitor . finalSchema (); schemas . put ( clazz , mapper . writerWithDefaultPrettyPrinter (). writeValueAsString ( schema )); I'm generating several schemas via the above code. One of the pojos has an internal embedded enum to limit the possible

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I am an excel newb, wondering if excel has built in functionality to do the following. Turn this data: Name | Activity | Option Bob | Monday Activities | Golf Bob | Tuesday Activities| Kayak Marge | Monday Activities | Spa John | Tuesday Activities| Soccer Liz | Tuesday Activities| Spa Into: Name | Monday Activities | Tuesday Activities Bob | Golf | Kayak Marge | Spa | John | | Soccer Liz | | Spa 回答1: Pivot tables allow you to manipulate and display numerical data, not strings such as "kayak", so while you can create a pivot table very

　　 Scala中的 for 表达式适用于迭代的热屎君道，它让你可以以不同的方式组合一些简单的因子来表达各种各样的迭代，它可以帮你处理诸如遍历整数序列的常见任务，也可以通过高级的表达式来遍历多个不同种类的集合，根据多个不同的条件过滤元素，产生新的集合。 遍历集合 　　 用 for 能做的最简单的事是遍历某个集合的所有元素。 val weeks = "Monday" :: "Tuesday" :: "Wednesday" :: "Thursday" :: "Friday" :: "Saturday" :: "Sunday" :: Nil weeks: List[String] = List(Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday) for(week <- weeks) println(week); 　　通过 "week <- weeks" 这样的生成器语法，我们将遍历 weeks 的元素。每一次迭代，一个新的名为 week 的 val 都会被初始化成一个元素的值。 　　 for 表达式的语法可以用于任何种类的集合，而不仅仅是数组。 Range (区间)是一类特殊的用例。可以用 “ 1 to 5 ”这样的语法来创建 Range ，并用 for 来遍历它们。 for(i <- 0 to 5)

　　 List和数组是非常相似的。列表的所有元素都具有相同的类型。二者的区别在于：1、List是不可变的，这意味着List的元素无法通过分配进行更改。List代表一个链表，而数组是平面的。 创建一个List： // 字符串列表 val stringList: List[String] = List("Monday", "Tuesday", "Wednesday") // 整型列表 val numList: List[Int] = List(1, 2, 3, 4) // 二维列表 val dim: List[List[Int]] = List( List(1, 0, 0), List(0, 1, 0), List(0, 0, 1) ) 　　两种方式创建一个空列表： 1、 scala> val emptyList = Nil emptyList: scala.collection.immutable.Nil.type = List() 2、 scala> val emptyList2 = List() emptyList2: List[Nothing] = List() 　　用字符串创建列表： scala> var a = "a"::"b"::"c"::Nil; a: List[String] = List(a, b, c) 　　使用 ::: 创建一个新列表 scala>