首页 \ 问答 \ 快速错误处理：范围(swift errorhandling: scope)

快速错误处理：范围(swift errorhandling: scope)

 在swift 2中，作用域的错误处理应该有多大？ 我理解错误处理（我认为），但我也怀疑实现细节。 如果我有一个调用抛出函数的函数，我应该如何处理我的do-try-catch的范围？ 
 

 我应该这样做： 
 

func doLotsOfErrorProneWork()
{
  //everything in scope
 do {
    try f1()
    f2()
    f3()
    try f4()
    f5()
 } catch {}
}

 

 或这个： 
 

func doLotsOfErrorProneWork()
{
  //smallest scope possible
  do {
    try f1()
  } catch {}

    f2()
    f3()

  do {
    try f4()
  } catch {}

  f5()
}

 

 这有什么不同吗？ 有什么不同？ 一个区别可能是f1和f4之间抛出的错误差异。 但是假设只有一个投掷功能。 
 

func doLotsOfErrorProneWork()
{
  //everything in scope
  do {
    try f1()
    f2()
  } catch {}

}

 

 或这个 
 

func doLotsOfErrorProneWork()
{
  //smallest scope possible
  do {
    try f1()
  } catch {}
    f2()
}

 

 或者只是风格问题？ 

How large should the scope be for error handling in swift 2? I understand error handling (I think), but I am also doubting on the implementation details. If I have a function that calls throwing functions, how should I handle the scope of my do-try-catch?
 

Should I do this:
 

func doLotsOfErrorProneWork()
{
  //everything in scope
 do {
    try f1()
    f2()
    f3()
    try f4()
    f5()
 } catch {}
}

 

or this:
 

func doLotsOfErrorProneWork()
{
  //smallest scope possible
  do {
    try f1()
  } catch {}

    f2()
    f3()

  do {
    try f4()
  } catch {}

  f5()
}

 

Does it make a difference? What is the difference? One difference is probably the difference in error that is thrown between f1 and f4. But suppose there is only one throwing function. 
 

func doLotsOfErrorProneWork()
{
  //everything in scope
  do {
    try f1()
    f2()
  } catch {}

}

 

or this
 

func doLotsOfErrorProneWork()
{
  //smallest scope possible
  do {
    try f1()
  } catch {}
    f2()
}

 

Or is it just a matter of style?

原文：https://stackoverflow.com/questions/35502504
更新时间：2023-10-31 14:10

最满意答案

 问题是R不允许矩阵填充日期。 "matrix"和"Date"都是类。 R中Date对象的底层表示只是一个整数。 因此，当您创建矩阵y ，它会从x （它是一个整数数组）中获取基础数据，添加一个维度属性，并使类为"matrix" 。 
 

 这个问题没有任何干净的解决方法。 尽管如此，你可以使用一些黑客技巧。 例如，你可以显式强制y是类"Date"以及class(y)<-c("matrix","Date") 。 y仍然会打印出日期向量，但您可以使用矩阵坐标来操作它： 
 

> class(y)<-c("matrix","Date")
> head(y)
 [1] "2013-09-02" "2013-09-09" "2013-09-16" "2013-09-23" "2013-09-30"
 [6] "2013-10-07" "2013-09-03" "2013-09-10" "2013-09-17" "2013-09-24"
[11] "2013-10-01" "2013-10-08" "2013-09-04" "2013-09-11" "2013-09-18"
[16] "2013-09-25" "2013-10-02" "2013-10-09" "2013-09-05" "2013-09-12"
[21] "2013-09-19" "2013-09-26" "2013-10-03" "2013-10-10" "2013-09-06"
[26] "2013-09-13" "2013-09-20" "2013-09-27" "2013-10-04" "2013-10-11"
[31] "2013-09-07" "2013-09-14" "2013-09-21" "2013-09-28" "2013-10-05"
[36] "2013-10-12" "2013-09-08" "2013-09-15" "2013-09-22" "2013-09-29"
[41] "2013-10-06" "2013-10-13"
> y[1,]
[1] "2013-09-02" "2013-09-03" "2013-09-04" "2013-09-05" "2013-09-06"
[6] "2013-09-07" "2013-09-08"
> y[1,]<-y[1,]+1
> y[1,]
[1] "2013-09-03" "2013-09-04" "2013-09-05" "2013-09-06" "2013-09-07"
[6] "2013-09-08" "2013-09-09"

 

 您也可以使用数据框而不是矩阵： 
 

> y<-data.frame(y)
> y<-data.frame(lapply(y,function(x) {class(x)<-"Date";x}))
> head(y)
          X1         X2         X3         X4         X5         X6         X7
1 2013-09-02 2013-09-03 2013-09-04 2013-09-05 2013-09-06 2013-09-07 2013-09-08
2 2013-09-09 2013-09-10 2013-09-11 2013-09-12 2013-09-13 2013-09-14 2013-09-15
3 2013-09-16 2013-09-17 2013-09-18 2013-09-19 2013-09-20 2013-09-21 2013-09-22
4 2013-09-23 2013-09-24 2013-09-25 2013-09-26 2013-09-27 2013-09-28 2013-09-29
5 2013-09-30 2013-10-01 2013-10-02 2013-10-03 2013-10-04 2013-10-05 2013-10-06
6 2013-10-07 2013-10-08 2013-10-09 2013-10-10 2013-10-11 2013-10-12 2013-10-13

 

 第三种可能性是以字符格式保存y中的元素，然后在需要时使用as.Date将它们转换为日期： 
 

> y<-matrix(as.character(x),nrow=20,byrow=T)
> head(y)
     [,1]         [,2]         [,3]         [,4]         [,5]        
[1,] "2013-09-02" "2013-09-03" "2013-09-04" "2013-09-05" "2013-09-06"
[2,] "2013-09-09" "2013-09-10" "2013-09-11" "2013-09-12" "2013-09-13"
[3,] "2013-09-16" "2013-09-17" "2013-09-18" "2013-09-19" "2013-09-20"
[4,] "2013-09-23" "2013-09-24" "2013-09-25" "2013-09-26" "2013-09-27"
[5,] "2013-09-30" "2013-10-01" "2013-10-02" "2013-10-03" "2013-10-04"
[6,] "2013-10-07" "2013-10-08" "2013-10-09" "2013-10-10" "2013-10-11"
     [,6]         [,7]        
[1,] "2013-09-07" "2013-09-08"
[2,] "2013-09-14" "2013-09-15"
[3,] "2013-09-21" "2013-09-22"
[4,] "2013-09-28" "2013-09-29"
[5,] "2013-10-05" "2013-10-06"
[6,] "2013-10-12" "2013-10-13"
> as.Date(y[1,])
[1] "2013-09-02" "2013-09-03" "2013-09-04" "2013-09-05" "2013-09-06"
[6] "2013-09-07" "2013-09-08"


The problem is R doesn't allow matrices to be populated with dates. Both "matrix" and "Date" are classes. The underlying representation of a Date object in R is just an integer. So when you create your matrix y, it takes the underlying data from x (which is an array of integers), adds a dimension attribute, and makes the class "matrix". 
 

There's not any clean way around this problem. There are a few hacks you can use, though. For example, you could explicitly force y to be of class "Date" as well as "matrix" with class(y)<-c("matrix","Date"). y would still print out as just a vector of dates, but you would be able to manipulate it using matrix cordinates:
 

> class(y)<-c("matrix","Date")
> head(y)
 [1] "2013-09-02" "2013-09-09" "2013-09-16" "2013-09-23" "2013-09-30"
 [6] "2013-10-07" "2013-09-03" "2013-09-10" "2013-09-17" "2013-09-24"
[11] "2013-10-01" "2013-10-08" "2013-09-04" "2013-09-11" "2013-09-18"
[16] "2013-09-25" "2013-10-02" "2013-10-09" "2013-09-05" "2013-09-12"
[21] "2013-09-19" "2013-09-26" "2013-10-03" "2013-10-10" "2013-09-06"
[26] "2013-09-13" "2013-09-20" "2013-09-27" "2013-10-04" "2013-10-11"
[31] "2013-09-07" "2013-09-14" "2013-09-21" "2013-09-28" "2013-10-05"
[36] "2013-10-12" "2013-09-08" "2013-09-15" "2013-09-22" "2013-09-29"
[41] "2013-10-06" "2013-10-13"
> y[1,]
[1] "2013-09-02" "2013-09-03" "2013-09-04" "2013-09-05" "2013-09-06"
[6] "2013-09-07" "2013-09-08"
> y[1,]<-y[1,]+1
> y[1,]
[1] "2013-09-03" "2013-09-04" "2013-09-05" "2013-09-06" "2013-09-07"
[6] "2013-09-08" "2013-09-09"

 

You could also use a data frame instead of a matrix: 
 

> y<-data.frame(y)
> y<-data.frame(lapply(y,function(x) {class(x)<-"Date";x}))
> head(y)
          X1         X2         X3         X4         X5         X6         X7
1 2013-09-02 2013-09-03 2013-09-04 2013-09-05 2013-09-06 2013-09-07 2013-09-08
2 2013-09-09 2013-09-10 2013-09-11 2013-09-12 2013-09-13 2013-09-14 2013-09-15
3 2013-09-16 2013-09-17 2013-09-18 2013-09-19 2013-09-20 2013-09-21 2013-09-22
4 2013-09-23 2013-09-24 2013-09-25 2013-09-26 2013-09-27 2013-09-28 2013-09-29
5 2013-09-30 2013-10-01 2013-10-02 2013-10-03 2013-10-04 2013-10-05 2013-10-06
6 2013-10-07 2013-10-08 2013-10-09 2013-10-10 2013-10-11 2013-10-12 2013-10-13

 

A third possiblity is to keep the elements in y in character format, and then convert them to dates using as.Date when you need them:
 

> y<-matrix(as.character(x),nrow=20,byrow=T)
> head(y)
     [,1]         [,2]         [,3]         [,4]         [,5]        
[1,] "2013-09-02" "2013-09-03" "2013-09-04" "2013-09-05" "2013-09-06"
[2,] "2013-09-09" "2013-09-10" "2013-09-11" "2013-09-12" "2013-09-13"
[3,] "2013-09-16" "2013-09-17" "2013-09-18" "2013-09-19" "2013-09-20"
[4,] "2013-09-23" "2013-09-24" "2013-09-25" "2013-09-26" "2013-09-27"
[5,] "2013-09-30" "2013-10-01" "2013-10-02" "2013-10-03" "2013-10-04"
[6,] "2013-10-07" "2013-10-08" "2013-10-09" "2013-10-10" "2013-10-11"
     [,6]         [,7]        
[1,] "2013-09-07" "2013-09-08"
[2,] "2013-09-14" "2013-09-15"
[3,] "2013-09-21" "2013-09-22"
[4,] "2013-09-28" "2013-09-29"
[5,] "2013-10-05" "2013-10-06"
[6,] "2013-10-12" "2013-10-13"
> as.Date(y[1,])
[1] "2013-09-02" "2013-09-03" "2013-09-04" "2013-09-05" "2013-09-06"
[6] "2013-09-07" "2013-09-08"

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