Question

我正在搜索R函数的所有帮助，这些函数会将timepan，例如“15分钟”或“1小时”或“6秒”或“1天”转换为日期时间对象，如“00:15:00”或“01:00:00”或“00:00:06”或“1960-01-02 00:00:00”（对此不确定）。我确信这样的函数存在，或者有一种简洁的方法可以避免编程...

更具体地说，我想做这样的事情（使用组合函数名称transform.span.to.time）：

library(chron)

times(transform.span.to.time("15 min"))

应该产生与

相同的结果

times("00:15:00")

返回类似于“00:15:00”的transform.span.to.time（“15分钟”）之类的函数是否存在，或者是否存在如何做到这一点的技巧？

Answer 1

我们将假设一个空格分隔数字和单位，并且在“secs”单位之后也没有尾随空格。这将处理混合单位：

test <- "0 hours 15 min 0 secs"

 transform.span <- function(test){ 
     testh <-         if(!grepl( " hour | hours ", "0 hours 15 min 0 secs")){ 
             # First consequent if no hours
                                      sub("^", "0:", test)} else { 
                                      sub(" hour | hours ", ":", test)}
     testm <- if(!grepl( " min | minutes ", testh)) {    
             #  first consequent if no minutes
                                      sub(" min | minutes ", "0:", testh)} else{
                                      sub(" min | minutes ", ":", testh)  }

     test.s <- if(!grepl( " sec| secs| seconds", testm)) { 
              # first consequent if no seconds
                                       sub(" sec| secs| seconds", "0", testm)} else{ 
                                       sub(" sec| secs| seconds", "", testm)}

     return(times(test.s)) }
###  Use
> transform.span(test)
[1] 00:15:00
> test2 <- "21 hours 15 min 38 secs"
> transform.span(test2)
[1] 21:15:38

Answer 2

第一个解决方案在gsubfn包中使用strapply并转换为天数，例如1小时是一天的1/24。第二个解决方案转换为R表达式，计算天数然后对其进行评估。

library(gsubfn)
library(chron)

unit2days <- function(d, u) 
     as.numeric(d) * switch(tolower(u), s = 1, m = 60, h = 3600)/(24 * 3600)
transform.span.to.time <- function(x) 
     sapply(strapply(x, "(\\d+) *(\\w)", unit2days), sum)

这是第二个解决方案：

library(chron)

transform.span.to.time2 <- function(x) {
    x <- paste(x, 0)
    x <- sub("h\\w*", "*3600+", x, ignore.case = TRUE)
    x <- sub("m\\w*", "*60+", x, ignore.case = TRUE)
    x <- sub("s\\w*", "+", x, ignore.case = TRUE)
    unname(sapply(x, function(x) eval(parse(text = x)))/(24*3600))
}

试验：

> x <- c("12 hours 3 min 1 sec", "22h", "18 MINUTES 23 SECONDS")
>
> times(transform.span.to.time(x))
[1] 12:03:01 22:00:00 00:18:23
>
> times(transform.span.to.time2(x))
[1] 12:03:01 22:00:00 00:18:23

Answer 3

基函数?cut.POSIXt对breaks的一组指定值有效：

breaks: a vector of cut points _or_ number giving the number of
      intervals which ‘x’ is to be cut into *_or_ an interval
      specification, one of ‘"sec"’, ‘"min"’, ‘"hour"’, ‘"day"’,
      ‘"DSTday"’, ‘"week"’, ‘"month"’, ‘"quarter"’ or ‘"year"’,
      optionally preceded by an integer and a space, or followed by
      ‘"s"’.  For ‘"Date"’ objects only ‘"day"’, ‘"week"’,
      ‘"month"’, ‘"quarter"’ and ‘"year"’ are allowed.*

通过输入cut.POSIXt查看源代码，相关部分以此开头：

else if (is.character(breaks) && length(breaks) == 1L) {

您可以采用本节中的代码来满足您的需求。

Answer 4

您可以使用difftime定义时间跨度：

span2time <- function(span,  units = c('mins', 'secs', 'hours')) {
  span.dt <- as.difftime(span,  units = match.arg(units))
  format(as.POSIXct("1970-01-01") + span.dt, "%H:%M:%S")
}

例如：

> span2time(15)
[1] "00:15:00"

编辑：修改为生成chron的times可接受的字符串。

Answer 5

@DWin：谢谢。

基于DWin示例，我重新排列了一下，结果如下：

transform.span<-function(timeSpan) { 
    timeSpanH <- if(!grepl(" hour | hours | hour| hours|hour |hours |hour|hours", timeSpan)) { 
                # First consequent if no hours
                sub("^", "00:", timeSpan)
            } else { 
                sub(" hour | hours | hour| hours|hour |hours |hour|hours", ":", timeSpan)
            }
    timeSpanM <- if(!grepl( " min | minutes | min| minutes|min |minutes |min|minutes", timeSpanH)) {    
                #  first consequent if no minutes
                paste("00:", timeSpanH, sep="")
            } else{
                sub(" min | minutes | min| minutes|min |minutes |min|minutes", ":", timeSpanH)  
            }
    timeSpanS <- if(!grepl( " sec| secs| seconds|sec|secs|seconds", timeSpanM)) { 
                # first consequent if no seconds
                paste(timeSpanM, "00", sep="")
            } else{ 
                sub(" sec| secs| seconds|sec|secs|seconds", "", timeSpanM)
            }

    return(timeSpanS) 
}
###  Use
test <- "1 hour 2 min 1 sec"
times(transform.span(test))
test1hour <- "1 hour"
times(transform.span(test1hour))
test15min <- "15 min"
times(transform.span(test15min))
test4sec <- "4 sec"
times(transform.span(test4sec))

从时间跨度（例如“15分钟”或“2秒”）到“00:15:00”或“00:00:02”

5 个答案: