润滑固定时间单位

Question

当我们取两个时间的差时，那里会自动发生一些事情。

> ymd_hms("2016-05-09 15:17:03") - ymd_hms("2016-05-09 15:17:04")
Time difference of -1 secs
> ymd_hms("2016-05-09 16:17:03") - ymd_hms("2016-05-09 15:17:04")
Time difference of 59.98333 mins
> ymd_hms("2016-05-10 16:17:03") - ymd_hms("2016-05-09 15:17:04")
Time difference of 1.041655 days

如何在不使用 difftime 函数的情况下修复单位。

所以我可以执行以下操作：

VECTOR = c(ymd_hms("2016-05-10 16:17:03"),
           ymd_hms("2016-05-10 17:19:33"),
           ymd_hms("2016-05-10 19:55:03")
         )

diffs = diff(VECTOR)

IntervalsInHours = toHours(diffs)

此外，有什么方法可以知道 lubridate 时间对象中使用的单位。例如，

> ymd_hms("2016-05-09 15:17:03") - ymd_hms("2016-05-09 15:17:04")
Time difference of -1 secs

这里使用的单位是seconds.

Answer 1

请尝试以下将时差转换为小时。

library(lubridate)
x=ymd_hms("2016-05-09 16:17:03") 
y=ymd_hms("2016-05-19 15:17:04")
diffs=as.duration(x-y)
IntervalsInHours=as.numeric(abs(diffs))/3600;IntervalsInHours

Answer 2

或者你可以这样使用：

library(lubridate)
x=ymd_hms("2016-05-09 16:17:03") 
y=ymd_hms("2016-05-19 16:17:04")
diffs=as.duration(x-y);
IntervalsInHours=abs(diffs)/dhours(1);IntervalsInHours

Answer 3

"you want to use diff function to take the time differences between a VECTOR elements, only in the units specified"

请尝试以下代码：（通过 int_diff 函数）

> VECTOR = c(ymd_hms("2016-05-10 16:17:03"),
+            ymd_hms("2016-05-10 16:17:04"),
+            ymd_hms("2016-05-10 17:19:33"),
+            ymd_hms("2016-05-10 19:55:03")
+ )
> as.numeric(int_diff(VECTOR))
[1]    1 3749 9330
> round(as.numeric(int_diff(VECTOR))/3600,2)
[1] 0.00 1.04 2.59

看，无论时间间隔最小单位是秒还是非秒，它总是按秒缩放，如下所示。

> VECTOR = c(ymd_hms("2016-05-10 16:17:03"),
+            #ymd_hms("2016-05-10 16:17:04"),
+            ymd_hms("2016-05-10 17:19:33"),
+            ymd_hms("2016-05-10 19:55:03")
+ )
> as.numeric(int_diff(VECTOR))
[1] 3750 9330
> round(as.numeric(int_diff(VECTOR))/3600,2)
[1] 1.04 2.59

Answer 4

我写了两个函数以防有人觉得有用。

timeDiffUnitConvert = function(Diffs, to="day", roundingN = 1){      
  if(to == "day"){
    R = round(as.numeric(as.duration(Diffs))/3600/24,roundingN)
  } else if (to == "hour") {
    R = round(as.numeric(as.duration(Diffs))/3600, roundingN)
  } else if (to == "min") {
    R = round(as.numeric(as.duration(Diffs))/60, roundingN)
  } else if (to == "sec"){
    R = round(as.numeric(as.duration(Diffs)), roundingN)
  } else {
    stop("to which unit? it must be `day`, `hour`, `min` or `sec`.")
  }
  R
}

timeDiffVector = function(TimeVector, to="day", roundingN = 1, attachNaMode = "none"){
  R = timeDiffUnitConvert(diff(TimeVector), to = to, roundingN = roundingN)
  if(attachNaMode == "leading"){
    R = c(NA,R)
  } else if(attachNaMode == "trailing"){
    R = c(R,NA)
  } else{
    stop("check your attachNaMode: shall be either `leading` or `trailing`")
  }
  R
}

润滑固定时间单位

Lubridate fix the time units

r

lubridate