merge.zoo 删除时区 [英] merge.zoo removes time zone

问题描述

merge.zoo 的结果与其输入的时区不同.

The result of merge.zoo does not have the same time zone as its input.

考虑下面的例子

library(zoo)
zoo_a=zoo(data.frame(a=1:5),
          seq(as.POSIXct("2014-01-01 00:00:01",tz="UTC"),
              as.POSIXct("2014-01-01 00:00:05",tz="UTC"),
              by=1)
          )
zoo_b=zoo(data.frame(a=1:4),
          seq(as.POSIXct("2014-01-01 00:00:01",tz="UTC"),
              as.POSIXct("2014-01-01 00:00:05",tz="UTC"),
              by=1)
          )

zoo_merged=merge(zoo_a,zoo_b)
time(zoo_merged)[1]
#2013-12-31 19:00:01 EST
time(zoo_a)[1]
#2014-01-01 00:00:01 UTC
time(zoo_b)[1]
#2014-01-01 00:00:01 UTC

与zoo_merged 关联的时区实际上不是EST 而是

The time zone associated with zoo_merged is not really EST but

library(lubridate)
tz(time(zoo_merged)[1])
#""

时区属性似乎已被删除，R 可能使用某种默认时区来显示数据.

The time zone attribute seems to have been removed and R is probably using some sort of default timezone to display the data.

我可以通过lubridate通过

time(zoo_merged)=with_tz(time(zoo_merged),tz="UTC")
time(zoo_merged)[1]
#2014-01-01 00:00:01 UTC

有什么办法可以正确解决这个问题，即之后不必更改时区?我正在考虑更改 merge.zoo 的代码，但相应代码中没有一行注释......

Is there any way to fix this properly, i.e. without having to change the timezone afterwards? I was thinking of changing the code for merge.zoo but there's not a single line of comments in the respective code...

推荐答案

G. Grothendieck 建议的解决方案

Solution as suggested by G. Grothendieck

library(xts)

merge2=function(x,y) {
  as.zoo(merge(as.xts(x), as.xts(y)))
}

time(merge2(zoo_a,zoo_b))[1]
#[1] "2014-01-01 00:00:01 UTC"

或者按照 42-

merge3=function(x,y) {
  if ((tmp_tzone<-attr(time(x),"tzone"))!=attr(time(y),"tzone")) {
    message("Timezone attributes of x and y are not the same. Using default tz.")
    return(merge(x,y))
  } else {
    tmp=merge(x,y)
    attr(time(tmp),"tzone")=tmp_tzone
    return(tmp)
  }
}

#input with same tzones 
time(merge3(zoo_a,zoo_b))[1]
#[1] "2014-01-01 00:00:01 UTC"

#input with different tzones
zoo_c=zoo(data.frame(a=1:4),
          seq(as.POSIXct("2014-01-01 00:00:01",tz="EDT"),
              as.POSIXct("2014-01-01 00:00:05",tz="EDT"),
              by=1)
)

time(merge3(zoo_a,zoo_c))[1]
#Timezone attributes of x and y are not the same. Using default tz.
#[1] "2014-01-01 01:00:01 CET"

这篇关于merge.zoo 删除时区的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！