0

我正在尝试使用 to.period 将毫秒数据转换为常规秒级数据-但是最终输出具有尾随毫秒

dput(head(x$Mid,20))
structure(c(3228.5, 3228.5, 3228.5, 3229, 3229, 3229, 3229.5, 
3229.5, 3229.5, 3229, 3229.5, 3229.5, 3229.5, 3230, 3228.5, 3231.5, 
3231, 3231, 3230, 3231), class = c("xts", "zoo"), index = structure(c(1543804206.41, 
1543804206.55, 1543804206.63, 1543804206.68, 1543804206.75, 1543804206.8, 
1543804206.85, 1543804206.93, 1543804206.97, 1543804207.03, 1543804207.07, 
1543804207.1, 1543804207.14, 1543804207.22, 1543804207.29, 1543804207.44, 
1543804207.7, 1543804207.75, 1543804207.88, 1543804207.93), tzone = "", tclass = c("POSIXct", 
"POSIXt")), .Dim = c(20L, 1L), .Dimnames = list(NULL, "Mid"))


dput(to.period(head(x$Mid,20),"seconds"))
structure(c(3228.5, 3229, 3229.5, 3231.5, 3228.5, 3228.5, 3229.5, 
3231), .Dim = c(2L, 4L), .Dimnames = list(NULL, c("head(x$Mid, 20).Open", 
"head(x$Mid, 20).High", "head(x$Mid, 20).Low", "head(x$Mid, 20).Close"
)), index = structure(c(1543804206.97, 1543804207.93), tzone = "", tclass = c("POSIXct", 
"POSIXt")), class = c("xts", "zoo"))

如何使时间戳上的 to.period 输出舍入到秒。

4

1 回答 1

1

to.period以您正在做的方式使用endpoints和使用将采用第二个的最后一个值。不四舍五入值。如果你愿意，你可以使用 lubridate 的or函数来得到你想要的。to.periodto.periodround_datefloor_date

基于您的 x$MID 的一个小示例：

t <- to.period(x, "seconds", OHLC = FALSE)
t
                          Mid
2018-12-03 03:30:06.97 3229.5
2018-12-03 03:30:07.93 3231.0

library(lubridate)
index(t) <- floor_date(index(t), "1s") # replace index with floor value

t
                       Mid
2018-12-03 03:30:06 3229.5
2018-12-03 03:30:07 3231.0
于 2021-03-12T10:10:08.460 回答