为什么sum和average在使用lubridate提取的小时数上不起作用，当使用+和除法的简单加法单独工作时。
数据集
Category Time1 Time2 hr1 hr2
1 A 0:30:00 24:00:00 30M 0S 24H 0M 0S
2 B 1:00:00 23:23:00 1H 0M 0S 23H 23M 0S
3 C 2:30:00 23:00:59 2H 30M 0S 23H 0M 59S
4 D 3:00:00 45:00:00 3H 0M 0S 45H 0M 0S
> dput(t1)
structure(list(Category = c("A", "B", "C", "D"), Time1 = c("0:30:00",
"1:00:00", "2:30:00", "3:00:00"), Time2 = c("24:00:00", "23:23:00",
"23:00:59", "45:00:00"), hr1 = structure(c(0, 0, 0, 0), year = c(0,
0, 0, 0), month = c(0, 0, 0, 0), day = c(0, 0, 0, 0), hour = c(0,
1, 2, 3), minute = c(30, 0, 30, 0), class = structure("Period", package = "lubridate")),
hr2 = structure(c(0, 0, 59, 0), year = c(0, 0, 0, 0), month = c(0,
0, 0, 0), day = c(0, 0, 0, 0), hour = c(24, 23, 23, 45), minute = c(0,
23, 0, 0), class = structure("Period", package = "lubridate"))), .Names = c("Category",
"Time1", "Time2", "hr1", "hr2"), row.names = c(NA, -4L), class = "data.frame")
RCODE
t1<-read.csv("time1.csv", header=TRUE, sep=",",stringsAsFactors=FALSE)
library(lubridate)
t1$hr1<-hms(t1$Time1)
t1$hr2<-hms(t1$Time2)
#This WORKS
> t1$hr1[4]+t1$hr2[3]
[1] "26H 0M 59S"
> (t1$hr1[4]+t1$hr2[3])/2
[1] "13H 0M 29.5S"
但这不是
> sum(t1$hr1[4]+t1$hr2[3])
[1] 59
> mean(t1$hr1[4]+t1$hr2[3])
[1] 59
答案 0 :(得分：0)
您可以使用sum转换为秒，执行mean或seconds_to_period，然后使用period转换回library(lubridate)
t1$hr1 = hms(t1$Time1)
t1$hr2 = hms(t1$Time2)
fsum = function(...){
seconds_to_period(sum(period_to_seconds(...)))
}
fmean = function(...){
seconds_to_period(mean(period_to_seconds(...)))
}
tohms = function(...){
ans = c(...)
ans@hour = ans@day*24 + ans@hour
ans@day = 0
ans
}
t2 = fsum(t1$hr1[4]+t1$hr2[3])
t2
#> [1] "1d 2H 0M 59S"
tohms(t2)
#> [1] "26H 0M 59S"
。
sum如果您预计mean或t1 = structure(list(Category = c("A", "B", "C", "D"),
Time1 = c("0:30:00", "1:00:00", "2:30:00", "3:00:00"),
Time2 = c("24:00:00", "23:23:00", "23:00:59", "45:00:00")),
class = "data.frame", row.names = c("1", "2", "3", "4"))
会超过一年（约365.25天），则可能需要进行进一步的调整。
数据
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