r - R查找持续时间
问题描述
我已经使用该库lubridate
来查找开始时间和结束时间之间的差异。这是我的代码:
library(lubridate)
a <- read.csv("C:/Users/Desktop/Dump.csv")
b <- a$StartTime
c <- a$EndTime
hm(c)- hm(b)
这是我的代码结果:
[1] "1H -18M 0S" "1H 5M 0S" "1H 16M 0S" "1H 16M 0S" "1H 5M 0S" "1H 16M 0S" "1H 18M 0S"
[8] "44M 0S" "1H 0M 0S" "1H 23M 0S" "2H -41M 0S" "1H -18M 0S" "1H -18M 0S" "1H -16M 0S"
[15] "1H 0M 0S" "41M 0S" "1H 0M 0S" "24M 0S" "1H -20M 0S" "1H 0M 0S" "1H -19M 0S"
[22] "1H 16M 0S" "1H 25M 0S" "1H 10M 0S" "1H 16M 0S" "44M 0S" "1H 18M 0S" "1H 23M 0S"
[29] "1H -19M 0S" "1H 6M 0S" "1H 16M 0S" "1H -19M 0S" "1H -17M 0S" "1H -20M 0S" "1H 0M 0S"
[36] "2H -40M 0S" "2H -40M 0S" "1H -20M 0S" "1H -20M 0S" "1H 5M 0S" "1H 17M 0S" "1H 11M 0S"
[43] "1H -23M 0S" "1H -30M 0S" "1H -29M 0S" "1H -19M 0S" "45M 0S" "1H -19M 0S" "1H 1M 0S"
[50] "1H -6M 0S" "30M 0S" "1H 28M 0S" "1H -21M 0S" "30M 0S" "1H 1M 0S" "1H -10M 0S"
[57] "1H 0M 0S" "1H -28M 0S" "42M 0S" "51M 0S" "52M 0S" "1H 17M 0S" "2H -43M 0S"
[64] "1H -19M 0S" "1H 1M 0S" "1H -26M 0S" "50M 0S" "1H -19M 0S" "1H 0M 0S" "1H -23M 0S"
[71] "1H 2M 0S" "40M 0S" "30M 0S" "1H -15M 0S" "1H -6M 0S" "30M 0S" "1H -33M 0S"
[78] "1H -20M 0S" "1H 4M 0S" "41M 0S" "1H 13M 0S" "1H 0M 0S" "1H 0M 0S" "1H -30M 0S"
[85] "29M 0S" "1H 10M 0S" "1H 5M 0S" "1H 0M 0S" "1H -27M 0S" "51M 0S" "1H -5M 0S"
[92] "31M 0S" "1H -30M 0S" "33M 0S" "1H 0M 0S" "36M 0S" "2H -43M 0S" "1H -30M 0S"
[99] "34M 0S" "50M 0S" "1H -31M 0S" "41M 0S" "50M 0S" "1H -15M 0S" "1H -15M 0S"
这是我的结果,但分钟数为负数,这意味着正确答案是“小时 - 分钟”。任何人都知道我的代码在哪里错了,或者我是否忘记编写任何代码。
数据的格式是因素。
解决方案
您可能有信号时间数据库(给出的代码不会重现这种结果)。不过,请尝试以正确的格式计算经过的时间(time2 - time3 或 time2%--%time3)和使用持续时间
time2 <- c("12:23:11", "09:45:31", "12:05:22")
time3 <- c("2:14", "2:16", "3:35")
time2 <- hms(time2)
## [1] "12H 23M 11S" "9H 45M 31S" "12H 5M 22S"
time3 <- hm(time3)
## [1] "2H 14M 0S" "2H 16M 0S" "3H 35M 0S"
elapsed.time <- time2 - time3
as.duration(elapsed.time)
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