我的数据集记录了某些蝙蝠在不同夜晚的行为。我想找到这些时间的平均值，这些时间采用 hh: mm（24 小时制）格式。
Onset:
23:42,
21:40,
21:20,
21:30,
22:15,
23:40,
23:30,
02:10,
00:40,
01:35,
01:28,
01:00,
01:00,
00:55,
01:35.
对于可能的R解决方案，样本数据：
onset <- c("23:42","21:40","21:20","21:30","22:15","23:40","23:30",
"02:10","00:40","01:35","01:28","01:00","01:00","00:55","01:35")
【问题讨论】：
您可以在 1 行中使用 lubridate：
seconds_to_period(mean(period_to_seconds(hm(Onset))))
产生
[1] "11H 12M 0S"
【讨论】：
不是一个班轮（以下代码已调整以解决 OP 在 cmets 中提出的夜间与早晨问题）：
onset <- c("23:42","21:40","21:20","21:30","22:15","23:40","23:30",
"02:10","00:40","01:35","01:28","01:00","01:00","00:55","01:35")
library(tibble)
onset.df <- t(data.frame(strsplit(onset, ":"), stringsAsFactors=F))
colnames(onset.df) <- c("hours", "minutes")
onset.df <- as_tibble(onset.df)
onset.df$hours <- as.numeric(onset.df$hours)
onset.df$minutes <- as.numeric(onset.df$minutes)
onset.df$minutes.fraction <- onset.df$minutes/60
onset.df$hours.fraction <- onset.df$hours+onset.df$minutes.fraction
mean(onset.df$hours.fraction)
[1] 11.2
# alternative approach to account for night / morning
onset.df$hours <- ifelse(onset.df$hours < 12,
onset.df$hours+24, onset.df$hours)
onset.df$hours.fraction <- onset.df$hours+onset.df$minutes.fraction
onset.df
# A tibble: 15 x 4
hours minutes minutes.fraction hours.fraction
<dbl <dbl <dbl <dbl>
1 23.0 42.0 0.700 23.7
2 21.0 40.0 0.667 21.7
3 21.0 20.0 0.333 21.3
4 21.0 30.0 0.500 21.5
5 22.0 15.0 0.250 22.2
6 23.0 40.0 0.667 23.7
7 23.0 30.0 0.500 23.5
8 26.0 10.0 0.167 26.2
9 24.0 40.0 0.667 24.7
10 25.0 35.0 0.583 25.6
11 25.0 28.0 0.467 25.5
12 25.0 0 0 25.0
13 25.0 0 0 25.0
14 24.0 55.0 0.917 24.9
15 25.0 35.0 0.583 25.6
onset.mean.raw <- mean(onset.df$hours.fraction)
onset.mean.format <- ifelse(onset.mean.raw >= 24,
onset.mean.raw-24, onset.mean.raw)
onset.mean.format.hour <- round(onset.mean.format, 0)
onset.mean.format.minutes <- round((onset.mean.format-onset.mean.format.hour)*60, 0)
paste("Average time of onset:", onset.mean.format.hour, "hours and",
onset.mean.format.minutes, "minutes")
[1] "Average time of onset: 0 hours and 0 minutes"
我只是使用tibble 来去掉行名并使表格在R 控制台中更易于阅读。
【讨论】：
很少使用的as.difftime 在这里会很方便，以匹配@Hack-R 的结果：
mean(as.difftime(dat$Onset, format="%H:%M", units="hours"))
#Time difference of 11.2 hours
不过，由于时间已经过去，我认为你需要变得更狡猾：
out <- as.numeric(as.difftime(dat$Onset, format="%H:%M", units="hours"))
mean(ifelse(out < 12, out + 24, out))
# [1] 24
...可以解释为午夜。
【讨论】：
您可以编写一个函数来进行计算。唯一的问题是方法mean.POSIXct 需要它的参数属于"POSIXct" 或"POSIXt" 类。由于您只有 HH:MM，因此该函数会为您执行强制转换。
meanHour <- function(h, format = "%H:%M"){
hh <- as.POSIXct(paste(Sys.Date(), h), "%Y-%m-%d %H:%M")
hmean <- mean(hh)
format(hmean, format = format)
}
meanHour(Onset)
#[1] "11:12"
编辑。
在 OP 的 cmets 之后，我编写了一个处理午夜过后几个小时的函数。
meanHour2 <- function(h){
f <- function(x){
x[1] <- ifelse(x[1] < 12, x[1] + 24, x[1])
60*x[1] + x[2]
}
hh <- strsplit(h, ":")
hh <- lapply(hh, as.integer)
hh <- sapply(hh, f)
hmean <- mean(hh)
H <- hmean %/% 60
M <- hmean %% 60
sprintf("%02d:%02d", H, M)
}
meanHour2(h)
#[1] "24:00"
数据。
Onset <- scan(what = character(),
text = "23:42, 21:40, 21:20, 21:30, 22:15, 23:40,
23:30, 02:10, 00:40, 01:35, 01:28, 01:00,
01:00, 00:55, 01:35",
sep = ",")
【讨论】：