如何计算R中变量的每小时平均值

Question

我无法计算几天内每小时的变量“计数”平均值。我有一个名为 data 的数据集，如下所示：

    Time                count
1   2019-06-30 05:00:00 17
2   2019-06-30 06:00:00 18
3   2019-06-30 07:00:00 26
4   2019-06-30 08:00:00 15
5   2019-07-01 00:00:00 13
6   2019-07-01 01:00:00 23
7   2019-07-01 02:00:00 13
8   2019-07-01 03:00:00 22

它包含几天内每小时的值。现在我想计算每个小时的值，这是该小时在所有日子里的平均值。像这样的：

    Time        count
1   00:00       22
2   01:00       13
3   02:00       11
4   03:00       9

我是 R 新手，只计算每日平均值：

DF2 <- data.frame(data, Day = as.Date(format(data$Time)))
aggregate(cbind(count) ~ Day, DF2, mean)

    Time        count
1   2019-06-30  22
2   2019-07-01  13
3   2019-07-02  11
4   2019-07-03  9

但我无法让它与每小时平均值一起工作。我试图在其他帖子中找到解决方案，但它们要么不起作用，要么似乎需要大量独特的计算。在 R 中必须有一种简单的方法来做到这一点。

这里是 dput(droplevels(head(data, 4))) 的输出：

structure(list(Time = structure(1:4, .Label = c("2019-06-30 05:00:00", 
"2019-06-30 06:00:00", "2019-06-30 07:00:00", "2019-06-30 08:00:00"
), class = "factor"), count = c(17L, 18L, 26L, 15L)), row.names = c(NA, 
4L), class = "data.frame")

有什么建议吗？提前谢谢！

马克西

Answer 1

只需花费substrings 和aggregate 的时间就可以了。

d$hour <- substring(d$time, 12)
d.2 <- aggregate(count ~ substring(d$time, 12), d, mean)
head(d.2)
#        hour count
# 1  00:00:00 35.00
# 2  01:00:00 73.50
# 3  02:00:00 45.50
# 4  03:00:00 61.75
# 5  04:00:00 65.25
# 6  05:00:00 40.00

或使用ave 获取每小时平均值作为新列。

d <- transform(d, h.average=ave(count, substring(time, 12)))
head(d)
#                  time count h.average
# 1 2019-06-30 00:00:00    40    35.00
# 2 2019-06-30 01:00:00    67    73.50
# 3 2019-06-30 02:00:00    34    45.50
# 4 2019-06-30 03:00:00    49    61.75
# 5 2019-06-30 04:00:00    67    65.25
# 6 2019-06-30 05:00:00    43    40.00

数据

d <- structure(list(time = structure(c(1561845600, 1561849200, 1561852800, 
1561856400, 1561860000, 1561863600, 1561867200, 1561870800, 1561874400, 
1561878000, 1561881600, 1561885200, 1561888800, 1561892400, 1561896000, 
1561899600, 1561903200, 1561906800, 1561910400, 1561914000, 1561917600, 
1561921200, 1561924800, 1561928400, 1561932000, 1561935600, 1561939200, 
1561942800, 1561946400, 1561950000, 1561953600, 1561957200, 1561960800, 
1561964400, 1561968000, 1561971600, 1561975200, 1561978800, 1561982400, 
1561986000, 1561989600, 1561993200, 1561996800, 1562000400, 1562004000, 
1562007600, 1562011200, 1562014800, 1562018400, 1562022000, 1562025600, 
1562029200, 1562032800, 1562036400, 1562040000, 1562043600, 1562047200, 
1562050800, 1562054400, 1562058000, 1562061600, 1562065200, 1562068800, 
1562072400, 1562076000, 1562079600, 1562083200, 1562086800, 1562090400, 
1562094000, 1562097600, 1562101200, 1562104800, 1562108400, 1562112000, 
1562115600, 1562119200, 1562122800, 1562126400, 1562130000, 1562133600, 
1562137200, 1562140800, 1562144400, 1562148000, 1562151600, 1562155200, 
1562158800, 1562162400, 1562166000, 1562169600, 1562173200, 1562176800, 
1562180400, 1562184000, 1562187600), class = c("POSIXct", "POSIXt"
), tzone = ""), count = c(40L, 67L, 34L, 49L, 67L, 43L, 58L, 
37L, 22L, 97L, 3L, 78L, 16L, 74L, 27L, 72L, 87L, 9L, 99L, 98L, 
38L, 98L, 48L, 30L, 89L, 94L, 73L, 37L, 81L, 20L, 98L, 67L, 17L, 
88L, 75L, 8L, 39L, 53L, 20L, 92L, 61L, 23L, 56L, 33L, 60L, 19L, 
80L, 50L, 10L, 74L, 19L, 77L, 87L, 40L, 53L, 39L, 60L, 39L, 37L, 
65L, 51L, 56L, 98L, 50L, 23L, 38L, 53L, 36L, 61L, 12L, 6L, 81L, 
1L, 59L, 56L, 84L, 26L, 57L, 83L, 56L, 3L, 45L, 19L, 50L, 84L, 
95L, 14L, 98L, 97L, 22L, 46L, 7L, 47L, 55L, 38L, 43L)), row.names = c(NA, 
-96L), class = "data.frame")

Answer 2

使用 lubridate 和 dplyr：按时间值分组

生成数据

library(dplyr)
library(lubridate)

df <- data.frame(Time=seq(as.POSIXct('2019-06-30 00:00:00'), as.POSIXct('2019-07-03 23:00:00'), by=3600),
  count = floor(runif(96, 12,71))
) 

按小时值分组，取平均值，打印漂亮

df %>% mutate(hour = lubridate::hour(Time)) %>%
  group_by(hour) %>% summarise(count=mean(count)) %>%
  # pretty print
  mutate(hour = sprintf("%02d:00", hour)) %>%
  print(n=24)