缺失值的平均函数R，循环5行

Question

我想在df中计算每5行的平均值。这是我的df:

<表类> 时间价值tbody><<tr>03/06/2021 06:15:00NA03/06/2021 06:16:00NA03/06/2021 06:17:002003/06/2021 06:18:002203/06/2021 06:19:002503/06/2021 06:20:00NA03/06/2021 06:21:0031日03/06/2021 06:22:002303/06/2021 06:23:001903/06/2021 06:24:002503/06/2021 06:25:003403/06/2021 06:26:004203/06/2021 06:27:00NA03/06/2021 06:28:001903/06/2021 06:29:001703/06/2021 06:30:0025

你可以通过引入一个虚拟变量来解决这个问题，该变量将你的观察结果分成五组，然后计算组内的平均值。下面是基于整理宇宙的MWE，它假设您的数据位于名为df的数据帧中。

library(tidyverse)
df %>% 
mutate(Group= 1 + floor((row_number()-1) / 5)) %>% 
group_by(Group) %>% 
summarise(Mean=mean(value, na.rm=TRUE), .groups="drop")
# A tibble: 4 × 2
Group  Mean
<dbl> <dbl>
1     1  22.3
2     2  24.5
3     3  28  
4     4  25  

基于purrr::map_dfr的解决方案:

library(purrr)
df <- data.frame(
stringsAsFactors = FALSE,
time = c("03/06/2021 06:15:00","03/06/2021 06:16:00",
"03/06/2021 06:17:00",
"03/06/2021 06:18:00","03/06/2021 06:19:00",
"03/06/2021 06:20:00","03/06/2021 06:21:00",
"03/06/2021 06:22:00","03/06/2021 06:23:00",
"03/06/2021 06:24:00","03/06/2021 06:25:00",
"03/06/2021 06:26:00",
"03/06/2021 06:27:00","03/06/2021 06:28:00",
"03/06/2021 06:29:00","03/06/2021 06:30:00"),
value = c(NA,NA,20L,22L,
25L,NA,31L,23L,19L,25L,34L,42L,NA,19L,17L,
25L)
)
map_dfr(1:(nrow(df)-5),
~ data.frame(Group =.x, Mean = mean(df$value[.x:(.x+5)],na.rm=T)))
#>    Group     Mean
#> 1      1 22.33333
#> 2      2 24.50000
#> 3      3 24.20000
#> 4      4 24.00000
#> 5      5 24.60000
#> 6      6 26.40000
#> 7      7 29.00000
#> 8      8 28.60000
#> 9      9 27.80000
#> 10    10 27.40000
#> 11    11 27.40000

如果您想取每5分钟的平均值，您可以使用lubridate的floor_date/ceiling_date功能来计算时间。

library(dplyr)
library(lubridate)
df %>%
mutate(time = mdy_hms(time), 
time = floor_date(time, '5 mins')) %>%
group_by(time) %>%
summarise(value = mean(value, na.rm = TRUE))
#  time                value
#  <dttm>              <dbl>
#1 2021-03-06 06:15:00  22.3
#2 2021-03-06 06:20:00  24.5
#3 2021-03-06 06:25:00  28  
#4 2021-03-06 06:30:00  25