在 R 中计算两个不同的均值

Question

我正在尝试从“R”中的以下数据集中计算两种不同的方法

Plot   Date     Time Canopyheight     mean    pre       post    Diff
103B1 11/12/2019 1    50
103B1 11/12/2019 4    50
103B1 11/12/2019 6    78
103B1 11/12/2019 22   100            69.5
103B1 11/13/2019 1     60
103B1 11/13/2019 4     70
103B1 11/13/2019 6     80 
103B1 11/13/2019 22   100            77.5     73.5
103B1 11/14/2019 1    50
103B1 11/14/2019 4    50
103B1 11/14/2019 6    78
103B1 11/14/2019 22  100            69.5
103B1 11/15/2019 1    60
103B1 11/15/2019 4    80
103B1 11/15/2019 6    90
103B1 11/15/2019 22  120            87.5               78.5     5.0

我能够获得平均值，但我无法获得前值和后值。

预期结果

使用代码，我们应该可以得到 '73.5' 的值，它是 '69.5 和 77.5' 的平均值，其他值也是这样计算的。差值将计算为 Pre 和 Post 值之间的差值。

代码

Prepost <- Prepost %>% group_by(Plot, Date) %>% 
  mutate(meancanopyheight = mean(Canopyheight, na.rm = T))
Prepost$Preharvest <- lapply(Prepost$Date, function(m) mean(Prepost$meanCanopyheight[Prepost$Date >= m |Prepost$Date <= m+4| Prepost$Date == m+8], na.rm = TRUE))

我尝试计算但无法计算，我已在此处添加代码供您参考。

感谢您的帮助。

Answer 1

你可以像这样使用dplyr：

library(dplyr)

df %>% 
  group_by(Date) %>% 
  summarize(mean = mean(Canopyheight)) %>%
  mutate(group = rep(c("pre", "post"), each = 2)) %>%
  group_by(group) %>%
  summarize(mean = mean(mean))
#> # A tibble: 2 x 2
#>   group  mean
#>   <chr> <dbl>
#> 1 post   78.5
#> 2 pre    73.5

^{由reprex package (v0.3.0) 于 2020 年 2 月 20 日创建}

基于来自 OP 的进一步数据，使该解决方案更通用：

library(dplyr)


df <- structure(list(Plot = c("TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", 
"TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", 
"TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", 
"TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", 
"TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", 
"TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", 
"TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", "TF_103B1", 
"TF_103B1", "TF_103B1"), Date = structure(c(18217, 18217, 18217, 
18217, 18218, 18218, 18218, 18218, 18219, 18219, 18219, 18219, 
18220, 18221, 18221, 18221, 18221, 18222, 18222, 18222, 18222, 
18246, 18246, 18246, 18246, 18247, 18247, 18247, 18247, 18248, 
18248, 18248, 18248, 18249, 18250, 18250, 18250, 18250, 18251, 
18251, 18251, 18251), class = "Date"), Time = c("1", "4", "6", 
"22", "1", "4", "6", "22", "1", "4", "6", "22", "22", "1", "4", 
"6", "22", "1", "4", "6", "22", "1", "4", "6", "22", "1", "4", 
"6", "22", "1", "4", "6", "22", "22", "1", "4", "6", "22", "1", 
"4", "6", "22"), Canopyheight = c(2064.55, 2064.51, 2063.03, 
2063.62, 2065.94, 2064.83, 2061.58, 2064.07, 2066.97, 2063.99, 
2065.37, 2064.7, 2067.8, 2065.6, 2067.05, 2064.95, 2075.76, 2073.06, 
2079.23, 2072.75, 2068.81, 2065.66, 2065.85, 2065.65, 2063.65, 
2063.44, 2068.05, 2072.38, 2067.2, 2068.1, 2067.26, 2069.27, 
2063.05, 2088.45, 2086.24, 2088.91, 2092.04, 2092, 2092.67, 2090.7, 
2091.59, 2090.99)), row.names = c(NA, 42L), class = "data.frame")

  df <- df   %>% 
  group_by(Date) %>% 
  summarize(mean = mean(Canopyheight)) %>%
  mutate(prepost = rep(rep(c("pre", "post"), each = 3), length.out = n()))

  df$start_date <- rep(df$Date[seq(nrow(df)) %% 6 == 0], each = 6)

  df %>%
  group_by(start_date, prepost) %>%
  summarize(mean = mean(mean))
#> # A tibble: 4 x 3
#> # Groups:   start_date [2]
#>   start_date prepost  mean
#>   <date>     <chr>   <dbl>
#> 1 2019-11-22 post    2070.
#> 2 2019-11-22 pre     2064.
#> 3 2019-12-21 post    2090.
#> 4 2019-12-21 pre     2067.

^{由reprex package (v0.3.0) 于 2020 年 2 月 21 日创建}