在 R 中计算几年序列中特定日期范围的平均值

Question

我的数据框如下

data.frame(
  stringsAsFactors = FALSE,
              Date = c("01/04/1999","30/09/1999",
                       "01/10/1999","31/03/2000","01/04/2000","30/09/2000",
                       "01/10/2000","31/03/2001"),
           Germany = c(8, 8, 7, 8, 10, 6, 8, 5),
        Luxembourg = c(2, 1, 9, 7, 1, 7, 3, 10),
           Belgium = c(2, 7, 10, 2, 9, 7, 2, 7)
)->df

在这种情况下，我必须计算夏季和冬季的平均值：

1日表示：从4月1日到9月30日
2日：10月1日至次年3月31日

最终得到以下结果：

我的原始数据框是过去 25 年的每日数据

Answer 1

您需要找到一种方法来创建独特的分组。 这应该可以解决问题：

library(dplyr)
library(lubridate)


df <- data.frame(
  stringsAsFactors = FALSE,
  Date = c("01/04/1999","30/09/1999",
           "01/10/1999","31/03/2000","01/04/2000","30/09/2000",
           "01/10/2000","31/03/2001"),
  Germany = c(8, 8, 7, 8, 10, 6, 8, 5),
  Luxembourg = c(2, 1, 9, 7, 1, 7, 3, 10),
  Belgium = c(2, 7, 10, 2, 9, 7, 2, 7)
)

df %>%
  mutate(Date = as.Date(Date, "%d/%m/%Y"),
         Month = month(Date),
         Season = if_else(Month >= 4 & Month <= 9, "Summer", "Winter"),
         Year = year(Date),
         YearGroup = case_when(Month >= 4 & Month <= 9 ~ as.character(Year),
                               Month <= 3 ~ paste(as.character(Year - 1), as.character(Year)),
                               Month >= 10 ~ paste(as.character(Year), as.character(Year + 1)))
         ) %>%
  select(!c(Date, Year)) %>%
  group_by(YearGroup, Season) %>%
  summarise_all(mean)
#> # A tibble: 4 x 6
#> # Groups:   YearGroup [4]
#>   YearGroup Season Germany Luxembourg Belgium Month
#>   <chr>     <chr>    <dbl>      <dbl>   <dbl> <dbl>
#> 1 1999      Summer     8          1.5     4.5   6.5
#> 2 1999 2000 Winter     7.5        8       6     6.5
#> 3 2000      Summer     8          4       8     6.5
#> 4 2000 2001 Winter     6.5        6.5     4.5   6.5

Answer 2

更新由于亲爱的@PLY 分享的想法，我对我的代码进行了轻微修改以获得 output 安排。

library(dplyr)
library(tidyr)
library(lubridate)
library(glue)

df %>%
  mutate(Date = dmy(Date),
         Year = year(Date), 
         Month = month(Date)) %>%
  mutate(Season = ifelse(Month >= 4 & Month < 10, glue("summer {Year}"), 
                         ifelse(Month <= 3, glue("Winter {Year-1}-{Year}"), 
                                ifelse(Month >= 10, glue("Winter {Year}-{Year + 1}"), 
                                       Month)))) %>%
  select(-c(Date, Year)) %>%
  separate(Season, c("Season", "Years"), " ") %>%
  group_by(Years, Season) %>%
  summarise(across(Germany:Belgium, ~ mean(.x, na.rm = TRUE))) %>%
  unite(Seasons, Season:Years)


# A tibble: 4 x 4
  Seasons          Germany Luxembourg Belgium
  <chr>              <dbl>      <dbl>   <dbl>
1 summer_1999          8          1.5     4.5
2 Winter_1999-2000     7.5        8       6  
3 summer_2000          8          4       8  
4 Winter_2000-2001     6.5        6.5     4.5

我需要分开季节和年份间隔，以便行的顺序正确。 否则，所有夏季行都将位于所有冬季行之上。

数据：

df <- data.frame(
  stringsAsFactors = FALSE,
  Date = c("01/04/1999","30/09/1999",
           "01/10/1999","31/03/2000","01/04/2000","30/09/2000",
           "01/10/2000","31/03/2001"),
  Germany = c(8, 8, 7, 8, 10, 6, 8, 5),
  Luxembourg = c(2, 1, 9, 7, 1, 7, 3, 10),
  Belgium = c(2, 7, 10, 2, 9, 7, 2, 7)
)

Answer 3

data.table选项

setDT(df)[
  ,
  Date := as.Date(Date, format = "%d/%m/%Y")
][
  ,
  lapply(.SD, mean),
  .(Season = findInterval(Date, seq(first(Date), last(Date), by = "6 months")))
]

给

   Season Germany Luxembourg Belgium
1:      1     8.0        1.5     4.5
2:      2     7.5        8.0     6.0
3:      3     8.0        4.0     8.0
4:      4     6.5        6.5     4.5