總結后如何考慮組內較大的日期

我正在按 3 乘 3 分組取平均值。為此，我正在使用該summarise功能。在這種情況下，我想從構成平均值的四個日期中選擇最后一個日期。

我試圖選擇最大值，但這樣我只是為整個組選擇了最高日期。

test = data.frame(my_groups = c("A", "A", "A", "B", "B", "C", "C", "C",  "A", "A", "A"),
                  measure = c(10, 20, 5, 2, 62 ,2, 5, 4, 6, 7, 25),
                  time= c("20-09-2020", "25-09-2020", "19-09-2020", "20-05-2020", "20-06-2021", 
                                      "11-01-2021", "13-01-2021", "13-01-2021", "15-01-2021", "15-01-2021", "19-01-2021"))
# > test
#    my_groups measure       time
# 1          A      10 20-09-2020
# 2          A      20 25-09-2020
# 3          A       5 19-09-2020
# 4          B       2 20-05-2020
# 5          B      62 20-06-2021
# 6          C       2 11-01-2021
# 7          C       5 13-01-2021
# 8          C       4 13-01-2021
# 9          A       6 15-01-2021
# 10         A       7 15-01-2021
# 11         A      25 19-01-2021

test %>%
  arrange(time) %>%
  group_by(my_groups) %>%
  summarise(mean_3 = rollapply(measure, 3, mean, by = 3, align = "left", partial = F),
            final_data = max(time))

# my_groups mean_3 final_data
#   <chr>       <dbl> <chr>     
# 1 A           12.7  25-09-2020
# 2 A           11.7  25-09-2020
# 3 C           3.67 13-01-2021

在第二行中，我希望日期是19-01-2021，而不是 group 的全域最大值A( 25-09-2020)。

關于我如何做到這一點的任何提示？

uj5u.com熱心網友回復：

我有 2 種 dplyr 方法供您使用。對此不滿意，因為當rollapplywithmax和dates 在 B 組中找不到任何內容時，它默認使用 double ，這與 A 和 C 組中的字符不匹配。

變異：

test %>%
  arrange(time) %>%
  group_by(my_groups) %>% 
  mutate(final = rollapply(time, 3, max, by = 3, fill = NA, align = "left", partial = F),
         mean_3 = rollapply(measure, 3, mean, by = 3, fill = NA, align = "left", partial = F)) %>% 
  filter(!is.na(final)) %>% 
  select(my_groups, final, mean_3) %>% 
  arrange(my_groups)

# A tibble: 3 x 3
# Groups:   my_groups [2]
  my_groups final      mean_3
  <chr>     <chr>       <dbl>
1 A         19-01-2021  12.7 
2 A         25-09-2020  11.7 
3 C         13-01-2021   3.67

總結這不是總結，但在代碼中更清晰一點：

test %>%
  arrange(time) %>%
  group_by(my_groups) %>% 
  summarise(final = rollapply(time, 3, max, by = 3, fill = NA, align = "left", partial = F),
         mean_3 = rollapply(measure, 3, mean, by = 3, fill = NA, align = "left", partial = F)) %>% 
  filter(!is.na(final))

`summarise()` has grouped output by 'my_groups'. You can override using the `.groups` argument.
# A tibble: 3 x 3
# Groups:   my_groups [2]
  my_groups final      mean_3
  <chr>     <chr>       <dbl>
1 A         19-01-2021  12.7 
2 A         25-09-2020  11.7 
3 C         13-01-2021   3.67

編輯：

從評論中添加了isa的解決方案。Partial = TRUE訣竅是：

test %>% 
  arrange(time) %>% 
  group_by(my_groups) %>% 
  summarise(mean_3 = rollapply(measure, 3, mean, by = 3, align = "left", partial = F), 
            final_data = rollapply(time, 3, max, by = 3, align = "left", partial = T))

`summarise()` has grouped output by 'my_groups'. You can override using the `.groups` argument.
# A tibble: 3 x 3
# Groups:   my_groups [2]
  my_groups mean_3 final_data
  <chr>      <dbl> <chr>     
1 A          12.7  19-01-2021
2 A          11.7  25-09-2020
3 C           3.67 13-01-2021

uj5u.com熱心網友回復：

另一種可能的解決方案：

library(tidyverse)

test = data.frame(my_groups = c("A", "A", "A", "B", "B", "C", "C", "C",  "A", "A", "A"),
                  measure = c(10, 20, 5, 2, 62 ,2, 5, 4, 6, 7, 25),
                  time= c("20-09-2020", "25-09-2020", "19-09-2020", "20-05-2020", "20-06-2021", 
                          "11-01-2021", "13-01-2021", "13-01-2021", "15-01-2021", "15-01-2021", "19-01-2021"))

test %>% 
  group_by(data.table::rleid(my_groups)) %>% 
  filter(n() == 3) %>% 
  summarise(
    groups = unique(my_groups), 
    mean_3 = mean(measure), final_data = max(time), .groups = "drop") %>%
  select(-1)

#> # A tibble: 3 × 3
#>   groups mean_3 final_data
#>   <chr>   <dbl> <chr>     
#> 1 A       11.7  25-09-2020
#> 2 C        3.67 13-01-2021
#> 3 A       12.7  19-01-2021

標籤：r dplyr

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