Заполните пробелы, используя среднее значение группы в R

Question

У меня есть набор данных, который имеет пробелы в одном из столбцов (темп).Я пытаюсь заполнить пробелы, используя «временные» данные от «датчика» или среднее значение «датчиков» в рамках одной и той же «обработки» и, конечно, той же отметки даты.Я пытаюсь сделать это с помощью tidyverse / lubridate.

date    treatment   sensor  temp
1/01/2019   1   A   30
2/01/2019   1   A   29.1
3/01/2019   1   A   21.2
4/01/2019   1   A   NA
1/01/2019   1   B   20.5
2/01/2019   1   B   19.8
3/01/2019   1   B   35.1
4/01/2019   1   B   23.5
1/01/2019   2   C   31.2
2/01/2019   2   C   32.1
3/01/2019   2   C   28.1
4/01/2019   2   C   31.2
1/01/2019   2   D   NA
2/01/2019   2   D   26.5
3/01/2019   2   D   27.9
4/01/2019   2   D   28

Это то, что я ожидаю:

date    treatment   sensor  temp
1/01/2019   1   A   30
2/01/2019   1   A   29.1
3/01/2019   1   A   21.2
4/01/2019   1   A   23.5
1/01/2019   1   B   20.5
2/01/2019   1   B   19.8
3/01/2019   1   B   35.1
4/01/2019   1   B   23.5
1/01/2019   2   C   31.2
2/01/2019   2   C   32.1
3/01/2019   2   C   28.1
4/01/2019   2   C   31.2
1/01/2019   2   D   31.2
2/01/2019   2   D   26.5
3/01/2019   2   D   27.9
4/01/2019   2   D   28

Большое спасибо за вашу помощь.

Answer 1

Другой вариант с na.aggregate из zoo

library(dplyr)
library(zoo)
df %>% 
   group_by(date, treatment) %>%
   mutate(temp = na.aggregate(temp))
# A tibble: 16 x 4
# Groups:   date, treatment [8]
#   date      treatment sensor  temp
#   <fct>         <int> <fct>  <dbl>
# 1 1/01/2019         1 A       30  
# 2 2/01/2019         1 A       29.1
# 3 3/01/2019         1 A       21.2
# 4 4/01/2019         1 A       23.5
# 5 1/01/2019         1 B       20.5
# 6 2/01/2019         1 B       19.8
# 7 3/01/2019         1 B       35.1
# 8 4/01/2019         1 B       23.5
# 9 1/01/2019         2 C       31.2
#10 2/01/2019         2 C       32.1
#11 3/01/2019         2 C       28.1
#12 4/01/2019         2 C       31.2
#13 1/01/2019         2 D       31.2
#14 2/01/2019         2 D       26.5
#15 3/01/2019         2 D       27.9
#16 4/01/2019         2 D       28  

данные

df <- structure(list(date = structure(c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 
4L, 1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L), .Label = c("1/01/2019", 
"2/01/2019", "3/01/2019", "4/01/2019"), class = "factor"), treatment = c(1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), 
    sensor = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 
    3L, 3L, 3L, 4L, 4L, 4L, 4L), .Label = c("A", "B", "C", "D"
    ), class = "factor"), temp = c(30, 29.1, 21.2, NA, 20.5, 
    19.8, 35.1, 23.5, 31.2, 32.1, 28.1, 31.2, NA, 26.5, 27.9, 
    28)), class = "data.frame", row.names = c(NA, -16L))

Answer 2

Вот один вариант использования map2_dbl из purrr. Мы group_by treatment и заменим NA temp первым не-NA temp тем же date в группе.

library(dplyr)
library(purrr)

df %>%
  group_by(treatment) %>%
  mutate(temp = map2_dbl(temp, date, ~if (is.na(.x)) 
                    temp[which.max(date == .y & !is.na(temp))] else .x))

#   date      treatment sensor  temp
#   <fct>         <int> <fct>  <dbl>
# 1 1/01/2019         1 A       30  
# 2 2/01/2019         1 A       29.1
# 3 3/01/2019         1 A       21.2
# 4 4/01/2019         1 A       23.5
# 5 1/01/2019         1 B       20.5
# 6 2/01/2019         1 B       19.8
# 7 3/01/2019         1 B       35.1
# 8 4/01/2019         1 B       23.5
# 9 1/01/2019         2 C       31.2
#10 2/01/2019         2 C       32.1
#11 3/01/2019         2 C       28.1
#12 4/01/2019         2 C       31.2
#13 1/01/2019         2 D       31.2
#14 2/01/2019         2 D       26.5
#15 3/01/2019         2 D       27.9
#16 4/01/2019         2 D       28  

данные

df <- structure(list(date = structure(c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 
4L, 1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L), .Label = c("1/01/2019", 
"2/01/2019", "3/01/2019", "4/01/2019"), class = "factor"), treatment = 
c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), 
sensor = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 
3L, 3L, 3L, 4L, 4L, 4L, 4L), .Label = c("A", "B", "C", "D"
), class = "factor"), temp = c(30, 29.1, 21.2, NA, 20.5, 
19.8, 35.1, 23.5, 31.2, 32.1, 28.1, 31.2, NA, 26.5, 27.9, 
28)), class = "data.frame", row.names = c(NA, -16L))

Answer 3

Как насчет этого:

df <- df %>%
group_by(date, treatment) %>%
mutate(
  fill = mean(temp, na.rm=TRUE), # value to fill in blanks
  temp2 = case_when(!is.na(temp) ~ temp,
                    TRUE ~ fill)
  )