Агрегирование и перекрестная таблица наблюдений за животными

Question

Я выполняю базовую c агрегацию 30-летних наблюдений за отдельными животными, и у меня есть файл сырых данных, где есть одна строка для каждого отдельного наблюдения за животным, где индивидуальное имя и время наблюдения являются параметрами.

Этот скрипт дает длинный формат таблицы, и мне нужен длинный формат:

table(dis2$who, lubridate::year(dis2$tid))

Вот данные:

> dput(dis2)
structure(list(who = c("KA-3", "KA-4", "KA-4", "KA-5", "KA-6", 
"KA-6", "KA-7", "KA-7", "KA-8", "KA-8", "KA-12", "KA-12", "KA-12", 
"KB-4", "KB-4", "KB-4", "KB-4", "KB-4", "KB-4", "KB-4", "KB-6", 
"KB-6", "KB-6", "KB-6", "KB-6", "KB-6", "KB-6", "KB-6", "KB-6", 
"KB-6", "KB-6", "KB-6", "KB-6", "KB-6", "KB-7", "KB-7", "KB-7", 
"KB-7", "KB-11", "KB-11", "KB-11", "KB-11", "KB-11", "KB-11", 
"KB-11", "KB-11", "KB-11", "KB-11", "KB-11", "KB-11", "KC-1", 
"KC-1", "KC-1", "KC-3", "KC-3", "KE-1", "KE-1", "KE-1", "KE-1", 
"KE-1", "KE-1", "KE-3", "KF-1", "KF-1", "KF-1", "KF-1", "KF-1", 
"KG-6", "KG-6", "KG-6", "KH-2", "KH-2", "KH-2", "KH-3", "KH-3", 
"KH-4", "KH-4", "KH-5", "KH-5", "KH-6", "KH-6", "KH-7", "KH-7", 
"KH-7", "KI-1", "KI-1", "KI-1", "KI-1", "KI-1", "KI-1", "KI-1", 
"KI-1", "KI-1", "KI-1", "KI-1", "KI-1", "KI-1", "KI-1", "KI-1"
), film = c(14, 10, 14, 18, 10, 18, 14, 14, 10, 14, 15, 18, 19, 
20, 30, 37, 47, 48, 48, 49, 23, 23, 30, 32, 35, 38, 41, 43, 47, 
48, 48, 48, 49, 49, 36, 37, 43, 49, 36, 36, 36, 36, 37, 38, 39, 
42, 43, 47, 48, 49, 50, 51, 998, 50, 50, 73, 73, 73, 74, 74, 
74, 74, 8, 8, 8, 8, 52, 88, 90, 91, 84, 84, 92, 92, 93, 92, 104, 
92, 93, 84, 93, 93, 93, 94, 999, 999, 95, 95, 96, 96, 97, 97, 
97, 98, 98, 98, 99, 99, 102), frame = c(36, 15, 34, 16, 18, 15, 
35, 37, 13, 35, 27, 18, 14, 17, 24, 17, 11, 22, 24, 34, 29, 31, 
25, 15, 18, 12, 5, 23, 19, 9, 10, 14, 16, 17, 37, 15, 8, 14, 
29, 33, 35, 37, 18, 22, 22, 33, 10, 8, 21, 36, 30, 8, 1, 2, 3, 
2, 19, 36, 3, 5, 21, 8, 4, 8, 11, 14, 2, 19, 16, 16, 24, 26, 
6, 21, 19, 27, 8, 30, 26, 25, 30, 20, 33, 19, 1, 2, 1, 24, 19, 
22, 15, 17, 19, 5, 9, 27, 1, 8, 1), tid = structure(c(570067200, 
570067200, 570067200, 570412800, 570067200, 570412800, 570067200, 
570067200, 570067200, 570067200, 570067200, 570412800, 570412800, 
633744000, 634435200, 634608000, 635126400, 635126400, 635126400, 
635126400, 633744000, 633744000, 634435200, 634435200, 634521600, 
634694400, 635040000, 635040000, 635126400, 635126400, 635126400, 
635126400, 635126400, 635126400, 634608000, 634608000, 635040000, 
635126400, 634608000, 634608000, 634608000, 634608000, 634608000, 
634694400, 635040000, 635040000, 635040000, 635126400, 635126400, 
635126400, 640483200, 640483200, 640483200, 640483200, 640483200, 
659664000, 659664000, 659664000, 659664000, 659664000, 659664000, 
659664000, 569721600, 569721600, 569721600, 569721600, 656035200, 
668476800, 668476800, 668476800, 668476800, 668476800, 668476800, 
668476800, 668476800, 668476800, 668736000, 668476800, 668476800, 
668476800, 668476800, 668476800, 668476800, 668476800, 575424000, 
655948800, 668563200, 668563200, 668563200, 668563200, 668563200, 
668563200, 668563200, 668563200, 668563200, 668563200, 668563200, 
668563200, 668649600), class = c("POSIXct", "POSIXt"), tzone = "UTC")), 
class = "data.frame", row.names = c(NA,-99L))

Answer 1

Может быть, просто используйте table?

table(df3$who, lubridate::year(df3$tid))
#>       
#>        1991 1992 1993 1994 1995 1996
#>   NP-3    0    9    3    4    4    1
#>   NQ-1    0   17    0    0    0    0
#>   NQ-2    0    5    1    0    0    0
#>   NQ-5    0    2    0    0    0    0
#>   NR-3    0    3    0    0    0    0
#>   NT-1    0    1    2    3    0    0
#>   NT-6    0    2    0    0    0    0
#>   NU-1    0   16    0    0    0    0
#>   NU-2    0    6    0    0    0    0
#>   NU-3    0   11    6    0    0    0
#>   NU-5    0   17    0    0    0    0
#>   NU-6    3   14    8    0    0    0
#>   NU-9    0    4    0    0    0    0

EDIT

Обновите исправленные данные:

table(dis2$who, lubridate::year(dis2$tid))
#>        
#>         1988 1990 1991
#>   KA-12    3    0    0
#>   KA-3     1    0    0
#>   KA-4     2    0    0
#>   KA-5     1    0    0
#>   KA-6     2    0    0
#>   KA-7     2    0    0
#>   KA-8     2    0    0
#>   KB-11    0   12    0
#>   KB-4     0    7    0
#>   KB-6     0   14    0
#>   KB-7     0    4    0
#>   KC-1     0    3    0
#>   KC-3     0    2    0
#>   KE-1     0    6    0
#>   KE-3     0    1    0
#>   KF-1     4    1    0
#>   KG-6     0    0    3
#>   KH-2     0    0    3
#>   KH-3     0    0    2
#>   KH-4     0    0    2
#>   KH-5     0    0    2
#>   KH-6     0    0    2
#>   KH-7     0    0    3
#>   KI-1     1    1   13

Answer 2

Мы могли бы сделать group_by 'год' и 'кто' и summarise с sum из 'frame', а затем преобразовать его в 'широкий' формат с помощью pivot_wider

library(dplyr)
library(tidyr)
library(lubridate)
df3 %>% 
    group_by(who, year = year(tid)) %>%
    summarise(frame = sum(frame)) %>%
    pivot_wider(names_from = year, values_from = frame)

Если нам просто нужна частота

df3 %>%
     count(who, year = year(tid)) %>%
     pivot_wider(names_from = year, values_from = n, values_fill = list(n = 0))
# A tibble: 13 x 7
#   who   `1992` `1993` `1994` `1995` `1996` `1991`
#   <chr>  <int>  <int>  <int>  <int>  <int>  <int>
# 1 NP-3       9      3      4      4      1      0
# 2 NQ-1      17      0      0      0      0      0
# 3 NQ-2       5      1      0      0      0      0
# 4 NQ-5       2      0      0      0      0      0
# 5 NR-3       3      0      0      0      0      0
# 6 NT-1       1      2      3      0      0      0
# 7 NT-6       2      0      0      0      0      0
# 8 NU-1      16      0      0      0      0      0
# 9 NU-2       6      0      0      0      0      0
#10 NU-3      11      6      0      0      0      0
#11 NU-5      17      0      0      0      0      0
#12 NU-6      14      8      0      0      0      3
#13 NU-9       4      0      0      0      0      0