根據分組變量返回數字

Question

我正在使用一個鳥類數據集，其中每個人（ID）都有他們出生的領土（TERR），出生年份（YOB）以及他們出生年份（DOB）之后的天數。

通常有多個人具有相同的TERR，YEAR和DOB。 可能會有更多的人出生在相同的TERR和YEAR中，但這些人將有不同的DOB（在此數據集中，第一組個人將具有比第二組個人更低的DOB）。

我想插入一個新列'n'，其中每年第一組個體返回'1'，第二組個體返回'2'，第三組個體返回'3'。 當它是下一年時，該數字將恢復為“1”。

例如

ID TERR YOB  DOB N
1  A1   1982 148 1
2  A1   1982 148 1
3  A1   1982 148 1
4  A1   1982 185 2
5  A1   1982 185 2
6  A1   1985 137 1
7  A1   1985 137 1
8  BIAN 1989 132 1
9  BIAN 1989 132 1
10 BIAN 1989 132 1
11 BIAN 1992 155 1
12 BIAN 1992 155 1
13 BIAN 1992 155 1
14 BIAN 1992 254 2
15 BIAN 1992 254 2
16 BIAN 1992 254 2
17 BIAN 1994 164 1
18 BIAN 1994 164 1
19 GATE 1998 119 1
20 GATE 1998 119 1
21 GATE 1998 172 2
22 GATE 1998 172 2
23 GATE 1998 172 2
24 GATE 1999 153 1
25 GATE 1999 153 1

我對R很新，所以任何幫助都非常感謝。 我一直在嘗試使用if_else函數，但沒有使用它。

Answer 1

在通過'TERR'，'YOB'進行分組后，獲得'DOB'與'DOB'的unique元素的match

library(dplyr)
out <- df1 %>%
         group_by(TERR, YOB) %>% 
         mutate(N1 = match(DOB, unique(DOB)))
identical(out$N, out$N1)
#[1] TRUE

out
# A tibble: 25 x 6
# Groups:   TERR, YOB [7]
#      ID TERR    YOB   DOB     N    N1
#   <int> <chr> <int> <int> <int> <int>
# 1     1 A1     1982   148     1     1
# 2     2 A1     1982   148     1     1
# 3     3 A1     1982   148     1     1
# 4     4 A1     1982   185     2     2
# 5     5 A1     1982   185     2     2
# 6     6 A1     1985   137     1     1
# 7     7 A1     1985   137     1     1
# 8     8 BIAN   1989   132     1     1
# 9     9 BIAN   1989   132     1     1
#10    10 BIAN   1989   132     1     1
# ... with 15 more rows

---

或者將'DOB'轉換為factor並將其強制轉換為numeric

df1 %>%
   group_by(TERR, YOB) %>% 
   mutate(N1 = as.integer(factor(DOB, levels = unique(DOB))))

---

在具有ave base R可以使用相同的方法

with(df1, ave(DOB, TERR, YOB, FUN = function(x) match(x, unique(x))))

數據

df1 <- structure(list(ID = 1:25, TERR = c("A1", "A1", "A1", "A1", "A1", 
"A1", "A1", "BIAN", "BIAN", "BIAN", "BIAN", "BIAN", "BIAN", "BIAN", 
"BIAN", "BIAN", "BIAN", "BIAN", "GATE", "GATE", "GATE", "GATE", 
"GATE", "GATE", "GATE"), YOB = c(1982L, 1982L, 1982L, 1982L, 
1982L, 1985L, 1985L, 1989L, 1989L, 1989L, 1992L, 1992L, 1992L, 
1992L, 1992L, 1992L, 1994L, 1994L, 1998L, 1998L, 1998L, 1998L, 
1998L, 1999L, 1999L), DOB = c(148L, 148L, 148L, 185L, 185L, 137L, 
137L, 132L, 132L, 132L, 155L, 155L, 155L, 254L, 254L, 254L, 164L, 
164L, 119L, 119L, 172L, 172L, 172L, 153L, 153L), N = c(1L, 1L, 
1L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 
1L, 1L, 2L, 2L, 2L, 1L, 1L)), .Names = c("ID", "TERR", "YOB", 
"DOB", "N"), class = "data.frame", row.names = c(NA, -25L))

Answer 2

這是一個data.table的解決方案：

library("data.table")
DT <- fread(
"ID TERR YOB  DOB N
1  A1   1982 148 1
2  A1   1982 148 1
3  A1   1982 148 1
4  A1   1982 185 2
5  A1   1982 185 2
6  A1   1985 137 1
7  A1   1985 137 1
8  BIAN 1989 132 1
9  BIAN 1989 132 1
10 BIAN 1989 132 1
11 BIAN 1992 155 1
12 BIAN 1992 155 1
13 BIAN 1992 155 1
14 BIAN 1992 254 2
15 BIAN 1992 254 2
16 BIAN 1992 254 2
17 BIAN 1994 164 1
18 BIAN 1994 164 1
19 GATE 1998 119 1
20 GATE 1998 119 1
21 GATE 1998 172 2
22 GATE 1998 172 2
23 GATE 1998 172 2
24 GATE 1999 153 1
25 GATE 1999 153 1")
DT[, N2:=rleidv(DOB), .(TERR, YOB)][]
# > DT[, N2:=rleidv(DOB), .(TERR, YOB)][]
#    ID TERR  YOB DOB N N2
# 1:  1   A1 1982 148 1  1
# 2:  2   A1 1982 148 1  1
# 3:  3   A1 1982 148 1  1
# 4:  4   A1 1982 185 2  2
# 5:  5   A1 1982 185 2  2
# 6:  6   A1 1985 137 1  1
# 7:  7   A1 1985 137 1  1
# 8:  8 BIAN 1989 132 1  1
# 9:  9 BIAN 1989 132 1  1
# 10: 10 BIAN 1989 132 1  1
# 11: 11 BIAN 1992 155 1  1
# 12: 12 BIAN 1992 155 1  1
# 13: 13 BIAN 1992 155 1  1
# 14: 14 BIAN 1992 254 2  2
# 15: 15 BIAN 1992 254 2  2
# 16: 16 BIAN 1992 254 2  2
# 17: 17 BIAN 1994 164 1  1
# 18: 18 BIAN 1994 164 1  1
# 19: 19 GATE 1998 119 1  1
# 20: 20 GATE 1998 119 1  1
# 21: 21 GATE 1998 172 2  2
# 22: 22 GATE 1998 172 2  2
# 23: 23 GATE 1998 172 2  2
# 24: 24 GATE 1999 153 1  1
# 25: 25 GATE 1999 153 1  1
#     ID TERR  YOB DOB N N2