Question

我正在处理一个鸟类数据集，其中每个个体（ID）都具有其出生地（TERR），出生年份（YOB）以及他们出生年份开始后的天数（DOB）

通常有多个人具有相同的TERR，YEAR和DOB。在相同的TERR和YEAR中出生的人可能会更多，但这些人的DOB有所不同（在此数据集中，第一组人的DOB比第二组人的“低” DOB）。

我想插入一个新列“ n”，其中每年第一组个人返回“ 1”，第二组个人返回“ 2”，第三组个人返回“ 3”。第二年时，该数字将恢复为“ 1”。

例如

private CancellationTokenSource cts = new CancellationTokenSource();
...

private async Task<bool> Wait()
{
    try
    {
        await Task.Delay(1000, cts.Token);
        return true;
    }
    catch
    {
        return false;
    }
}

// cancel Wait() from another method
cts.Cancel();

我对R很陌生，因此非常感谢您的帮助。我一直在尝试使用if_else函数，但步步为营。

Answer 1

按“ TERR”，“ YOB”分组后，使用“ DOB”的match个元素获得“ DOB”的unique

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

或将'DOB'转换为factor，并将其强制转换为numeric

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

在base R和ave中可以使用相同的方法

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

数据

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

这是data.table的解决方案：

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

根据分组变量返回数字

2 个答案:

数据