R 在组内创建 ID [英] R create ID within a group
问题描述
我有以下数据集:
df<-structure(list(IDFAM = c("2010 7599 2996 1", "2010 7599 3071 1",
"2010 7599 3071 1", "2010 7599 3660 1", "2010 7599 4736 1", "2010 7599 6235 1",
"2010 7599 6299 1", "2010 7599 9903 1", "2010 7599 11013 1",
"2010 7599 11778 1", "2010 7599 11778 1", "2010 7599 12248 1",
"2010 7599 13127 1", "2010 7599 14261 1", "2010 7599 16280 1",
"2010 7599 16280 1", "2010 7599 16280 1", "2010 7599 16280 1",
"2010 7599 16280 1", "2010 7599 17382 1"), AGED = c(45L, 47L,
24L, 46L, 46L, 44L, 43L, 43L, 43L, 16L, 43L, 46L, 44L, 47L, 43L,
16L, 20L, 18L, 18L, 43L)), .Names = c("IDFAM", "AGED"), row.names = c("5614",
"5748", "5753", "6864", "8894", "11761", "11884", "18738", "20896",
"22351", "22353", "23267", "24939", "27072", "30946", "30947",
"30949", "30950", "30952", "33034"), class = "data.frame")
我想为每个具有相同 IDFAM 值的观察分配一个 ID，范围从 1 到 n，其中 n 是具有相同 IDFAM.这将导致下表:
I would like to assign an ID to each observation having the same IDFAM value ranging from 1 to n, where n is the number of observations with the same value of IDFAM. This would result in the following table:
IDFAM AGED ID
2010 7599 2996 1 45 1
2010 7599 3071 1 47 1
2010 7599 3071 1 24 2
2010 7599 3660 1 46 1
2010 7599 4736 1 46 1
2010 7599 6235 1 44 1
2010 7599 6299 1 43 1
2010 7599 9903 1 43 1
2010 7599 11013 1 43 1
2010 7599 11778 1 16 1
2010 7599 11778 1 43 2
2010 7599 12248 1 46 1
2010 7599 13127 1 44 1
2010 7599 14261 1 47 1
2010 7599 16280 1 43 1
2010 7599 16280 1 16 2
2010 7599 16280 1 20 3
2010 7599 16280 1 18 4
2010 7599 16280 1 18 5
2010 7599 17382 1 43 1
我该怎么做?谢谢.
推荐答案
有几种方法.
在基础 R 中，使用 ave:
In base R, use ave:
with(df, ave(rep(1, nrow(df)), IDFAM, FUN = seq_along))
# [1] 1 1 2 1 1 1 1 1 1 1 2 1 1 1 1 2 3 4 5 1
<小时>
使用data.table"包，使用sequence(.N):
library(data.table)
DT <- as.data.table(df)
DT[, ID := sequence(.N), by = IDFAM]
<小时>
使用dplyr"包，尝试:
With the "dplyr" package, try:
df %>% group_by(IDFAM) %>% mutate(count = sequence(n()))
或(如 Hadley 在评论中推荐的):
or (as recommended by Hadley in the comments):
df %>% group_by(IDFAM) %>% mutate(count = row_number(IDFAM))
<小时>
更新
由于这似乎是相对频繁地要求的东西，因此此功能已作为函数 (getanID) 添加到我的splitstackshape"包中.它基于上面的data.table"方法.
Update
Since this seems to be something that is asked for relatively frequently, this feature has been added as a function (getanID) in my "splitstackshape" package. It is based on the "data.table" approach above.
library(splitstackshape)
getanID(df, id.vars = "IDFAM")
# IDFAM AGED .id
# 1: 2010 7599 2996 1 45 1
# 2: 2010 7599 3071 1 47 1
# 3: 2010 7599 3071 1 24 2
# 4: 2010 7599 3660 1 46 1
# 5: 2010 7599 4736 1 46 1
# 6: 2010 7599 6235 1 44 1
# 7: 2010 7599 6299 1 43 1
# 8: 2010 7599 9903 1 43 1
# 9: 2010 7599 11013 1 43 1
# 10: 2010 7599 11778 1 16 1
# 11: 2010 7599 11778 1 43 2
# 12: 2010 7599 12248 1 46 1
# 13: 2010 7599 13127 1 44 1
# 14: 2010 7599 14261 1 47 1
# 15: 2010 7599 16280 1 43 1
# 16: 2010 7599 16280 1 16 2
# 17: 2010 7599 16280 1 20 3
# 18: 2010 7599 16280 1 18 4
# 19: 2010 7599 16280 1 18 5
# 20: 2010 7599 17382 1 43 1
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