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- Numbering rows within groups in a data frame 5 answers
- 在数据框5中为组中的行编号
How can we generate unique id numbers within each group of a dataframe? Here's some data grouped by "personid":
如何在每个dataframe组中生成唯一的id号?下面是一些由“personid”分组的数据:
personid date measurement
1 x 23
1 x 32
2 y 21
3 x 23
3 z 23
3 y 23
I wish to add an id column with a unique value for each row within each subset defined by "personid", always starting with 1. This is my desired output:
我希望在由“personid”定义的每个子集中为每一行添加一个id列，其值都是唯一的，总是从1开始。这是我想要的输出:
personid date measurement id
1 x 23 1
1 x 32 2
2 y 21 1
3 x 23 1
3 z 23 2
3 y 23 3
I appreciate any help.
我很感谢任何帮助。
6 个解决方案
#1
24
The misleadingly named ave() function, with argument FUN=seq_along, will accomplish this nicely -- even if your personid column is not strictly ordered.
带有参数FUN=seq_along的命名错误的ave()函数将很好地实现这一点——即使您的personid列没有严格排序。
df <- read.table(text = "personid date measurement
1 x 23
1 x 32
2 y 21
3 x 23
3 z 23
3 y 23", header=TRUE)
## First with your data.frame
ave(df$personid, df$personid, FUN=seq_along)
# [1] 1 2 1 1 2 3
## Then with another, in which personid is *not* in order
df2 <- df[c(2:6, 1),]
ave(df2$personid, df2$personid, FUN=seq_along)
# [1] 1 1 1 2 3 2
#2
18
Some dplyr alternatives, using convenience functions row_number and n.
一些dplyr替代方案，使用方便函数row_number和n。
library(dplyr)
df %>% group_by(personid) %>% mutate(id = row_number())
df %>% group_by(personid) %>% mutate(id = 1:n())
df %>% group_by(personid) %>% mutate(id = seq_len(n()))
df %>% group_by(personid) %>% mutate(id = seq_along(personid))
You may also use getanID from package splitstackshape. Note that the input dataset is returned as a data.table.
您也可以使用getanID从包splitstackshape。注意，输入数据集作为数据表返回。
getanID(data = df, id.vars = "personid")
# personid date measurement .id
# 1: 1 x 23 1
# 2: 1 x 32 2
# 3: 2 y 21 1
# 4: 3 x 23 1
# 5: 3 z 23 2
# 6: 3 y 23 3
#3
12
Using data.table, and assuming you wish to order by date within the personid subset
使用数据。表，并假设您希望在personid子集内按日期排序
library(data.table)
DT <- data.table(Data)
DT[,id := order(date), by = personid]
## personid date measurement id
## 1: 1 x 23 1
## 2: 1 x 32 2
## 3: 2 y 21 1
## 4: 3 x 23 1
## 5: 3 z 23 3
## 6: 3 y 23 2
If you wish do not wish to order by date
如果您不希望按日期订货的话
DT[, id := 1:.N, by = personid]
## personid date measurement id
## 1: 1 x 23 1
## 2: 1 x 32 2
## 3: 2 y 21 1
## 4: 3 x 23 1
## 5: 3 z 23 2
## 6: 3 y 23 3
Any of the following would also work
以下任何一种都可以
DT[, id := seq_along(measurement), by = personid]
DT[, id := seq_along(date), by = personid]
The equivalent commands using plyr
使用plyr的等效命令
library(plyr)
# ordering by date
ddply(Data, .(personid), mutate, id = order(date))
# in original order
ddply(Data, .(personid), mutate, id = seq_along(date))
ddply(Data, .(personid), mutate, id = seq_along(measurement))
#4
7
I think there's a canned command for this, but I can't remember it. So here's one way:
我想这是一个固定的命令，但我记不得了。这是一个方法:
> test <- sample(letters[1:3],10,replace=TRUE)
> cumsum(duplicated(test))
[1] 0 0 1 1 2 3 4 5 6 7
> cumsum(duplicated(test))+1
[1] 1 1 2 2 3 4 5 6 7 8
This works because duplicated returns a logical vector. cumsum evalues numeric vectors, so the logical gets coerced to numeric.
这是因为重复返回一个逻辑向量。cumsum表示数值向量，因此逻辑被强制为数值。
You can store the result to your data.frame as a new column if you want:
您可以将结果存储到您的数据中。
dat$id <- cumsum(duplicated(test))+1
#5
5
Assuming your data are in a data.frame named Data, this will do the trick:
假设您的数据位于一个名为data的数据框架中，这将实现以下功能:
# ensure Data is in the correct order
Data <- Data[order(Data$personid),]
# tabulate() calculates the number of each personid
# sequence() creates a n-length vector for each element in the input,
# and concatenates the result
Data$id <- sequence(tabulate(Data$personid))
#6
2
You can use sqldf
您可以使用sqldf
df<-read.table(header=T,text="personid date measurement
1 x 23
1 x 32
2 y 21
3 x 23
3 z 23
3 y 23")
library(sqldf)
sqldf("SELECT a.*, COUNT(*) count
FROM df a, df b
WHERE a.personid = b.personid AND b.ROWID <= a.ROWID
GROUP BY a.ROWID"
)
# personid date measurement count
#1 1 x 23 1
#2 1 x 32 2
#3 2 y 21 1
#4 3 x 23 1
#5 3 z 23 2
#6 3 y 23 3
#1
24
The misleadingly named ave() function, with argument FUN=seq_along, will accomplish this nicely -- even if your personid column is not strictly ordered.
带有参数FUN=seq_along的命名错误的ave()函数将很好地实现这一点——即使您的personid列没有严格排序。
df <- read.table(text = "personid date measurement
1 x 23
1 x 32
2 y 21
3 x 23
3 z 23
3 y 23", header=TRUE)
## First with your data.frame
ave(df$personid, df$personid, FUN=seq_along)
# [1] 1 2 1 1 2 3
## Then with another, in which personid is *not* in order
df2 <- df[c(2:6, 1),]
ave(df2$personid, df2$personid, FUN=seq_along)
# [1] 1 1 1 2 3 2
#2
18
Some dplyr alternatives, using convenience functions row_number and n.
一些dplyr替代方案，使用方便函数row_number和n。
library(dplyr)
df %>% group_by(personid) %>% mutate(id = row_number())
df %>% group_by(personid) %>% mutate(id = 1:n())
df %>% group_by(personid) %>% mutate(id = seq_len(n()))
df %>% group_by(personid) %>% mutate(id = seq_along(personid))
You may also use getanID from package splitstackshape. Note that the input dataset is returned as a data.table.
您也可以使用getanID从包splitstackshape。注意，输入数据集作为数据表返回。
getanID(data = df, id.vars = "personid")
# personid date measurement .id
# 1: 1 x 23 1
# 2: 1 x 32 2
# 3: 2 y 21 1
# 4: 3 x 23 1
# 5: 3 z 23 2
# 6: 3 y 23 3
#3
12
Using data.table, and assuming you wish to order by date within the personid subset
使用数据。表，并假设您希望在personid子集内按日期排序
library(data.table)
DT <- data.table(Data)
DT[,id := order(date), by = personid]
## personid date measurement id
## 1: 1 x 23 1
## 2: 1 x 32 2
## 3: 2 y 21 1
## 4: 3 x 23 1
## 5: 3 z 23 3
## 6: 3 y 23 2
If you wish do not wish to order by date
如果您不希望按日期订货的话
DT[, id := 1:.N, by = personid]
## personid date measurement id
## 1: 1 x 23 1
## 2: 1 x 32 2
## 3: 2 y 21 1
## 4: 3 x 23 1
## 5: 3 z 23 2
## 6: 3 y 23 3
Any of the following would also work
以下任何一种都可以
DT[, id := seq_along(measurement), by = personid]
DT[, id := seq_along(date), by = personid]
The equivalent commands using plyr
使用plyr的等效命令
library(plyr)
# ordering by date
ddply(Data, .(personid), mutate, id = order(date))
# in original order
ddply(Data, .(personid), mutate, id = seq_along(date))
ddply(Data, .(personid), mutate, id = seq_along(measurement))
#4
7
I think there's a canned command for this, but I can't remember it. So here's one way:
我想这是一个固定的命令，但我记不得了。这是一个方法:
> test <- sample(letters[1:3],10,replace=TRUE)
> cumsum(duplicated(test))
[1] 0 0 1 1 2 3 4 5 6 7
> cumsum(duplicated(test))+1
[1] 1 1 2 2 3 4 5 6 7 8
This works because duplicated returns a logical vector. cumsum evalues numeric vectors, so the logical gets coerced to numeric.
这是因为重复返回一个逻辑向量。cumsum表示数值向量，因此逻辑被强制为数值。
You can store the result to your data.frame as a new column if you want:
您可以将结果存储到您的数据中。
dat$id <- cumsum(duplicated(test))+1
#5
5
Assuming your data are in a data.frame named Data, this will do the trick:
假设您的数据位于一个名为data的数据框架中，这将实现以下功能:
# ensure Data is in the correct order
Data <- Data[order(Data$personid),]
# tabulate() calculates the number of each personid
# sequence() creates a n-length vector for each element in the input,
# and concatenates the result
Data$id <- sequence(tabulate(Data$personid))
#6
2
You can use sqldf
您可以使用sqldf
df<-read.table(header=T,text="personid date measurement
1 x 23
1 x 32
2 y 21
3 x 23
3 z 23
3 y 23")
library(sqldf)
sqldf("SELECT a.*, COUNT(*) count
FROM df a, df b
WHERE a.personid = b.personid AND b.ROWID <= a.ROWID
GROUP BY a.ROWID"
)
# personid date measurement count
#1 1 x 23 1
#2 1 x 32 2
#3 2 y 21 1
#4 3 x 23 1
#5 3 z 23 2
#6 3 y 23 3