Question

数据集包含三个变量：id，sex和grade（factor）。

mydata <- data.frame(id=c(1,1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,4), sex=c(1,1,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,1),
                     grade=c("a","b","c","d","e", "x","y","y","x", "q","q","q","q", "a", "a", "a", NA, "b"))

对于每个ID，我需要查看我们拥有多少个唯一等级，然后创建一个新列（调用N）来记录成绩频率。例如，对于ID = 1，我们有＆＃34; grade＆＃34;的五个唯一值，因此N = 4;对于ID = 2，我们有＆＃34;等级＆＃34;的两个唯一值，所以N = 2;对于ID = 4，我们有两个独特的值＆＃34; grade＆＃34; （忽略NA），所以N = 2。

最终数据集是

mydata <- data.frame(id=c(1,1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,4), sex=c(1,1,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,1),
                     grade=c("a","b","c","d","e", "x","y","y","x", "q","q","q","q", "a", "a", "a", NA, "b"))
mydata$N <- c(5,5,5,5,5,2,2,2,2,1,1,1,1,2,2,2,2,2)

Answer 1

新答案：

uniqueN - data.table的函数有一个na.rm参数，我们可以按如下方式使用：

library(data.table)
setDT(mydata)[, n := uniqueN(grade, na.rm = TRUE), by = id]

给出：

> mydata
    id sex grade n
 1:  1   1     a 5
 2:  1   1     b 5
 3:  1   1     c 5
 4:  1   1     d 5
 5:  1   1     e 5
 6:  2   0     x 2
 7:  2   0     y 2
 8:  2   0     y 2
 9:  2   0     x 2
10:  3   0     q 1
11:  3   0     q 1
12:  3   0     q 1
13:  3   0     q 1
14:  4   1     a 2
15:  4   1     a 2
16:  4   1     a 2
17:  4   1    NA 2
18:  4   1     b 2

旧回答：

使用 data.table ，您可以按以下方式执行此操作：

library(data.table)
setDT(mydata)[, n := uniqueN(grade[!is.na(grade)]), by = id]

或：

setDT(mydata)[, n := uniqueN(na.omit(grade)), by = id]

Answer 2

您可以使用包data.table：

library(data.table)
setDT(mydata)

#I have removed NA's, up to you how to count them
mydata[,N_u:=length(unique(grade[!is.na(grade)])),by=id]

非常简短，可读且快速。它也可以在base-R中完成：

#lapply(split(grade,id),...: splits data into subsets by id
#unlist: creates one vector out of multiple vectors
#rep: makes sure each ID is repeated enough times

mydata$N <- unlist(lapply(split(mydata$grade,mydata$id),function(x){
  rep(length(unique(x[!is.na(x)])),length(x))
}
))

因为有关于什么更快的讨论，让我们做一些基准测试。

给定数据集：

> test1
Unit: milliseconds
          expr      min       lq     mean   median       uq       max neval cld
 length_unique 3.043186 3.161732 3.422327 3.286436 3.477854 10.627030   100   b
       uniqueN 2.481761 2.615190 2.763192 2.738354 2.872809  3.985393   100  a

更大的数据集：（10000个观测值，1000个id）

> test2
Unit: milliseconds
          expr      min       lq     mean   median       uq       max neval cld
 length_unique 11.84123 24.47122 37.09234 30.34923 47.55632  97.63648   100  a 
       uniqueN 25.83680 50.70009 73.78757 62.33655 97.33934 210.97743   100   b

Answer 3

使用dplyr::n_distinct及其na.rm参数的dplyr选项：

library(dplyr)
mydata %>% group_by(id) %>% mutate(N = n_distinct(grade, na.rm = TRUE))
#Source: local data frame [18 x 4]
#Groups: id [4]
#
#      id   sex  grade     N
#   (dbl) (dbl) (fctr) (int)
#1      1     1      a     5
#2      1     1      b     5
#3      1     1      c     5
#4      1     1      d     5
#5      1     1      e     5
#6      2     0      x     2
#7      2     0      y     2
#8      2     0      y     2
#9      2     0      x     2
#10     3     0      q     1
#11     3     0      q     1
#12     3     0      q     1
#13     3     0      q     1
#14     4     1      a     2
#15     4     1      a     2
#16     4     1      a     2
#17     4     1     NA     2
#18     4     1      b     2

Answer 4

看起来我们对data.table有多个投票，但你也可以使用基础R函数ave()：

mydata$N <- ave(as.character(mydata$grade),mydata$id,
                FUN = function(x) length(unique(x[!is.na(x)])))

Answer 5

使用tapply和查找表

mydata <- data.frame(id=c(1,1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,4), 
                     sex=c(1,1,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,1),
                     grade=c("a","b","c","d","e", "x","y","y","x", "q",
                             "q","q","q", "a", "a", "a", NA, "b"))
uniqN <- tapply(mydata$grade, mydata$id, function(x) sum(!is.na(unique(x))))
mydata$N <- uniqN[mydata$id]

Answer 6

这是一个dplyr方法。由于整洁的原因，我将摘要表分开。

library(dplyr)

summary =
  mydata %>%
  distinct(id, grade) %>%
  filter(grade %>% is.na %>% `!`) %>%
  count(id)

mydata %>%
  left_join(summary)

按ID的行数

6 个答案: