如何按组计算唯一值的数量？ [重复]

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Counting unique / distinct values by group in a data frame 10 answers
在数据框中按组计算唯一/不同的值10个答案

ID= c('A', 'A', 'A', 'B', 'B', 'B')
color=c('white', 'green', 'orange', 'white', 'green', 'green')

d = data.frame (ID, color)

My desired result is

我想要的结果是

unique_colors=c(3,3,3,2,2,2)
d = data.frame (ID, color, unique_colors)

or more clear in a new dataframe c

或者在新的数据框架中更清楚c

ID= c('A','B')
unique_colors=c(3,2)
c = data.frame (ID,unique_colors)

I've tried different combinations of aggregate and ave as well as by and with and I suppose it is a combination of those functions.

我尝试了不同的聚合和ave以及by和with的组合，我想它是这些函数的组合。

The solution would include:

解决方案包括：

length(unique(d$color))

to calculate the number of unique elements.

计算唯一元素的数量。

1 个解决方案

#1

I think you've got it all wrong here. There is no need neither in plyr or <- when using data.table.

我想你在这里弄错了。在使用data.table时，plyr或< - 都不需要。

Recent versions of data.table, v >= 1.9.6, have a new function uniqueN() just for that.

data.table的最新版本，v> = 1.9.6，只有一个新函数uniqueN（）。

library(data.table) ## >= v1.9.6
setDT(d)[, .(count = uniqueN(color)), by = ID]
#    ID count
# 1:  A     3
# 2:  B     2

If you want to create a new column with the counts, use the := operator

如果要使用计数创建新列，请使用：=运算符

setDT(d)[, count := uniqueN(color), by = ID]

Or with dplyr use the n_distinct function

或者使用dplyr使用n_distinct函数

library(dplyr)
d %>%
  group_by(ID) %>%
  summarise(count = n_distinct(color))
# Source: local data table [2 x 2]
# 
#   ID count
# 1  A     3
# 2  B     2

Or (if you want a new column) use mutate instead of summary

或者（如果你想要一个新列）使用mutate而不是summary

d %>%
  group_by(ID) %>%
  mutate(count = n_distinct(color))

#1