计算 r 中的独特因素

Question

我想知道在记录的每个出生日期出生的独特水坝的数量。我的数据框与此类似：

dam <- c("2A11","2A11","2A12","2A12","2A12","4D23","4D23","1X23")
bdate <- c("2009-10-01","2009-10-01","2009-10-01","2009-10-01",
           "2009-10-01","2009-10-03","2009-10-03","2009-10-03")
mydf <- data.frame(dam,bdate)
mydf
#    dam      bdate
# 1 2A11 2009-10-01
# 2 2A11 2009-10-01
# 3 2A12 2009-10-01
# 4 2A12 2009-10-01
# 5 2A12 2009-10-01
# 6 4D23 2009-10-03
# 7 4D23 2009-10-03
# 8 1X23 2009-10-03

我使用了aggregate(dam ~ bdate, data=mydf, FUN=length)，但它计算了在特定日期出生的所有水坝

bdate dam
1 2009-10-01   5
2 2009-10-03   3

，相反，我需要有这样的事情：

mydf2
  bdate      dam
1 2009-10-01  2
2 2009-10-03  2

非常感谢您的帮助！

Answer 1

怎么样：

aggregate(dam ~ bdate, data=mydf, FUN=function(x) length(unique(x)))

Answer 2

您还可以先对数据运行 unique：

aggregate(dam ~ bdate, data=unique(mydf[c("dam","date")]), FUN=length)

然后您也可以使用 table 而不是 aggregate，尽管输出略有不同。

> table(unique(mydf[c("dam","date")])$bdate)

2009-10-01 2009-10-03 
         2          2 

Answer 3

这只是如何思考问题以及如何解决问题的方法之一的示例。

split.mydf <- with(mydf, split(x = mydf, f = bdate)) #each list element has only one date.
# it's just a matter of counting unique dams
unique.mydf <- lapply(X = split.mydf, FUN = unique)
#and then count the number of unique elements
unilen.mydf <- lapply(unique.mydf, length)
#you can do these two last steps in one go like so
lapply(split.mydf, FUN = function(x) length(unique(x)))

as.data.frame(unlist(unilen.mydf)) #data.frame is just a special list, so this is water to your mill

           unlist(unilen.mydf)
2009-10-01                   2
2009-10-03                   2

Answer 4

在dplyr中，您可以使用n_distinct：

library(tidyverse)
mydf %>%
  group_by(bdate) %>%
  summarize(dam = n_distinct(dam))