分组和计数以获得接近值

我想计算每country的次数status is open以及次数status is closed。然后计算closerate per country.

Data:

customer <- c(1,2,3,4,5,6,7,8,9)
country <- c('BE', 'NL', 'NL','NL','BE','NL','BE','BE','NL')
closeday <- c('2017-08-23', '2017-08-05', '2017-08-22', '2017-08-26', 
'2017-08-25', '2017-08-13', '2017-08-30', '2017-08-05', '2017-08-23')
closeday <- as.Date(closeday)

df <- data.frame(customer,country,closeday)

Adding status:

df$status <- ifelse(df$closeday < '2017-08-20', 'open', 'closed') 

  customer country   closeday status
1        1      BE 2017-08-23 closed
2        2      NL 2017-08-05   open
3        3      NL 2017-08-22 closed
4        4      NL 2017-08-26 closed
5        5      BE 2017-08-25 closed
6        6      NL 2017-08-13   open
7        7      BE 2017-08-30 closed
8        8      BE 2017-08-05   open
9        9      NL 2017-08-23 closed

计算closerate

closerate <- length(which(df$status == 'closed')) / 
(length(which(df$status == 'closed')) + length(which(df$status == 'open')))

[1] 0.6666667

显然，这就是closerate为总数。挑战在于获得closerate per country。我尝试添加closerate计算为df by:

df$closerate <- length(which(df$status == 'closed')) / 
(length(which(df$status == 'closed')) + length(which(df$status == 'open')))

但它给所有行一个closerate0.66，因为我没有分组。我相信我不应该使用长度函数，因为计数可以通过分组来完成。我读了一些有关使用的信息dplyr计算每组的逻辑输出，但这没有成功。

这是所需的输出：

aggregate(list(output = df$status == "closed"),
          list(country = df$country),
          function(x)
              c(close = sum(x),
                open = length(x) - sum(x),
                rate = mean(x)))
#  country output.close output.open output.rate
#1      BE         3.00        1.00        0.75
#2      NL         3.00        2.00        0.60

有一个解决方案使用table评论里好像已经被删除了。无论如何，你也可以使用table

output = as.data.frame.matrix(table(df$country, df$status))
output$closerate = output$closed/(output$closed + output$open)
output
#   closed open closerate
#BE      3    1      0.75
#NL      3    2      0.60