Question

我想按country计算status open的次数和status为closed的次数。然后根据closerate计算country。

数据：

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)

添加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，这是总数的closerate。面临的挑战是每closerate获得country。我尝试将closerate计算添加到df：

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

但它给所有行closerate 0.66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。我阅读了一些有关使用dplyr来计算每组逻辑输出的信息，但这并没有成功。

这是所需的输出：

Grouped by country

Answer 1

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

在评论中使用table的解决方案似乎已被删除。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

Answer 2

您可以使用tapply：

data.frame(open=tapply(df$status=="open", df$country, sum),
           closed=tapply(df$status=="closed", df$country, sum)
           closerate=tapply(df$status=="closed", df$country, mean))`

Answer 3

data.table方法。

library(data.table)
setDT(df)[, {temp <- status=="closed"; # store temporary logical variable
            .(closed=sum(temp), open=sum(!temp), closeRate=mean(temp))}, # calculate stuff
          by=country] # by country

返回

   country closed open closeRate
1:      BE      3    1      0.75
2:      NL      3    2      0.60

Answer 4

这是一个A * B = [[30, 45], [10, 30]]解决方案。

dplyr

...返回

output <- df %>%
  count(country, status) %>%
  group_by(country) %>%
  mutate(total = sum(n)) %>%
  mutate(percent = n/total)

Answer 5

这是tidyverse的快速解决方案：

library(dplyr)
df %>% group_by(country) %>% 
  mutate(status =ifelse(closeday < '2017-08-20', 'open', 'closed'),
         closerate=mean(status=="closed"))

返回：

# A tibble: 9 x 5
# Groups:   country [2]
  customer country   closeday status closerate
     <dbl>  <fctr>     <date>  <chr>     <dbl>
1        1      BE 2017-08-23 closed      0.75
2        2      NL 2017-08-05   open      0.60
3        3      NL 2017-08-22 closed      0.60
4        4      NL 2017-08-26 closed      0.60
5        5      BE 2017-08-25 closed      0.75
6        6      NL 2017-08-13   open      0.60
7        7      BE 2017-08-30 closed      0.75
8        8      BE 2017-08-05   open      0.75
9        9      NL 2017-08-23 closed      0.60

这里我将TRUE / FALSE的向量放入mean()函数时将逻辑强制转换为整数。

或者，使用data.table：

library(data.table)
setDT(df)[,status:=ifelse(closeday < '2017-08-20', 'open', 'closed')]
df[, .(closerate=mean(status=="closed")), by=country]

分组和计数以获得贴近

5 个答案: