Question

我在r

中有以下数据框

   Service      Container_Pick_Day
    ABC              0
    ABC              1
    ABC              1
    ABC              2
    ABC              NA
    ABC              0
    ABC              1
    DEF              NA
    DEF              0
    DEF              1
    DEF              1
    DEF              1
    DEF              2
    DEF              1

列Container_Pick_Day是数字，由NA值组成。我想要做的是计算Service在0th day,after 1 day,2 day and so on上忽略NA值而选择的容器的明智百分比

所需的数据框将是

 Service      Container_Pick_Day      Percentage
    ABC              0                (2/6)*100 = 33.33 
    ABC              1                (3/6)*100 = 50  
    ABC              2                (1/6)*100 = 16.67
    DEF              0                (1/6)*100 = 16.67
    DEF              1                (3/6)*100 = 50
    DEF              2                (1/6)*100 = 16.67

我在R中跟随，但它在输出中生成NA值

  df%>% 
     group_by(Service) %>%
     summarise(pick_day_perc = n()/sum(Container_Pick_Day),na.rm=T) %>% 
     as.data.frame()

我是否必须按Service and Container_Pick_Day分组？

Answer 1

根据@nicola，@ karun和我自己提供的上述所有评论添加答案

library(dplyr)

#nicola
df %>% 
 filter(!is.na(Container_Pick_Day)) %>% 
 group_by(Service,Container_Pick_Day) %>% 
 summarise(Percentage=n()) %>% 
 group_by(Service) %>% 
 mutate(Percentage=Percentage/sum(Percentage)*100)

#akrun
df %>% 
 filter(complete.cases(Container_Pick_Day)) %>% 
 count(Service, Container_Pick_Day) %>% 
 group_by(Service) %>% 
 transmute(Container_Pick_Day, Percentage=n/sum(n)*100)

#Sotos
df %>% 
 na.omit() %>% 
 group_by_all() %>% 
 summarise(ptg = n()) %>% 
 group_by(Service) %>% 
 mutate(ptg = prop.table(ptg)*100)

所有结果，

Service Container_Pick_Day Percentage
   <fctr>              <int>      <dbl>
1     ABC                  0   33.33333
2     ABC                  1   50.00000
3     ABC                  2   16.66667
4     DEF                  0   16.66667
5     DEF                  1   66.66667
6     DEF                  2   16.66667

在dplyr中分组并计算百分比

1 个答案: