calculate the proportion in r-CodePudding

id=1:10
age=c(100,7,23,66,34,67,45,50,99,7)
data=data.frame(id,age)
data$age_stage=cut(data$age,breaks=c(0,50,100))

  id age age_stage
1   1 100  (50,100]
2   2   7    (0,50]
3   3  23    (0,50]
4   4  66  (50,100]
5   5  34    (0,50]
6   6  67  (50,100]
7   7  45    (0,50]
8   8  50    (0,50]
9   9  99  (50,100]
10 10   7    (0,50]

I want to calculate the proportion of people who are older than 50. How should I do this?

CodePudding user response：

You can check proportion using prop.table.

prop.table(table(data$age_stage))

  (0,50] (50,100] 
     0.6      0.4

CodePudding user response：

Dplyr solution:


library(tidyverse)

id=1:10
age=c(100,7,23,66,34,67,45,50,99,7)
data=data.frame(id,age)
data$age_stage=cut(data$age,breaks=c(0,50,100))


data %>% # getting data 
  group_by(age_stage) %>% # grouping by age_stage
  summarise(cnt = n()) %>% # counting how many by group of age_stage
  mutate(props = round(cnt / sum(cnt), 2)) # getting frequency


#> # A tibble: 2 x 3
#>   age_stage   cnt props
#>   <fct>     <int> <dbl>
#> 1 (0,50]        6   0.6
#> 2 (50,100]      4   0.4

CodePudding user response：

#I will prefer to use a function that I can reuse or just use lapply if #I change my data.

older_than_50 <- function(data){
             proportion <- mean(age > 50)
           
           return (proportion)
             }

older_than_50(data)