I have the following example table where I am required to find the median age of a herd of animals. Not only does it have a 0
, it is also has a grouped frequency of animals for a given age.
library(tidyverse)
a<-data.frame(Age=c(0,1,2,3,4,5,6,7,8,9),
Individuals=c(3655,2535,898,235,559,265,258,3659,7895,3655))
a%>%summarise(Age=as.numeric(Age),
Median=sort(as.numeric(Age)*Persons/sum(Individuals)))
I understand that the standard median()
option does not work. I tried to be clever and attempted something like: median(rep(a$Age, a$Individuals))
, but the memory consumption was too much. Besides, I think it will fail with a larger dataset.
CodePudding user response:
You can uncount
the original data frame and then use the standard median
function.
a %>% uncount(Individuals) %>% summarise(Median=median(Age))
Median
1 7
And to check:
> sum(a$Individuals)/2
[1] 11807
> sum(a$Individuals[1:7])
[1] 8405
> sum(a$Individuals[1:8])
[1] 12064
All good.
CodePudding user response:
You could be abit clever and do:
a %>%
arrange(Age) %>%
summarise(median = Age[findInterval(sum(Individuals)/2, cumsum(Individuals)) 1])
median
1 7