R 中的函数：如何在同一 Function 内返回均值、中值、标准差

Question

How can I return the mean, median, and standard deviation within same function in R? All that I can get to return is the last part of the function which calculated the standard deviation. I was thinking that by assigning summarystat(Tail_wags) to b that when I returned 'b' that I would have all three value. Added the result for the three values I need outside of the function after 'b' to see what values are supposed to be.

Dog_biscuits <- c(0,1,2,3,4,5,6,7,8,9,10) 
Tail_wags <- c(0,0,1,3,8,13,14,12,15,16,14)
dog_wags<-cbind(Dog_biscuits,Tail_wags)
dog_wags

summarystat<- function(x) {
  z1 <- mean(x)
  z2<-median(x)
  z3<-sd(x)
}
b<-summarystat(Tail_wags)
b

b
[1] 6.497552
> mean(Tail_wags)
[1] 8.727273
> median(Tail_wags)
[1] 12
> sd(Tail_wags)
[1] 6.497552

Answer 1

You can only return one object from a function. The trick to achieve what you want is to return a list:

summarystat<- function(x) {
  z1 <- mean(x)
  z2 <- median(x)
  z3 <- sd(x)
  return(list(mean=z1, median=z2, sd=z3))
}

Answer 2

You can combine and return the variables using the generic c() function.

summarystat<- function(x) {
  z1 <- mean(x, na.rm = TRUE)
  z2<-median(x, na.rm = TRUE)
  z3<-sd(x,na.rm = TRUE)

  return(c(mean=z1,median=z2,standard_dev=z3))
}
Tail_wags <- c(0,0,1,3,8,13,14,12,15,16,14)
summarystat(Tail_wags)
#     mean       median standard_dev 
# 8.727273    12.000000     6.497552 

Answer 3

Your are looking after something like:

summarystat <- function(x) {
  my_list <- list("mean" = mean(x), "median" = median(x), "sd" = sd(x))
  return(my_list) 
}

Usage:

vals <- summarystat(Tail_wags)
> a$mean
> a$sd
> a$median

Answer 4

Function and application:

do.call("rbind", lapply(dog_wags, function(x){

      list(mean_val = mean(x),

      median_val = median(x),

      sd_val = sd(x))

    }

  )

)

Data:

Dog_biscuits <- c(0,1,2,3,4,5,6,7,8,9,10) 
Tail_wags <- c(0,0,1,3,8,13,14,12,15,16,14)
dog_wags <- data.frame(cbind(Dog_biscuits,Tail_wags))

Answer 5

Alternatively, you can get rid of the function completely and use something like pastecs::stat.desc and then subtract the values you want

Dog_biscuits <- c(0,1,2,3,4,5,6,7,8,9,10) 
Tail_wags <- c(0,0,1,3,8,13,14,12,15,16,14)
dog_wags<-cbind(Dog_biscuits,Tail_wags)

pastecs::stat.desc(Tail_wags)[["mean"]]
# 8.727273

Check out this article for more summary functions.

Answer 6

A somewhat different approach that lets one choose the functions to return.

---

Code:

fooapply <- function(x, functions = c("mean", "median", "sd"), na.rm = T){
  func <- functions
  vec <- c()
  for(i in 1:length(func)){
    if(na.rm == T){
      eval(parse(text = paste0("vec[", i,"]", "<-", func[i], "(x, na.rm = T)")))
    } 
    else{
      eval(parse(text = paste0("vec[", i,"]", "<-", func[i], "(x)")))
    }
  }
  names(vec) <- functions
  return(vec)
}

Result

To obtain your desired result you can just your vector into the function. Per default, the function will omit NA's and calculate the mean, median and sd.

fooapply(Tail_wags)

    mean    median        sd 
 8.727273 12.000000  6.497552 

Additionally, one can also add or remove functions or swap them out:

fooapply(Tail_wags, c("mean", "median", "IQR"))

Note that some of the included functions will report an error when NA's are included without specifying na.rm = T , others will just report NA as result. The mean() function, for example, will return NA when calculated for a vector that includes NA. In contrast, IQR() will throw an error when NA's are included within the vector and hence requires the na.rm = T (which is set as TRUE by default) statement in order for fooapply() to work.