如何从不同的数据框中获取某些列的平均值？

Question

Problem

I want to use a function to get the mean numbers from different dataframes.

Example dataframes

WEEK.1 <- 1:52
DIF.4 <- runif(52, 0, 20)
df <- as.data.frame(cbind(WEEK.1, DIF.4))

WEEK.2 <- 1:52
DIF.5 <- runif(52, 0, 20)
df2 <- as.data.frame(cbind(WEEK.2, DIF.5))

WEEK.3 <- 1:52
DIF.6 <- runif(52, 0, 20)
df3 <- as.data.frame(cbind(WEEK.3, DIF.6))

Attempt

a) Without a function, I'd have to repeat the following three times:

colnames(df) <- str_replace_all(colnames(df), "\\.\\d", "")
df_ <- filter(df, WEEK >= 26)
mean(df_$DIF) 

b) My attempt to do so with a function:

meanChange <- function(x) {
  colnames(x) <- str_replace_all(colnames(x), "\\.\\d", "")
  x_ <- filter(x, WEEK >= 26)
  mean(x_$DIF) 
}

x <- c("df", "df2", "df3")
changes <- as.data.frame(lapply(x, meanChange))

Result

Error in `colnames<-`(`*tmp*`, value = character(0)) : 
  attempt to set 'colnames' on an object with less than two dimensions
Called from: `colnames<-`(`*tmp*`, value = character(0))
Browse[1]> Q

Help would be appreciated.

Answer 1

This is a possibile solution:

# create a list of your dataframes
dfl <- list(df, df2, df3)

# apply mean function 
lapply(1:3, function(i) mean(dfl[[i]][[2]][which(dfl[[i]][[1]] >= 26)]))

Index i selects the single dataframe into list dfl . [[1]] and [[2]] select respectively column 1 and column 2 of each dataframe.

---

Another possible solution is:

lapply(dfl, function(x) mean(x[[2]][which(x[[1]] >= 26)]))

Here lapply works directly with dfl without index.

---

NOTE : both solutions return a list as a result. If you want a vector use sapply rather than lapply

Answer 2

• You are passing y as an argument to the function but using x inside.
• For the last line I think you mean mean(x_$DIF)

library(tidyverse)

meanChange <- function(x) {
  colnames(x) <- str_replace_all(colnames(x), "\\.\\d", "")
  x_ <- filter(x, WEEK >= 26)
  mean(x_$DIF) 
}

• Put the dataframes in a list and use sapply to apply the function.

y <- list(df, df2, df3)
sapply(y, meanChange)