2

I would like to exchange all NA values in the columns for the respective medians

id <- c(1,2,3,4,5,6,7,8,9,10)
varA <- c(15,10,8,19,7,5,NA,11,12,NA)
varB <- c(NA,1,2,3,4,3,3,2,1,NA)
df <- data.frame(id, varA,varB)

median(df$varA, na.rm=TRUE)
median(df$varB, na.rm=TRUE)

df1 <- df

# Columns to be modified with Median in place of the NA

col <- c("varA", "varB")                           

df1[col] <- sapply(df1[col],  
                              function(x) replace(x, x %in% is.na(df1), median[col]))
df1

Error in [.default(df1, col) : invalid subscript type 'closure'

Ruam Pimentel
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Rodrigo Zambon
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    You're anonymous function `replace(x, x %in% is.na(df1), median[col]))` has all sorts of problems: Use `x` only, no `df1` or `col`. And you can't use `[` on functions, `median[col]` makes no sense and causes your error. Change it to ``replace(x, is.na(x), median(x, na.rm = TRUE))` and it has a chance of working. – Gregor Thomas Nov 27 '22 at 01:42

3 Answers3

2

We may use

library(zoo)
df[col] <-  na.aggregate(df[col], FUN = median)
akrun
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1

dplyr + tidyr solution

library(dplyr)
library(tidyr)

df %>% 
mutate(varA = replace_na(varA, median(varA, na.rm = TRUE)),
       varB = replace_na(varB, median(varB, na.rm = TRUE)))
Giulio Centorame
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0

Another option, which is similar to your original attempt.

df1[col] <- apply(df1[col], 2, \(x) ifelse(is.na(x), median(x, na.rm = TRUE), x) ) 
df1
#>    id varA varB
#> 1   1 15.0  2.5
#> 2   2 10.0  1.0
#> 3   3  8.0  2.0
#> 4   4 19.0  3.0
#> 5   5  7.0  4.0
#> 6   6  5.0  3.0
#> 7   7 10.5  3.0
#> 8   8 11.0  2.0
#> 9   9 12.0  1.0
#> 10 10 10.5  2.5
AndS.
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