Replace missing values with column mean

问题

I am not sure how to loop over each column to replace the NA values with the column mean. When I am trying to replace for one column using the following, it works well.

Column1[is.na(Column1)] <- round(mean(Column1, na.rm = TRUE))

The code for looping over columns is not working:

for(i in 1:ncol(data)){
    data[i][is.na(data[i])] <- round(mean(data[i], na.rm = TRUE))
}

the values are not replaced. Can someone please help me with this?

回答1:

A relatively simple modification of your code should solve the issue:

for(i in 1:ncol(data)){
  data[is.na(data[,i]), i] <- mean(data[,i], na.rm = TRUE)
}

回答2:

If DF is your data frame of numeric columns:

library(zoo)
na.aggregate(DF)

ADDED:

Using only the base of R define a function which does it for one column and then lapply to every column:

NA2mean <- function(x) replace(x, is.na(x), mean(x, na.rm = TRUE))
replace(DF, TRUE, lapply(DF, NA2mean))

The last line could be replaced with the following if it's OK to overwrite the input:

DF[] <- lapply(DF, NA2mean)

回答3:

To add to the alternatives, using @akrun's sample data, I would do the following:

d1[] <- lapply(d1, function(x) { 
  x[is.na(x)] <- mean(x, na.rm = TRUE)
  x
})
d1

回答4:

You could also try:

 cM <- colMeans(d1, na.rm=TRUE)
 indx <- which(is.na(d1), arr.ind=TRUE)
 d1[indx] <- cM[indx[,2]]
 d1

data

set.seed(42)
d1 <- as.data.frame(matrix(sample(c(NA,0:5), 5*10, replace=TRUE), ncol=10))

回答5:

lapply can be used instead of a for loop.

d1[] <- lapply(d1, function(x) ifelse(is.na(x), mean(x, na.rm = TRUE), x))

This doesn't really have any advantages over the for loop, though maybe it's easier if you have non-numeric columns as well, in which case

d1[sapply(d1, is.numeric)] <- lapply(d1[sapply(d1, is.numeric)], function(x) ifelse(is.na(x), mean(x, na.rm = TRUE), x))

is almost as easy.

回答6:

There is also quick solution using the imputeTS package:

library(imputeTS)
na_mean(yourDataFrame)

回答7:

# Lets say I have a dataframe , df as following -
df <- data.frame(a=c(2,3,4,NA,5,NA),b=c(1,2,3,4,NA,NA))

# create a custom function
fillNAwithMean <- function(x){
    na_index <- which(is.na(x))        
    mean_x <- mean(x, na.rm=T)
    x[na_index] <- mean_x
    return(x)
}

(df <- apply(df,2,fillNAwithMean))
   a   b
2.0 1.0
3.0 2.0
4.0 3.0
3.5 4.0
5.0 2.5
3.5 2.5

回答8:

Go simply with Zoo, it will simply replace all NA values with mean of the column values:

library(zoo)
na.aggregate(data)

回答9:

dplyr's mutate_all or mutate_at could be useful here:

library(dplyr)                                                             

set.seed(10)                                                               
df <- data.frame(a = sample(c(NA, 1:3)    , replace = TRUE, 10),           
                 b = sample(c(NA, 101:103), replace = TRUE, 10),                            
                 c = sample(c(NA, 201:203), replace = TRUE, 10))                            

df         

#>     a   b   c
#> 1   2 102 203
#> 2   1 102 202
#> 3   1  NA 203
#> 4   2 102 201
#> 5  NA 101 201
#> 6  NA 101 202
#> 7   1  NA 203
#> 8   1 101  NA
#> 9   2 101 203
#> 10  1 103 201

df %>% mutate_all(~ifelse(is.na(.x), mean(.x, na.rm = TRUE), .x))          

#>        a       b        c
#> 1  2.000 102.000 203.0000
#> 2  1.000 102.000 202.0000
#> 3  1.000 101.625 203.0000
#> 4  2.000 102.000 201.0000
#> 5  1.375 101.000 201.0000
#> 6  1.375 101.000 202.0000
#> 7  1.000 101.625 203.0000
#> 8  1.000 101.000 202.1111
#> 9  2.000 101.000 203.0000
#> 10 1.000 103.000 201.0000

df %>% mutate_at(vars(a, b),~ifelse(is.na(.x), mean(.x, na.rm = TRUE), .x))

#>        a       b   c
#> 1  2.000 102.000 203
#> 2  1.000 102.000 202
#> 3  1.000 101.625 203
#> 4  2.000 102.000 201
#> 5  1.375 101.000 201
#> 6  1.375 101.000 202
#> 7  1.000 101.625 203
#> 8  1.000 101.000  NA
#> 9  2.000 101.000 203
#> 10 1.000 103.000 201

回答10:

A one-liner using tidyr's replace_na is

library(tidyr)
replace_na(mtcars,as.list(colMeans(mtcars,na.rm=T)))

回答11:

Similar to the answer pointed out by @Thomas, This can also be done using ifelse() method of R:

for(i in 1:ncol(data)){
  data[,i]=ifelse(is.na(data[,i]),
                  ave(data[,i],FUN=function(y) mean(y, na.rm = TRUE)),
                  data[,i])
}

where, Arguments to ifelse(TEST, YES , NO) are:-

TEST- logical condition to be checked

YES- executed if the condition is True

NO- else when the condition is False

and ave(x, ..., FUN = mean) is method in R used for calculating averages of subsets of x[]

来源：https://stackoverflow.com/questions/25835643/replace-missing-values-with-column-mean

标签

missing-data

imputation