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?
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)
}
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)
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
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))
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.
# 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
There is also quick solution using the imputeTS package:
library(imputeTS)
na.mean(yourDataFrame)
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[]
A one-liner using tidyr's replace_na is
library(tidyr)
replace_na(mtcars,as.list(colMeans(mtcars,na.rm=T)))
Go simply with Zoo, it will simply replace all NA values with mean of the column values:
na.aggregate(data)
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
来源：https://stackoverflow.com/questions/25835643/replace-missing-values-with-column-mean