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Replace Missing Values by Column Mean in R DataFrame

Last Updated : 21 Dec, 2023
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In this article, we are going to see how to replace missing values with columns mean in R Programming Language. Missing values in a dataset are usually represented as NaN or NA. Such values must be replaced with another value or removed. This process of replacing another value in place of missing data is known as Data Imputation

Creating data frame with missing values

R




# creating a dataframe
data <- data.frame(marks1 = c(NA, 22, NA, 49, 75),
                   marks2 = c(81, 14, NA, 61, 12),
                   marks3 = c(78.5, 19.325, NA, 28, 48.002))
data


Output:

  marks1 marks2 marks3
1 NA 81 78.500
2 22 14 19.325
3 NA NA NA
4 49 61 28.000
5 75 12 48.002

Replace columns using mean() function

Let’s see how to impute missing values with each column’s mean using a dataframe and mean( ) function. mean() function is used to calculate the arithmetic mean of the elements of the numeric vector passed to it as an argument.

Syntax of mean() : mean(x, trim = 0, na.rm = TRUE, …)

Arguments:

  • x – any object
  • trim – observations to be trimmed from each end of x before the mean is computed
  • na.rm – TRUE to remove NA values

Replacing NA for all columns using mean( ) function

R




data$marks2[is.na(data$marks2)]<-mean(data$marks2,na.rm=TRUE)
 
data


Output:

  marks1 marks2 marks3
1 NA 81 78.500
2 22 14 19.325
3 NA 42 NA
4 49 61 28.000
5 75 12 48.002

In this code we fill the missing values of marks2 column with mean value.

Replacing Missing Data in all columns Using for-Loop

With the help of For loops in R we will Replacing Missing Data in all columns.

R




# replacing NA with each column's mean
for(i in colnames(data))
    data[,i][is.na(data[,i])] <- a[,i]
data


Output:

    marks1 marks2   marks3
1 48.66667 81 78.50000
2 22.00000 14 19.32500
3 48.66667 42 43.45675
4 49.00000 61 28.00000
5 75.00000 12 48.00200

Replace column using colMeans() function

colMeans() function is used to compute the mean of each column of a matrix or array

Syntax of colMeans() : colMeans(x, na.rm = FALSE, dims = 1 …)

Arguments:

  • x: object
  • dims: dimensions are regarded as ‘columns’ to sum over
  • na.rm: TRUE to ignore NA values

Here we are going to use colMeans function to replace the NA in columns.

R




# creating a dataframe
data <- data.frame(marks1 = c(NA, 22, NA, 49, 75),
                   marks2 = c(81, 14, NA, 61, 12),
                   marks3 = c(78.5, 19.325, NA, 28, 48.002))
data
# using colMeans()
mean_val <- colMeans(data,na.rm = TRUE)
 
# replacing NA with mean value of each column
for(i in colnames(data))
  data[,i][is.na(data[,i])] <- mean_val[i]
data


Output :

  marks1 marks2 marks3
1 NA 81 78.500
2 22 14 19.325
3 NA NA NA
4 49 61 28.000
5 75 12 48.002

data

marks1 marks2 marks3
1 48.66667 81 78.50000
2 22.00000 14 19.32500
3 48.66667 42 43.45675
4 49.00000 61 28.00000
5 75.00000 12 48.00200

Replacing NA using apply() function

In this method, we will use apply() function to replace the NA from the columns.

Syntax of apply() : apply(X, MARGIN, FUN, …)

Arguments:

  • X – an array, including a matrix
  • MARGIN – a vector
  • FUN – the function to be applied

R




# creating a dataframe
data <- data.frame(marks1 = c(NA, 22, NA, 49, 75),
                marks2 = c(81, 14, NA, 61, 12),
                marks3 = c(78.5, 19.325, NA, 28, 48.002))
data
 
# computing mean of all columns using apply()
all_column_mean <- apply(data, 2, mean, na.rm=TRUE)
 
# imputing NA with the mean calculated
for(i in colnames(data))
  data[,i][is.na(data[,i])] <- all_column_mean[i]
data


Output :

  marks1 marks2 marks3
1 NA 81 78.500
2 22 14 19.325
3 NA NA NA
4 49 61 28.000
5 75 12 48.002

data

marks1 marks2 marks3
1 48.66667 81 78.50000
2 22.00000 14 19.32500
3 48.66667 42 43.45675
4 49.00000 61 28.00000
5 75.00000 12 48.00200

Using na.aggregate() Function of zoo Package

We can also replace the missing values using na.aggregate Function of zoo Package in R.

R




# Install & load zoo package
install.packages("zoo")                                  
library("zoo")
 
# creating a dataframe
data <- data.frame(marks1 = c(NA, 22, NA, 49, 75),
                   marks2 = c(81, 14, NA, 61, 12),
                   marks3 = c(78.5, 19.325, NA, 28, 48.002))
data
# using na.aggregate function to replace missing values
data<- na.aggregate(data)  
 
data


Output:

  marks1 marks2 marks3
1 NA 81 78.500
2 22 14 19.325
3 NA NA NA
4 49 61 28.000
5 75 12 48.002
marks1 marks2 marks3
1 48.66667 81 78.50000
2 22.00000 14 19.32500
3 48.66667 42 43.45675
4 49.00000 61 28.00000
5 75.00000 12 48.00200



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