Replace Missing Values by Column Mean in R DataFrame
Last Updated :
21 Dec, 2023
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
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
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
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
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
mean_val <- colMeans (data,na.rm = TRUE )
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
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
all_column_mean <- apply (data, 2, mean, na.rm= TRUE )
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.packages ( "zoo" )
library ( "zoo" )
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
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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