How to replace missing values with linear interpolation method in an R vector?

The linear interpolation is a method of fitting a curve using linear polynomials and it helps us to create a new data points but these points lie within the range of the original values for which the linear interpolation is done. Sometimes these values may go a little far from the original values but not too far. In R, if we have some missing values then na.approx function of zoo package can be used to replace the NA with linear interpolation method.

Example1

Loading zoo package:

Live Demo

> library(zoo)
> x1<-sample(c(NA,2,5),10,replace=TRUE)
> x1

Output

[1] 2 2 2 5 2 2 5 NA 2 5

Replacing NA with linear interpolation:

Example

> na.approx(x1)

Output

[1] 2.0 2.0 2.0 5.0 2.0 2.0 5.0 3.5 2.0 5.0

Example2

Live Demo

> x2<-sample(c(NA,1:4),150,replace=TRUE)
> x2

Output

[1] 2 NA NA 2 1 1 NA 2 4 NA 1 2 1 4 3 3 1 3 1 4 4 2 3 1 3
[26] 1 4 2 4 2 1 2 1 3 NA 2 NA 3 1 2 3 3 3 2 4 4 3 3 4 3
[51] 1 4 3 1 4 NA NA NA 2 NA 3 4 NA 2 3 3 1 4 2 4 NA NA 4 3 2
[76] 3 NA 3 NA 4 3 2 3 NA 3 1 1 3 2 NA 1 3 3 NA 3 NA 2 NA 4 1
[101] NA 2 2 4 3 NA 4 NA 2 2 NA 3 2 NA NA 3 NA 3 1 NA 1 NA 1 NA 1
[126] 2 1 3 4 1 4 2 3 NA 3 NA NA 4 NA 2 NA 4 2 3 NA 1 2 1 3 4

Example

> na.approx(x2)

Output

 [1] 2.000000 2.000000 2.000000 2.000000 1.000000 1.000000 1.500000 2.000000
 [9] 4.000000 2.500000 1.000000 2.000000 1.000000 4.000000 3.000000 3.000000
[17] 1.000000 3.000000 1.000000 4.000000 4.000000 2.000000 3.000000 1.000000
[25] 3.000000 1.000000 4.000000 2.000000 4.000000 2.000000 1.000000 2.000000
[33] 1.000000 3.000000 2.500000 2.000000 2.500000 3.000000 1.000000 2.000000
[41] 3.000000 3.000000 3.000000 2.000000 4.000000 4.000000 3.000000 3.000000
[49] 4.000000 3.000000 1.000000 4.000000 3.000000 1.000000 4.000000 3.500000
[57] 3.000000 2.500000 2.000000 2.500000 3.000000 4.000000 3.000000 2.000000
[65] 3.000000 3.000000 1.000000 4.000000 2.000000 4.000000 4.000000 4.000000
[73] 4.000000 3.000000 2.000000 3.000000 3.000000 3.000000 3.500000 4.000000
[81] 3.000000 2.000000 3.000000 3.000000 3.000000 1.000000 1.000000 3.000000
[89] 2.000000 1.500000 1.000000 3.000000 3.000000 3.000000 3.000000 2.500000
[97] 2.000000 3.000000 4.000000 1.000000 1.500000 2.000000 2.000000 4.000000
[105] 3.000000 3.500000 4.000000 3.000000 2.000000 2.000000 2.500000 3.000000
[113] 2.000000 2.333333 2.666667 3.000000 3.000000 3.000000 1.000000 1.000000
[121] 1.000000 1.000000 1.000000 1.000000 1.000000 2.000000 1.000000 3.000000
[129] 4.000000 1.000000 4.000000 2.000000 3.000000 3.000000 3.000000 3.333333
[137] 3.666667 4.000000 3.000000 2.000000 3.000000 4.000000 2.000000 3.000000
[145] 2.000000 1.000000 2.000000 1.000000 3.000000 4.000000

Example3

Live Demo

> x3<-sample(c(NA,rnorm(5)),80,replace=TRUE)
> x3

Output

[1] -0.7419539 -0.7419539 -0.7419539 -0.7419539 NA -0.2225833
[7] -0.7240064 0.8134500 -0.2225833 -0.2225833 0.8134500 -0.7419539
[13] -0.7240064 -0.7419539 -0.7240064 -0.7419539 -0.7240064 0.7383318
[19] NA -0.7240064 0.7383318 0.7383318 NA 0.8134500
[25] -0.2225833 -0.7419539 -0.2225833 0.8134500 0.8134500 NA
[31] -0.2225833 -0.2225833 -0.7240064 -0.2225833 0.7383318 NA
[37] NA -0.7419539 -0.7240064 -0.7240064 -0.7419539 0.7383318
[43] 0.8134500 -0.7240064 0.7383318 0.8134500 0.7383318 0.8134500
[49] 0.7383318 -0.7240064 -0.2225833 -0.7240064 -0.7240064 -0.7240064
[55] 0.7383318 0.7383318 NA -0.2225833 -0.7419539 -0.7419539
[61] 0.8134500 -0.2225833 -0.2225833 0.7383318 -0.2225833 0.8134500
[67] -0.2225833 0.7383318 -0.7240064 0.7383318 NA -0.2225833
[73] 0.7383318 -0.7419539 0.8134500 -0.2225833 NA -0.7240064
[79] -0.2225833 -0.2225833

Example

> na.approx(x3)

Output

[1] -0.741953856 -0.741953856 -0.741953856 -0.741953856 -0.482268589
[6] -0.222583323 -0.724006386 0.813450002 -0.222583323 -0.222583323
[11] 0.813450002 -0.741953856 -0.724006386 -0.741953856 -0.724006386
[16] -0.741953856 -0.724006386 0.738331799 0.007162706 -0.724006386
[21] 0.738331799 0.738331799 0.775890900 0.813450002 -0.222583323
[26] -0.741953856 -0.222583323 0.813450002 0.813450002 0.295433340
[31] -0.222583323 -0.222583323 -0.724006386 -0.222583323 0.738331799
[36] 0.244903247 -0.248525304 -0.741953856 -0.724006386 -0.724006386
[41] -0.741953856 0.738331799 0.813450002 -0.724006386 0.738331799
[46] 0.813450002 0.738331799 0.813450002 0.738331799 -0.724006386
[51] -0.222583323 -0.724006386 -0.724006386 -0.724006386 0.738331799
[56] 0.738331799 0.257874238 -0.222583323 -0.741953856 -0.741953856
[61] 0.813450002 -0.222583323 -0.222583323 0.738331799 -0.222583323
[66] 0.813450002 -0.222583323 0.738331799 -0.724006386 0.738331799
[71] 0.257874238 -0.222583323 0.738331799 -0.741953856 0.813450002
[76] -0.222583323 -0.473294855 -0.724006386 -0.222583323 -0.222583323

Example4

Live Demo

> x4<-sample(c(NA,rpois(20,2)),100,replace=TRUE)
> x4

Output

[1] 3 3 0 2 NA 2 2 2 1 NA 0 1 3 3 3 3 1 1 3 3 1 2 1 1 2
[26] 3 5 5 0 2 1 1 3 2 1 3 2 NA 3 3 0 0 3 3 6 2 3 3 2 3
[51] 3 2 0 NA 2 NA 3 5 NA 0 3 1 5 2 1 NA 3 3 3 2 2 6 5 2 1
[76] 2 1 5 2 3 NA 0 0 2 2 2 0 5 2 3 6 0 3 3 3 3 2 2 3 1

Example

> na.approx(x4)

Output

[1] 3.0 3.0 0.0 2.0 2.0 2.0 2.0 2.0 1.0 0.5 0.0 1.0 3.0 3.0 3.0 3.0 1.0 1.0
[19] 3.0 3.0 1.0 2.0 1.0 1.0 2.0 3.0 5.0 5.0 0.0 2.0 1.0 1.0 3.0 2.0 1.0 3.0
[37] 2.0 2.5 3.0 3.0 0.0 0.0 3.0 3.0 6.0 2.0 3.0 3.0 2.0 3.0 3.0 2.0 0.0 1.0
[55] 2.0 2.5 3.0 5.0 2.5 0.0 3.0 1.0 5.0 2.0 1.0 2.0 3.0 3.0 3.0 2.0 2.0 6.0
[73] 5.0 2.0 1.0 2.0 1.0 5.0 2.0 3.0 1.5 0.0 0.0 2.0 2.0 2.0 0.0 5.0 2.0 3.0
[91] 6.0 0.0 3.0 3.0 3.0 3.0 2.0 2.0 3.0 1.0

Example5

Live Demo

> x5<-sample(c(NA,rpois(5,3)),100,replace=TRUE)
> x5

Output

[1] 3 1 3 6 5 3 5 NA 5 5 3 1 3 1 3 NA 3 5 6 NA 3 3 5 5 3
[26] 5 NA 3 3 3 5 5 NA 5 6 3 1 3 1 3 3 5 NA 5 6 1 3 6 5 5
[51] 1 5 NA 5 NA 1 5 3 1 6 NA 5 1 5 NA NA 6 6 5 1 5 5 NA 3 5
[76] 5 5 5 1 5 NA NA 1 6 5 5 5 5 5 1 5 NA 1 NA 3 NA 3 6 5 1

Example

> na.approx(x5)

Output

 [1] 3.000000 1.000000 3.000000 6.000000 5.000000 3.000000 5.000000 5.000000
 [9] 5.000000 5.000000 3.000000 1.000000 3.000000 1.000000 3.000000 3.000000
[17] 3.000000 5.000000 6.000000 4.500000 3.000000 3.000000 5.000000 5.000000
[25] 3.000000 5.000000 4.000000 3.000000 3.000000 3.000000 5.000000 5.000000
[33] 5.000000 5.000000 6.000000 3.000000 1.000000 3.000000 1.000000 3.000000
[41] 3.000000 5.000000 5.000000 5.000000 6.000000 1.000000 3.000000 6.000000
[49] 5.000000 5.000000 1.000000 5.000000 5.000000 5.000000 3.000000 1.000000
[57] 5.000000 3.000000 1.000000 6.000000 5.500000 5.000000 1.000000 5.000000
[65] 5.333333 5.666667 6.000000 6.000000 5.000000 1.000000 5.000000 5.000000
[73] 4.000000 3.000000 5.000000 5.000000 5.000000 5.000000 1.000000 5.000000
[81] 3.666667 2.333333 1.000000 6.000000 5.000000 5.000000 5.000000 5.000000
[89] 5.000000 1.000000 5.000000 3.000000 1.000000 2.000000 3.000000 3.000000
[97] 3.000000 6.000000 5.000000 1.000000

Nizamuddin Siddiqui

Updated on 19-Nov-2020 07:51:59

• Related Questions & Answers
• How to replace vector values less than 2 with 2 in an R vector?
• How to replace missing values with median in an R data frame column?
• How to replace missing values with row means in an R data frame?
• How to replace values in a vector with values in the same vector in R?
• How to replace missing values in a column with corresponding values in other column of an R data frame?
• How to replace a value in an R vector?
• Create a random sample by ignoring missing values in an R vector.
• How to replace one vector elements with another vector elements in R?
• How to replace missing values recorded with blank spaces in R with NA or any other value?
• How to convert a column with missing values to binary with 0 for missing values in R?
• Python Pandas - Fill NaN with Linear Interpolation
• How to replace numbers with ordinal strings for a survey in an R vector?
• How to use pandas series.fillna() to replace missing values?
• How to create the plot of a vector that contains missing values by adding the missing values in base R?
• How to replace NA values with zeros in an R data frame?