- Trending Categories
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
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:
> 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
> 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
> 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
> 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
> 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
- Related Articles
- 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?
- How to replace one vector elements with another vector elements in R?
- Create a random sample by ignoring missing values in an R vector.
- How to replace missing values recorded with blank spaces in R with NA or any other value?
- How to replace numbers with ordinal strings for a survey in an R vector?
- How to convert a column with missing values to binary with 0 for missing values in R?
- 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?
- How to find the number of unique values in a vector by excluding missing values in R?
- How to merge two unequal data frames and replace missing values to 0 in R?
Advertisements