Linear interpolation of data

Question

I have a data frame with the year (x) and an associated percentage (y)

data.frame(x = c(1997,2000,2003,2006,2009,2010,2013,2014),
           y = c(.02,.023,.025,.024,.026,.027,.029,.031)

Here is a line chart of this data frame :

I would like to interpolate my data to get the percentage of missing years based on a linear regression.

I could make a linear model of each piece of curve but it would be tedious.

Is there a simple way to do it with R?

INPUT :

df = data.frame(
  year=c(1997,2000,2003,2006,2009,2010,2013,2014),
  percent=c(0.020, 0.023, 0.025, 0.024, 0.026, 0.027, 0.029, 0.031)
)

OUTPUT (for a function f) :

f(2006)==0.024
f(2007)==0.024.666
f(2008)==0.025.333
f(2009)==0.026

Why don't do a linear model on all the data instead of doing by each piece of curve ? — Orhan Yazar, Aug 11 '17 at 12:51
Because I can only assume that the percentage is linear between 2 points. — Dan Chaltiel, Aug 11 '17 at 12:52
This question seems to be less related to R and more to non-linear modeling. — trosendal, Aug 11 '17 at 12:53
then why don't try a linear model of degree 2 ? `model <- lm(y~x+I(x)^2)` ? — Orhan Yazar, Aug 11 '17 at 12:53
It might be easier if you posted a [reproducible example](https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example). I have not had enough coffee to guess and enter data from your chart. — bouncyball, Aug 11 '17 at 12:53
@trosendal I disagree, I know non-linear modeling, the question is how to achieve simply this with R — Dan Chaltiel, Aug 11 '17 at 12:54
You are kind for trying to help me but I don't want any polynomial, I want a linear regression. Bouncyball is right, I'll provide it in few minutes. — Dan Chaltiel, Aug 11 '17 at 12:56
Do you want to extrapolate or interpolate? Your Q says the former, but it looks like the latter to me. Have a look at `?stats::approx`. — Axeman, Aug 11 '17 at 12:56
@DanChaltiel, lm means linear model try to use what i wrote `model <- lm(y~x+I(x)^2)` — Orhan Yazar, Aug 11 '17 at 12:57
You could try a spline fit: `spline(df$year, df$percent, n=18)` — Dave2e, Aug 11 '17 at 13:08

score 5 · Accepted Answer · answered Aug 11 '17 at 13:03

5

One way is to use linear inter polation with zoo:

library(tidyr)
library(zoo)

df_complete <- complete(df, year = full_seq(year, 1))
df_complete$percent <- na.approx(df_complete$percent)


plot(df_complete)

answered Aug 11 '17 at 13:03

Axeman

32,068
8
81
94

Perfect, thanks a lot ! So this is interpolation and not extrapolation, I understand why I couldn't find my answer. – Dan Chaltiel Aug 11 '17 at 13:07
1

Yes. Interpolation is inside the range of your data, extrapolation is outside the range of your data. – Axeman Aug 11 '17 at 13:08
1

seems obvious now that I know :-) – Dan Chaltiel Aug 11 '17 at 13:09

Linear interpolation of data

1 Answers1

Related