# Piecewise Linear Interpolation 22

Posted by **Loren Shure**,

`John D'Errico`is back today to talk about linear interpolation.

### Contents

### Introduction

You saw in my previous blog that high order polynomials can have some problems. Why not go to the opposite extreme? Use a piecewise version of `linear interpolation`? I like to call it `connect-the-dots`, after the child's game of that name. This is really the simplest interpolation of all.

In MATLAB, given a list of points, sampled from some functional relationship in one dimension, how would we perform piecewise

linear interpolation? There are really two steps.

For any point u, given a set of (x,y) pairs with a monotonic vector x (by monotonic, I mean that x(k) < x(k+1) ), first find

the index k, such that

Second, perform the linear interpolation to predict the value of y at x=u, between the pair of points (x(k),y(k)) and (x(k+1),y(k+1)).

Each data point in the list of points becomes a point where the slope of the piecewise linear interpolant changes to a new

value. However, the function is still continuous across those locations. So one might call these locations "knots" because

at those points consecutive polynomial segments are tied together. Knots is a common term applied to splines for these locations;

breaks is a common alternative name.

### Create Some Data to Interpolate

x = linspace(0,2*pi,10)'; y = sin(x); plot(x,y,'o') title 'A simple trig function' xlabel X ylabel Y

Suppose I want to interpolate this function at some intermediate point, perhaps u == 1.3?

u = 1.3;

### histc Solves the Binning Problem

The first step I describe above is what I call binning. `histc` solves that problem efficiently.

[k,k] = histc(u,x); k

k = 2

As an aside, look at the way I took the output from `histc`. Since I only need the second output from `histc` but not the first output, rather than clutter up my workspace with a variable that I did not need, the output style `[k,k]` returns only the information I need.

Next, it seems instructive to dive a little more deeply into binning, so let me offer a few alternatives to `histc`.

### Binning - A Loop With An Explicit Test

Just a test inside a loop suffices.

for k = 1:(length(x)-1) if (x(k) <= u) && (u < x(k+1)) break end end x k

x = 0 0.69813 1.3963 2.0944 2.7925 3.4907 4.1888 4.8869 5.5851 6.2832 k = 2

### Binning - A Semi-vectorized Test

Do the binning with a single vectorized test. This works reasonably as long as I have only a scalar value for `u`, so I call this only a semi-vectorized solution.

k = sum(u>=x)

k = 2

When I want to place multiple points into their respective bins, this `sum` fails.

There are other ways to place points in bins in MATLAB, including a `sort`, or with hash tables. You can find several different binning methods in `bindex` on the file exchange. It is useful mainly to those with older MATLAB releases, because `histc became` available with version 5.3 and later of MATLAB.

### Fully Vectorized Binning

Next, I may, and often do, have a list of points to interpolate. This is a common event, where I wish to more finely resample

a curve that is sampled only at some short list of points. This time, let me generate 1000 points at which to interpolate

the sampled function.

u = linspace(x(1),x(end),1000)'; [k,k] = histc(u,x);

This next line handles the very last point. Recall the definition of a bin as

Note the strict inequality on the left. So that very last point (at `x(end)`) might be technically said to fall into the `n|th bin when I have |n` break points.

On the other hand, it is more convenient to put anything that lies exactly at the very last break point into bin `(n-1)`.

By the way, any piecewise interpolation should worry about points that fall fully above or below the domain of the data. Will

the code extrapolate or not? Should you extrapolate?

n = length(x); k(k == n) = n - 1;

### Interpolation as a Linear Combination

The final step is to interpolate between two points. Long ago, I recall from high school what was called a point-slope form

for a line. If you know a pair of points that a line passes through, as `(x(k),y(k))` and `(x(k+1),y(k+1))`, then the slope of the line is simple to compute. An equation for our line as a function of the parameter `u` is just:

I can also rewrite this in a way that I like, by defining a parameter `t` as

Then the interpolant is a simple linear combination of the function values at each end of the interval.

### Do the Interpolation and Plot the Result

See how nicely this all worked, in a fully vectorized coding style.

t = (u - x(k))./(x(k+1) - x(k)); yu = (1-t).*y(k) + t.*y(k+1); plot(x,y,'bo',u,yu,'r-') xlabel X ylabel Y title 'The connect-the-dots interpolant'

### Use interp1 Instead

It is always better to use a built-in tool to solve your problem than to do it yourself, so I might just as well have used

`interp1` to accomplish this interpolation.

yinterp1 = interp1(x,y,u,'linear'); plot(x,y,'bo',u,yinterp1,'r-') xlabel X ylabel Y title 'The connect-the-dots interpolant, using interp1'

In my next blog I'll begin talking about piecewise cubic interpolants. Until then please tell me your ideas `here`. Are there some associated topics that I should cover?

Published with MATLAB® 7.6

**359**views (last 30 days) | |

**Category:**- Interpolation & Fitting

### Comments

To leave a comment, please click here to sign in to your MathWorks Account or create a new one.

## Recent Comments