a trend calculated from quick

a trend calculated from robust

The result of a trend prediction. Use this to make predictions using predictY.

Extract a line from a trend.

Given an x, predict y.

predictY { slope = 1, intercept = 0 } 1
    --> 1


predictY { slope = -1, intercept = 0 } 5.5
    --> -5.5

a single 2-dimensional point

Plot a line through a series of points (x, y). So given a perfect linear relationship:

 quick [ (1, 1), (2, 2), (3, 3), (4, 4) ]
     |> Result.map line
     --> Ok { slope = 1, intercept = 0 }

This is the fastest of the functions in this module, but it's also the most susceptible to being thrown off by outliers. Let's look at that line again, but with an outlier:

 quick [ (1, 1), (2, 2), (3, 3), (4, 0) ]
     |> Result.map line
     --> Ok { slope = -0.2, intercept = 2 }

The more outliers you have, the worse fit you'll get. You can figure out if this is happening by sending your Trend to goodnessOfFit.

Under the covers, this is an ordinary least squares regression.

Get the goodness of fit for a quick trend. This is a number between 0 to 1. A higher number generally indicates a better fit, but it doesn't know anything about what your data means. This means that you have to use some judgement in interpreting it!

quick [ (1, 1), (2, 2), (3, 3), (4, 4) ]
    |> Result.map goodnessOfFit
    --> Ok 1

And again with that outlier from quick:

 quick [ (1, 1), (2, 2), (3, 3), (4, 0) ]
     |> Result.map goodnessOfFit
     --> Ok 0.039999999999999813

This calculation is only valid for quick trends, since it measures how well a fit has minimized the square sum of error. That means it's only really useful for ordinary least squares, not the Theil-Sen estimator we use for robust

Maintainer's note: this will evaluate the fit for the original data. If you need to evaluate goodness of fit for new data given an existing Trend, we'll need to expose a new function. I don't have a concrete use case for this, so the function does not corrently exists. I want to make this library work for you, so please open an issue if you find yourself in this situation!

When your data has outliers, you'll want to use a robust estimator instead of the quick estimator. This is much slower (it runs roughly in O(n^2) time), but will still give good results in the face of corrupted data. Specifically, it will still work if up to ~29.3% of your data consists of outliers. And again, the easiest way to check this is to visualize it with terezka/elm-plot or something similar.

For good data, we have the same results as quick:

 robust [ (1, 1), (2, 2), (3, 3), (4, 4) ]
     |> Result.map line
     --> Ok { slope = 1, intercept = 0 }

But when we have outliers, we still get a good result:

 robust [ (1, 1), (2, 2), (3, 3), (4, 0) ]
     |> Result.map line
     --> Ok { slope = 1, intercept = 0 }

Under the covers, this is a Theil-Sen estimator (which is pretty cool and easy to get an intuitive grasp of what's going on; check it out!)

Calculate a confidence interval from a robust set of data. Consult WikiPedia for a thorough understanding of what this may mean for your data set. This function gives a 95% confidence interval.

Maintainer's note: We ought to be able to generate a confidence interval for quick trends too, but I'm not confident enough in my math skills to do it correctly. Help wanted here! If you know how to do that calculation, let's work together and add it.