Safe Haskell | Safe |
---|---|

Language | Haskell2010 |

Random useful stuff I didn't know where to put.

- wilson :: Double -> Int -> Int -> (Double, Double, Double)
- invnormcdf :: (Ord a, Floating a) => a -> a
- choose :: Integral a => a -> a -> a
- estimateComplexity :: (Integral a, Floating b, Ord b) => a -> a -> Maybe b
- showNum :: Show a => a -> String
- showOOM :: Double -> String
- log1p :: (Floating a, Ord a) => a -> a
- expm1 :: (Floating a, Ord a) => a -> a
- (<#>) :: (Floating a, Ord a) => a -> a -> a
- log1mexp :: (Floating a, Ord a) => a -> a
- log1pexp :: (Floating a, Ord a) => a -> a
- lsum :: (Floating a, Ord a) => [a] -> a
- llerp :: (Floating a, Ord a) => a -> a -> a -> a

# Documentation

wilson :: Double -> Int -> Int -> (Double, Double, Double) Source #

Calculates the Wilson Score interval.
If `(l,m,h) = wilson c x n`

, then `m`

is the binary proportion and
`(l,h)`

it's `c`

-confidence interval for `x`

positive examples out of
`n`

observations. `c`

is typically something like 0.05.

invnormcdf :: (Ord a, Floating a) => a -> a Source #

choose :: Integral a => a -> a -> a Source #

Binomial coefficient: \( \mbox{choose n k} = \frac{n!}{(n-k)! k!} \)

estimateComplexity :: (Integral a, Floating b, Ord b) => a -> a -> Maybe b Source #

Try to estimate complexity of a whole from a sample. Suppose we
sampled `total`

things and among those `singles`

occured only once.
How many different things are there?

Let the total number be `m`

. The copy number follows a Poisson
distribution with paramter `lambda`

. Let \( z := e^{\lambda} \), then
we have:

\[ P( 0 ) = e^{-\lambda} = \frac{1}{z} \\ P( 1 ) = \lambda e^{-\lambda} = \frac{\ln z}{z} \\ P(\ge 1) = 1 - e^{-\lambda} = 1 - \frac{1}{z} \\ \] \[ \mbox{singles} = m \frac{\ln z}{z} \\ \mbox{total} = m \left( 1 - \frac{1}{z} \right) \\ \] \[ D := \frac{\mbox{total}}{\mbox{singles}} = (1 - \frac{1}{z}) * \frac{z}{\ln z} \\ f := z - 1 - D \ln z = 0 \]

To get `z`

, we solve using Newton iteration and then substitute to
get `m`

:

\[ df/dz = 1 - D/z \\ z' = z - \frac{ z (z - 1 - D \ln z) }{ z - D } \\ m = \mbox{singles} * \frac{z}{\ln z} \]

It converges as long as the initial `z`

is large enough, and `10D`

(in the line for `zz`

below) appears to work well.

log1p :: (Floating a, Ord a) => a -> a Source #

Computes `log (1+x)`

to a relative precision of `10^-8`

even for
very small `x`

. Stolen from http://www.johndcook.com/cpp_log_one_plus_x.html

expm1 :: (Floating a, Ord a) => a -> a Source #

Computes \( e^x - 1 \) to a relative precision of `10^-10`

even for
very small `x`

. Stolen from http://www.johndcook.com/cpp_expm1.html

(<#>) :: (Floating a, Ord a) => a -> a -> a infixl 5 Source #

Computes \( \ln \left( e^x + e^y \right) \) without leaving the log domain and hence without losing precision.

log1mexp :: (Floating a, Ord a) => a -> a Source #

Computes \( \ln (1 - e^x) \), following Martin Mächler.

log1pexp :: (Floating a, Ord a) => a -> a Source #

Computes \( \ln (1 + e^x) \), following Martin Mächler.