Question

In the mathematical languages, you can create a vector as follows:

x = seq(0, 2*pi, length.out = 100)

This outputs:

[1] 0.00000000 0.06346652 0.12693304 0.19039955 0.25386607 0.31733259 0.38079911
  [8] 0.44426563 0.50773215 0.57119866 0.63466518 0.69813170 0.76159822 0.82506474
 [15] 0.88853126 0.95199777 1.01546429 1.07893081 1.14239733 1.20586385 1.26933037
 [22] 1.33279688 1.39626340 1.45972992 1.52319644 1.58666296 1.65012947 1.71359599
 [29] 1.77706251 1.84052903 1.90399555 1.96746207 2.03092858 2.09439510 2.15786162
 [36] 2.22132814 2.28479466 2.34826118 2.41172769 2.47519421 2.53866073 2.60212725
 [43] 2.66559377 2.72906028 2.79252680 2.85599332 2.91945984 2.98292636 3.04639288
 [50] 3.10985939 3.17332591 3.23679243 3.30025895 3.36372547 3.42719199 3.49065850
 [57] 3.55412502 3.61759154 3.68105806 3.74452458 3.80799110 3.87145761 3.93492413
 [64] 3.99839065 4.06185717 4.12532369 4.18879020 4.25225672 4.31572324 4.37918976
 [71] 4.44265628 4.50612280 4.56958931 4.63305583 4.69652235 4.75998887 4.82345539
 [78] 4.88692191 4.95038842 5.01385494 5.07732146 5.14078798 5.20425450 5.26772102
 [85] 5.33118753 5.39465405 5.45812057 5.52158709 5.58505361 5.64852012 5.71198664
 [92] 5.77545316 5.83891968 5.90238620 5.96585272 6.02931923 6.09278575 6.15625227
 [99] 6.21971879 6.28318531

How can this be achieved in Haskell?

I tried creating a lambda function and using it with map, but I could n’t get the same output.

Thanks

let myPi = (\x -> 2*pi)
map myPi [1..10]

Answer 1

Well, you can just do

[0, 2*pi/100 .. 2*pi]

Note that this is not ideal both performance- and floating-point-rounding–wise (because it translates to enumFromThenTo), Daniel Fischer’s version is better (it translates to enumFromTo). Thinking it over, GHC will probably compile both to almost equally-fast code, but I’m not sure. If it’s really performance-critical, it’s best not to use lists at all but e.g. Data.Vector.

---

As Jakub Hampl remarked, Haskell can deal with infinite lists. That’s probably not much use to you here, but it opens interesting possibilties – for instance, you might not be sure which resolution you actually need. You can let your list begin with a very low resolution, then fold back and start again with a higher one. One simple way to achieve this:

import Data.Fixed

multiResS₁ = [ log x `mod'` 2*pi | x<-[1 .. ] ]

using this to plot the sine function looks like this

Prelude Data.Fixed Graphics.Rendering.Chart.Simple> let domainS₁ = take 200 multiResS₁
Prelude Data.Fixed Graphics.Rendering.Chart.Simple> plotPNG “multiresS1.png” domainS₁ sin