Haskell create a list with specific increment

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]

Also in Haskell it need not be finite.

Jakub Hampl
– Jakub Hampl

2012-11-20 21:05:38 +00:00
Commented Nov 20, 2012 at 21:05 — Jakub Hampl
– Jakub Hampl, Commented Nov 20, 2012 at 21:05
Your lambda doesn't seem to use the x at all?

singpolyma
– singpolyma

2012-11-20 21:37:25 +00:00
Commented Nov 20, 2012 at 21:37 — singpolyma
– singpolyma, Commented Nov 20, 2012 at 21:37

leftaroundabout · Accepted Answer · 2012-11-20 21:27:32Z

11

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

enter image description here

edited Nov 20, 2012 at 21:27

answered Nov 20, 2012 at 21:03

leftaroundabout

121k6 gold badges185 silver badges337 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

user557240 Over a year ago

Thanks, I'm not worried about performance. I needed it to try and plot the sin function

user557240 Over a year ago

Also I'm trying to implement the graph with Graphics.Rendering.Cairo, once I've got these points do I need to scale them as they are not drawing on the canvas?

leftaroundabout Over a year ago

That's an unrelated question. Not sure what you actually mean, but it's likely that you do need to scale in some way.

dave4420 · Accepted Answer · 2012-11-20 21:20:24Z

7

Easiest is a list comprehension,

[(2*pi)*k/99 | k <- [0 .. 99]]

(the multiplication with k/99 mitigates the floating point rounding, so the last value is exactly 2*pi.)

edited Nov 20, 2012 at 21:20

dave4420

47.2k6 gold badges121 silver badges153 bronze badges

answered Nov 20, 2012 at 21:04

Daniel Fischer

184k19 gold badges319 silver badges436 bronze badges

1 Comment

singpolyma Over a year ago

Isn't that just map (\k -> (2*pi)*k/99) [0..99] ?

Collectives™ on Stack Overflow

Haskell create a list with specific increment

2 Answers 2

3 Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related