Python – Sorting a list of RGB triplets into a spectrum

color-spacehsvpythonpython-imaging-libraryspectrum

I have a list of RGB triplets, and I'd like to plot them in such a way that they form something like a spectrum.

I've converted them to HSV, which people seem to recommend.

from PIL import Image, ImageDraw
import colorsys

def make_rainbow_rgb(colors, width, height):
    """colors is an array of RGB tuples, with values between 0 and 255"""

    img = Image.new("RGBA", (width, height))
    canvas = ImageDraw.Draw(img)

    def hsl(x):
        to_float = lambda x : x / 255.0
        (r, g, b) = map(to_float, x)
        h, s, l = colorsys.rgb_to_hsv(r,g,b)
        h = h if 0 < h else 1 # 0 -> 1
        return h, s, l

    rainbow = sorted(colors, key=hsl)

    dx = width / float(len(colors)) 
    x = 0
    y = height / 2.0
    for rgb in rainbow:
        canvas.line((x, y, x + dx, y), width=height, fill=rgb)
        x += dx
    img.show()

However, the result doesn't look very much like a nice rainbow-y spectrum. I suspect I need to either convert to a different color space or handle the HSL triplet differently.

this doesn't look like a spectrum

Does anyone know what I need to do to make this data look roughly like a rainbow?

Update:

I was playing around with Hilbert curves and revisited this problem. Sorting the RGB values (same colors in both images) by their position along a Hilbert curve yields an interesting (if still not entirely satisfying) result:

RGB values sorted along a Hilbert curve.

Best Answer

You're trying to convert a three-dimensional space into a one-dimensional space. There's no guarantee that you can make a pleasing rainbow out of it, as Oli says.

What you can do is "bucket" the colors into a few different categories based on saturation and value/lightness, and then sort within the categories, to get several independent gradients. For example, high-saturation colors first for the classic rainbow, then mid-saturation high-value colors (pastels), then low-saturation (grays).

Alternately, if all you care about is the rainbow, convert to hsl, then slam saturation to 1.0 and value to 0.5, convert back to rgb and render that instead of the original color.

Caveats follow

Note that while this is perhaps the cleanest way to answer the question as asked, index is a rather weak component of the list API, and I can't remember the last time I used it in anger. It's been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about list.index follow. It is probably worth initially taking a look at the documentation for it:

list.index(x[, start[, end]])
Return zero-based index in the list of the first item whose value is equal to x. Raises a ValueError if there is no such item.

The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.

Linear time-complexity in list length

An index call checks every element of the list in order, until it finds a match. If your list is long, and you don't know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give index a hint. For instance, in this snippet, l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:

>>> import timeit
>>> timeit.timeit('l.index(999_999)', setup='l = list(range(0, 1_000_000))', number=1000)
9.356267921015387
>>> timeit.timeit('l.index(999_999, 999_990, 1_000_000)', setup='l = list(range(0, 1_000_000))', number=1000)
0.0004404920036904514

Only returns the index of the first match to its argument

A call to index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.

>>> [1, 1].index(1)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2

Most places where I once would have used index, I now use a list comprehension or generator expression because they're more generalizable. So if you're considering reaching for index, take a look at these excellent Python features.

Throws if element not present in list

A call to index results in a ValueError if the item's not present.

>>> [1, 1].index(2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: 2 is not in list

If the item might not be present in the list, you should either

Check for it first with item in my_list (clean, readable approach), or
Wrap the index call in a try/except block which catches ValueError (probably faster, at least when the list to search is long, and the item is usually present.)

Best Answer

Related Solutions

Python – How to check if a list is empty

Python – Finding the index of an item in a list

Caveats follow

Linear time-complexity in list length

Only returns the index of the first match to its argument

Throws if element not present in list

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