This article presents a Python script to map a spectrum of wavelengths to a representation of a colour. There is no unique way to do this, but the formulation used here is based on the CIE colour matching functions, $\bar{x}(\lambda)$, $\bar{y}(\lambda)$ and $\bar{z}(\lambda)$. These model the chromatic response of a "standard observer" by mapping a power spectrum of wavelengths, $P(\lambda)$, to a set of *tristimulus values*, $X$, $Y$ and $Z$, analogous to the actual response of the three types of cone cell in the human eye.

\begin{align*} X &= \int P(\lambda)\bar{x}(\lambda)\mathrm{d}\lambda,\\ Y &= \int P(\lambda)\bar{y}(\lambda)\mathrm{d}\lambda,\\ Z &= \int P(\lambda)\bar{z}(\lambda)\mathrm{d}\lambda, \end{align*}

$X$, $Y$ and $Z$ can be normalized by dividing by their sum (at the expense of losing information about the brightness of the light):

$$ x = \frac{X}{X+Y+Z}, \quad y = \frac{Y}{X+Y+Z}, \quad z = \frac{Z}{X+Y+Z} = 1 - x - y $$

In this way, only two parameters, $x$ and $y$ are needed to describe the colour (more accurately, the *chromaticity*) of the light. The CIE standard chromaticity diagram is shown below.

Original image by user Spigget, licensed under Creative Commons Attribution-Share Alike 3.0 Unported.

Further conversion of $(x, y)$ to RGB values for output by a display device requires transformation by the appropriate chromaticity matrix. Geometrically, this maps points in the above colour "tongue" onto the subset of points within the RGB "gamut", the indicated triangular region. A *colour system* may be defined by a matrix of three primary colour chromaticities (the vertices of the triangle) and a *white point*: a set of chromaticiy coordinates defining the "colour" white for some purpose.

$$ \left( \begin{array}{lll} x_r & x_g & x_b \\ y_r & y_g & y_b \\ z_r & z_g & z_b \end{array} \right) \left( \begin{array}{l} r \\ g \\ b \\ \end{array} \right) = \left( \begin{array}{l} x \\ y \\ z \\ \end{array} \right). $$

Multiplication of the vector of $(x,y,z)$ values by the inverse of this matrix therefore gives the RGB values describing the corresponding colour within the system being used.

Not all $(x, y)$ pairs map to points within the RGB gamut (they would give negative values for one or more component): one way to deal with this is to "desaturate" by raising the values of all components equally until they are all non-negative.

The code below defines a class, `ColourSystem`

, for representing and using colour systems, and instantiates a few particular examples. The CIE matching function is read in from the file cie-cmf.txt.

```
# colour_system.py
import numpy as np
def xyz_from_xy(x, y):
"""Return the vector (x, y, 1-x-y)."""
return np.array((x, y, 1-x-y))
class ColourSystem:
"""A class representing a colour system.
A colour system defined by the CIE x, y and z=1-x-y coordinates of
its three primary illuminants and its "white point".
TODO: Implement gamma correction
"""
# The CIE colour matching function for 380 - 780 nm in 5 nm intervals
cmf = np.loadtxt('cie-cmf.txt', usecols=(1,2,3))
def __init__(self, red, green, blue, white):
"""Initialise the ColourSystem object.
Pass vectors (ie NumPy arrays of shape (3,)) for each of the
red, green, blue chromaticities and the white illuminant
defining the colour system.
"""
# Chromaticities
self.red, self.green, self.blue = red, green, blue
self.white = white
# The chromaticity matrix (rgb -> xyz) and its inverse
self.M = np.vstack((self.red, self.green, self.blue)).T
self.MI = np.linalg.inv(self.M)
# White scaling array
self.wscale = self.MI.dot(self.white)
# xyz -> rgb transformation matrix
self.T = self.MI / self.wscale[:, np.newaxis]
def xyz_to_rgb(self, xyz, out_fmt=None):
"""Transform from xyz to rgb representation of colour.
The output rgb components are normalized on their maximum
value. If xyz is out the rgb gamut, it is desaturated until it
comes into gamut.
By default, fractional rgb components are returned; if
out_fmt='html', the HTML hex string '#rrggbb' is returned.
"""
rgb = self.T.dot(xyz)
if np.any(rgb < 0):
# We're not in the RGB gamut: approximate by desaturating
w = - np.min(rgb)
rgb += w
if not np.all(rgb==0):
# Normalize the rgb vector
rgb /= np.max(rgb)
if out_fmt == 'html':
return self.rgb_to_hex(rgb)
return rgb
def rgb_to_hex(self, rgb):
"""Convert from fractional rgb values to HTML-style hex string."""
hex_rgb = (255 * rgb).astype(int)
return '#{:02x}{:02x}{:02x}'.format(*hex_rgb)
def spec_to_xyz(self, spec):
"""Convert a spectrum to an xyz point.
The spectrum must be on the same grid of points as the colour-matching
function, self.cmf: 380-780 nm in 5 nm steps.
"""
XYZ = np.sum(spec[:, np.newaxis] * self.cmf, axis=0)
den = np.sum(XYZ)
if den == 0.:
return XYZ
return XYZ / den
def spec_to_rgb(self, spec, out_fmt=None):
"""Convert a spectrum to an rgb value."""
xyz = self.spec_to_xyz(spec)
return self.xyz_to_rgb(xyz, out_fmt)
illuminant_D65 = xyz_from_xy(0.3127, 0.3291)
cs_hdtv = ColourSystem(red=xyz_from_xy(0.67, 0.33),
green=xyz_from_xy(0.21, 0.71),
blue=xyz_from_xy(0.15, 0.06),
white=illuminant_D65)
cs_smpte = ColourSystem(red=xyz_from_xy(0.63, 0.34),
green=xyz_from_xy(0.31, 0.595),
blue=xyz_from_xy(0.155, 0.070),
white=illuminant_D65)
cs_srgb = ColourSystem(red=xyz_from_xy(0.64, 0.33),
green=xyz_from_xy(0.30, 0.60),
blue=xyz_from_xy(0.15, 0.06),
white=illuminant_D65)
```

Here's one application of the `ColourSystem`

class: to visualize the colour of a black body at a given temperature, and example given on this excellent page on the colour rendering of spectra. The spectral radiance of a black body is given by the *Planck function*:

$$ B(\lambda; T) = \frac{2hc^2}{\lambda^5}\frac{1}{\exp\left(\frac{hc}{\lambda k_\mathrm{B}T}\right) - 1} $$

Feeding $B(\lambda; T)$ as `spec`

to the function `ColourSystem.spec_to_rgb`

returns the RGB components corresponding to the colour of a black body at temperature $T$. This is visualized for different temperatures below.

```
import numpy as np
from scipy.constants import h, c, k
import matplotlib.pyplot as plt
from matplotlib.patches import Circle
from colour_system import cs_hdtv
cs = cs_hdtv
def planck(lam, T):
""" Returns the spectral radiance of a black body at temperature T.
Returns the spectral radiance, B(lam, T), in W.sr-1.m-2 of a black body
at temperature T (in K) at a wavelength lam (in nm), using Planck's law.
"""
lam_m = lam / 1.e9
fac = h*c/lam_m/k/T
B = 2*h*c**2/lam_m**5 / (np.exp(fac) - 1)
return B
fig, ax = plt.subplots()
# The grid of visible wavelengths corresponding to the grid of colour-matching
# functions used by the ColourSystem instance.
lam = np.arange(380., 781., 5)
for i in range(24):
# T = 500 to 12000 K
T = 500*i + 500
# Calculate the black body spectrum and the HTML hex RGB colour string
# it looks like
spec = planck(lam, T)
html_rgb = cs.spec_to_rgb(spec, out_fmt='html')
# Place and label a circle with the colour of a black body at temperature T
x, y = i % 6, -(i // 6)
circle = Circle(xy=(x, y*1.2), radius=0.4, fc=html_rgb)
ax.add_patch(circle)
ax.annotate('{:4d} K'.format(T), xy=(x, y*1.2-0.5), va='center',
ha='center', color=html_rgb)
# Set the limits and background colour; remove the ticks
ax.set_xlim(-0.5,5.5)
ax.set_ylim(-4.35, 0.5)
ax.set_xticks([])
ax.set_yticks([])
ax.set_facecolor('k')
# Make sure our circles are circular!
ax.set_aspect("equal")
plt.show()
```

## Comments

Comments are pre-moderated. Please be patient and your comment will appear soon.

## Axel Schindler 5 years, 6 months ago

Thanks for sharing. The link for the cie-cmf.txt file is not working, however

Link | Reply## christian 5 years, 6 months ago

Thanks, Axel -- I've fixed this broken link now.

Link | Reply## Mukyu 3 years, 3 months ago

Are there any reference about the xyz2rgb part ... ? The matlab seems to be different from it. Matlab only implement matrix transform and gamma correction, but the code you proposed included "desaturating" and "normalize a rgb vector" ...

Link | Reply## asad 2 years, 11 months ago

i have problem with ''from colour_system import cs_hdtv''

Link | Replyis colour_system a module?

i can't find something like it in internet?and python gives error

## christian 2 years, 11 months ago

colour_system.py is the filename you should give the code above the program you're running (the second block of code). I've added a comment line to make it a bit clearer.

Link | Reply## Claude Falbriard 2 years, 11 months ago

Under Anaconda Juypyter Python 3.6 I was able to instantiate the class with the following statement:

Link | Replycs = ColourSystem(red=xyz_from_xy(0.67, 0.33),

green=xyz_from_xy(0.21, 0.71),

blue=xyz_from_xy(0.15, 0.06),

white=illuminant_D65)

No need for class import as the python code is present in the previous notebook cell.

There is also an outdated matplotlib call, please use:

#ax.set_axis_bgcolor('k') # outdated

ax.set_facecolor('k')

Hope this also works for you. Regards, Claude

## christian 2 years, 11 months ago

Yup, if you run both blocks in a Jupyter Notebook you don't need to import anything.

Link | ReplyThanks for pointing out the updated matplotlib call: I wrote this before set_axis_bgcolor was deprecated. I will update.

## blackle 2 years, 8 months ago

for gamma correction I made this function:

Link | Replydef linear_srgb_to_rgb(rgb):

nonlinearity = np.vectorize(lambda x: 12.92*x if x < 0.0031308 else 1.055*(x**(1.0/2.4))-0.055)

return nonlinearity(rgb)

see https://en.wikipedia.org/wiki/SRGB#The_forward_transformation_(CIE_XYZ_to_sRGB) for the formula

## MD 1 year, 11 months ago

I find that if i insert a spectrum containing every wavelength at equal intensity, it returns light orange, instead of white. How does that happen? Equal intensities for every wavelength should appear as white light (255, 255, 255).

Link | Reply## fook 1 year, 9 months ago

Because the white point of sRGB displays is D65, not illuminant E.

Link | Reply## David 1 year, 8 months ago

Hi, I want to convert a reflectance spectrum into a colour. I guess that this can be done with that script.

Link | ReplyPerhaps, a silly question, but I am inexperienced with coding.

How can I include my spectrum into the script, for this to be converted in a colour and RGC colour values?

Many thanks! Cheer, David

## christian 1 year, 8 months ago

You should be OK to provide your spectrum as spec in a call to spec_to_rgb(), but it should be on the same grid of points as the colour-matching function, self.cmf: 380-780 nm in 5 nm steps. I hope it goes OK!

Link | Reply## David 1 year, 7 months ago

Hi Christian, many thanks for replying!

Link | ReplyOk, my spectrum is from 380 to 780 nm with 5 nm in step

Shall I call it in this line?

cmf = np.loadtxt('cie-cmf.txt', usecols=(1,2,3))

Obviously, changing the name to that of my spectrum?

Thanks and cheers,

David

## christian 1 year, 7 months ago

That's the colour matching function. You'll want to replace spec in this line: html_rgb = cs.spec_to_rgb(spec, out_fmt='html') with your spectrum, spec, as a suitable array.

Link | Reply## David 1 year, 7 months ago

Many thanks, Christian!

Link | Reply## Yannis 1 year, 5 months ago

Hi, thanks for providing the code for spectrum to RGB conversion. What I'd like to ask is how can I convert a spectrum in a different spectral range or with a different interval. Thank you in advance!

Link | Reply## christian 1 year, 5 months ago

I think in this case it might be best to interpolate / bin your spectrum to the same interval and wavelength spacing as the colour-matching function.

Link | Reply## Clay 1 year ago

How can I avoid the intensity normalization? I made normalization optional in spec_to_xyz() and rgb_to_xyz(), but it's not working, and I still can't get brown, gray, or any dark colors. I'm trying to approximate mixing three monochromatic light sources of adjustable frequency and amplitude. Thanks

Link | Reply## xiao7 6 months ago

whether a limited 2D plane can Reproduce all visible colors in theory？

Link | Reply## christian 6 months ago

Well, you have three different types of cone receptor each giving a different response to a given light stimulus, and if you take out the total brightness you might approximate that you are left with two independent variables...

Link | Reply## New Comment