Monday, December 28, 2015

The 10 colors of Pi

Picking up patterns

Or the lack thereof. The brain is pretty good at spotting patterns and anomalies. But we have to help it with something that can be easily abstracted. Numbers are not good for that.

Of reds and greens and blues

Colors are good helpers. Shades, hues. Unfortunately, many people are affected by colorblindness. It is said that in some segments of the population, up to 8% of men and 0.4% of women experience congenital color deficiency, with the most common being red-green color blindness.
In [1]:
%matplotlib inline
import matplotlib.pyplot as plt
from matplotlib.patches import Rectangle
import seaborn as sns
sns.set_context("talk")

Seaborn palettes

According to Seaborn's documentation (http://web.stanford.edu/~mwaskom/software/seaborn/tutorial/color_palettes.html), there is a ready made color palette called colorblind. There is not a lot of details on this, so I thought I'd experiment with this and ask for feedback. Unfortunately, only 6 colors are available until it starts recycling itself. That is clearly not good if we want to see patterns in numbers that range from 0 to 9 on each digit.
Another interesting choice is cubehelix. It works in color and grayscale. You can learn more about it here: http://www.mrao.cam.ac.uk/~dag/CUBEHELIX/
First thing first, let's set it as the default palette, and display it.
In [2]:
sns.set_palette(sns.color_palette("cubehelix", 10))
sns.palplot(sns.color_palette())

An infinite source of entertainment

At the very least, an infinite source of digits: pi
So we will be plotting a large grid, with each square representing a digit of pi and filled in the color corresponding to the color in the Seaborn palette. Let's grab a pi digit generator. There is one here that's been around since the days of python 2.5:
https://www.daniweb.com/programming/software-development/code/249177/pi-generator-update
In [3]:
def pi_generate():
    """
    generator to approximate pi
    returns a single digit of pi each time iterated
    """
    q, r, t, k, m, x = 1, 0, 1, 1, 3, 3
    while True:
        if 4 * q + r - t < m * t:
            yield m
            q, r, t, k, m, x = 10*q, 10*(r-m*t), t, k, (10*(3*q+r))//t - 10*m, x
        else:
            q, r, t, k, m, x = q*k, (2*q+r)*x, t*x, k+1, (q*(7*k+2)+r*x)//(t*x), x+2

The 10 colors of Pi

We are now ready to do our actual visualization of pi.
For legal size paper, 55x97 is a good size (plus the ratio is the square root of pi). For poster, I use 154 x 204 = 31416 digits :)
It'll run for a while, even at the reduced size, probably a good time to go and get your favorite drink...
In [4]:
width = 55  # 154
height = 97  # 204 
digit = pi_generate()

fig = plt.figure(figsize=((width + 2) / 3., (height + 2) / 3.))
ax = fig.add_axes((0.05, 0.05, 0.9, 0.9),
                  aspect='equal', frameon=False,
                  xlim=(-0.05, width + 0.05),
                  ylim=(-0.05, height + 0.05))

for axis in (ax.xaxis, ax.yaxis):
    axis.set_major_formatter(plt.NullFormatter())
    axis.set_major_locator(plt.NullLocator())
    
for j in range(height-1,-1,-1):
    for i in range(width):
        pi_digit = next(digit)
        ax.add_patch(Rectangle((i, j),
                               width=1,
                               height=1,
                               ec=sns.color_palette()[pi_digit],
                               fc=sns.color_palette()[pi_digit],
                              )
                    )
        ax.text(i + 0.5, j + 0.5,
                pi_digit, color='k',
                fontsize=10,
                ha='center', va='center')
ax.text(0,-1,"'THE 10 COLORS OF PI' by Francois Dion", fontsize=15)
Out[4]:
<matplotlib.text.Text at 0x10966f1d0>

The original jupyter notebook can be downloaded here:
https://github.com/fdion/infographics_research/blob/master/The%2010%20colors%20of%20Pi.ipynb

Francois Dion
@f_dion

4 comments:

Andrew Oakley said...

I'm heavily colour blind, and can confirm that the Seaborn Cubehelix 10 palette is pretty good. I do get the 4th and 5th colour confused, though, so it's not quite as distinct as the Seaborn Colorblind palette - but, as you say, that doesn't have enough hues. The Seaborn Cubehelix 8 palette is entirely distinguishable for me, so my solution would be... calculate Pi on octal ;-)

Francois Dion said...

Now that you mention it, I looked at cubehelix under a bit more scrutiny, and I see what you are talking about, using a filter to simulate protanopia.

I might have to try my hand at a home-brewed palette.

Andrew Oakley said...

You have to bear in mind that few people are as colour-blind as I am. For the vast, vast majority of colour blind viewers, the Seaborn Colorblind palette will be distinguishable. If you want to cater for the extremes, then labelling is required - at which point you might as well just use the digits 0-9. If you try to cater for the extremes of colour blindness, then you also need to ask yourself why you're catering for them and not for the similar, if not larger, number of actually blind people - and for extreme colour blind users such as myself, I find it is often easier and quicker to use the accessibility features designed, or suited, for totally blind people. For example I do a lot of web reading using the text-only elinks browser in a terminal that's set to monochrome (yellow on black works best for me, although emulating a 1970s green-screen is a good balance of usability and retro caché), as I find text browsers considerably less distracting and easier to distinguish detail than a full-colour graphical browser.

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