[python] How to generate random colors in matplotlib?

What's the trivial example of how to generate random colors for passing to plotting functions?

I'm calling scatter inside a loop and want each plot a different color.

for X,Y in data:
   scatter(X, Y, c=??)

c: a color. c can be a single color format string, or a sequence of color specifications of length N, or a sequence of N numbers to be mapped to colors using the cmap and norm specified via kwargs (see below). Note that c should not be a single numeric RGB or RGBA sequence because that is indistinguishable from an array of values to be colormapped. c can be a 2-D array in which the rows are RGB or RGBA, however.

This question is related to python matplotlib

The answer is


enter code here

import numpy as np

clrs = np.linspace( 0, 1, 18 )  # It will generate 
# color only for 18 for more change the number
np.random.shuffle(clrs)
colors = []
for i in range(0, 72, 4):
    idx = np.arange( 0, 18, 1 )
    np.random.shuffle(idx)
    r = clrs[idx[0]]
    g = clrs[idx[1]]
    b = clrs[idx[2]]
    a = clrs[idx[3]]
    colors.append([r, g, b, a])

for X,Y in data:
   scatter(X, Y, c=numpy.random.rand(3,))

Here is a more concise version of Ali's answer giving one distinct color per plot :

import matplotlib.pyplot as plt

N = len(data)
cmap = plt.cm.get_cmap("hsv", N+1)
for i in range(N):
    X,Y = data[i]
    plt.scatter(X, Y, c=cmap(i))

Since the question is How to generate random colors in matplotlib? and as I was searching for an answer concerning pie plots, I think it is worth to put an answer here (for pies)

import numpy as np
from random import sample
import matplotlib.pyplot as plt
import matplotlib.colors as pltc
all_colors = [k for k,v in pltc.cnames.items()]

fracs = np.array([600, 179, 154, 139, 126, 1185])
labels = ["label1", "label2", "label3", "label4", "label5", "label6"]
explode = ((fracs == max(fracs)).astype(int) / 20).tolist()

for val in range(2):
    colors = sample(all_colors, len(fracs))
    plt.figure(figsize=(8,8))
    plt.pie(fracs, labels=labels, autopct='%1.1f%%', 
            shadow=True, explode=explode, colors=colors)
    plt.legend(labels, loc=(1.05, 0.7), shadow=True)
    plt.show()

Output

enter image description here

enter image description here


elaborating @john-mee 's answer, if you have arbitrarily long data but don't need strictly unique colors:

for python 2:

from itertools import cycle
cycol = cycle('bgrcmk')

for X,Y in data:
    scatter(X, Y, c=cycol.next())

for python 3:

from itertools import cycle
cycol = cycle('bgrcmk')

for X,Y in data:
    scatter(X, Y, c=next(cycol))

this has the advantage that the colors are easy to control and that it's short.


Based on Ali's and Champitoad's answer:

If you want to try different palettes for the same, you can do this in a few lines:

cmap=plt.cm.get_cmap(plt.cm.viridis,143)

^143 being the number of colours you're sampling

I picked 143 because the entire range of colours on the colormap comes into play here. What you can do is sample the nth colour every iteration to get the colormap effect.

n=20 for i,(x,y) in enumerate(points): plt.scatter(x,y,c=cmap(n*i))


For some time I was really annoyed by the fact that matplotlib doesn't generate colormaps with random colors, as this is a common need for segmentation and clustering tasks.

By just generating random colors we may end with some that are too bright or too dark, making visualization difficult. Also, usually we need the first or last color to be black, representing the background or outliers. So I've wrote a small function for my everyday work

Here's the behavior of it:

new_cmap = rand_cmap(100, type='bright', first_color_black=True, last_color_black=False, verbose=True)

Generated colormap

Than you just use new_cmap as your colormap on matplotlib:

ax.scatter(X,Y, c=label, cmap=new_cmap, vmin=0, vmax=num_labels)

The code is here:

def rand_cmap(nlabels, type='bright', first_color_black=True, last_color_black=False, verbose=True):
    """
    Creates a random colormap to be used together with matplotlib. Useful for segmentation tasks
    :param nlabels: Number of labels (size of colormap)
    :param type: 'bright' for strong colors, 'soft' for pastel colors
    :param first_color_black: Option to use first color as black, True or False
    :param last_color_black: Option to use last color as black, True or False
    :param verbose: Prints the number of labels and shows the colormap. True or False
    :return: colormap for matplotlib
    """
    from matplotlib.colors import LinearSegmentedColormap
    import colorsys
    import numpy as np


    if type not in ('bright', 'soft'):
        print ('Please choose "bright" or "soft" for type')
        return

    if verbose:
        print('Number of labels: ' + str(nlabels))

    # Generate color map for bright colors, based on hsv
    if type == 'bright':
        randHSVcolors = [(np.random.uniform(low=0.0, high=1),
                          np.random.uniform(low=0.2, high=1),
                          np.random.uniform(low=0.9, high=1)) for i in xrange(nlabels)]

        # Convert HSV list to RGB
        randRGBcolors = []
        for HSVcolor in randHSVcolors:
            randRGBcolors.append(colorsys.hsv_to_rgb(HSVcolor[0], HSVcolor[1], HSVcolor[2]))

        if first_color_black:
            randRGBcolors[0] = [0, 0, 0]

        if last_color_black:
            randRGBcolors[-1] = [0, 0, 0]

        random_colormap = LinearSegmentedColormap.from_list('new_map', randRGBcolors, N=nlabels)

    # Generate soft pastel colors, by limiting the RGB spectrum
    if type == 'soft':
        low = 0.6
        high = 0.95
        randRGBcolors = [(np.random.uniform(low=low, high=high),
                          np.random.uniform(low=low, high=high),
                          np.random.uniform(low=low, high=high)) for i in xrange(nlabels)]

        if first_color_black:
            randRGBcolors[0] = [0, 0, 0]

        if last_color_black:
            randRGBcolors[-1] = [0, 0, 0]
        random_colormap = LinearSegmentedColormap.from_list('new_map', randRGBcolors, N=nlabels)

    # Display colorbar
    if verbose:
        from matplotlib import colors, colorbar
        from matplotlib import pyplot as plt
        fig, ax = plt.subplots(1, 1, figsize=(15, 0.5))

        bounds = np.linspace(0, nlabels, nlabels + 1)
        norm = colors.BoundaryNorm(bounds, nlabels)

        cb = colorbar.ColorbarBase(ax, cmap=random_colormap, norm=norm, spacing='proportional', ticks=None,
                                   boundaries=bounds, format='%1i', orientation=u'horizontal')

    return random_colormap

It's also on github: https://github.com/delestro/rand_cmap


If you want to ensure the colours are distinct - but don't know how many colours are needed. Try something like this. It selects colours from opposite sides of the spectrum and systematically increases granularity.

import math

def calc(val, max = 16):
    if val < 1:
        return 0
    if val == 1:
        return max

    l = math.floor(math.log2(val-1))    #level 
    d = max/2**(l+1)                    #devision
    n = val-2**l                        #node
    return d*(2*n-1)
import matplotlib.pyplot as plt

N = 16
cmap = cmap = plt.cm.get_cmap('gist_rainbow', N)

fig, axs = plt.subplots(2)
for ax in axs:
    ax.set_xlim([ 0, N])
    ax.set_ylim([-0.5, 0.5])
    ax.set_yticks([])

for i in range(0,N+1):
    v = int(calc(i, max = N))
    rect0 = plt.Rectangle((i, -0.5), 1, 1, facecolor=cmap(i))
    rect1 = plt.Rectangle((i, -0.5), 1, 1, facecolor=cmap(v))
    axs[0].add_artist(rect0)
    axs[1].add_artist(rect1)

plt.xticks(range(0, N), [int(calc(i, N)) for i in range(0, N)])
plt.show()

output

Thanks to @Ali for providing the base implementation.


When less than 9 datasets:

colors = "bgrcmykw"
color_index = 0

for X,Y in data:
    scatter(X,Y, c=colors[color_index])
    color_index += 1

Improving the answer https://stackoverflow.com/a/14720445/6654512 to work with Python3. That piece of code would sometimes generate numbers greater than 1 and matplotlib would throw an error.

for X,Y in data:
   scatter(X, Y, c=numpy.random.random(3))