python - Matplotlib scatter plot with legend

Question

I want to create a Matplotlib scatter plot, with a legend showing the colour for each class. For example, I have a list of x and y values, and a list of classes values. Each element in the x, y and classes lists corresponds to one point in the plot. I want each class to have its own colour, which I have already coded, but then I want the classes to be displayed in a legend. What paramaters do I pass to the legend() function to achieve this?

Here is my code so far:

import matplotlib.pyplot as plt
x = [1, 3, 4, 6, 7, 9]
y = [0, 0, 5, 8, 8, 8]
classes = ['A', 'A', 'B', 'C', 'C', 'C']
colours = ['r', 'r', 'b', 'g', 'g', 'g']
plt.scatter(x, y, c=colours)
plt.show()

Answer 1

First, I have a feeling you meant to use apostrophes, not backticks when declaring colours.

For a legend you need some shapes as well as the classes. For example, the following creates a list of rectangles called recs for each colour in class_colours.

import matplotlib.patches as mpatches

classes = ['A','B','C']
class_colours = ['r','b','g']
recs = []
for i in range(0,len(class_colours)):
    recs.append(mpatches.Rectangle((0,0),1,1,fc=class_colours[i]))
plt.legend(recs,classes,loc=4)

There is a second way of creating a legend, in which you specify the "Label" for a set of points using a separate scatter command for each set. An example of this is given below.

classes = ['A','A','B','C','C','C']
colours = ['r','r','b','g','g','g']
for (i,cla) in enumerate(set(classes)):
    xc = [p for (j,p) in enumerate(x) if classes[j]==cla]
    yc = [p for (j,p) in enumerate(y) if classes[j]==cla]
    cols = [c for (j,c) in enumerate(colours) if classes[j]==cla]
    plt.scatter(xc,yc,c=cols,label=cla)
plt.legend(loc=4)

The first method is the one I've personally used, the second I just found looking at the matplotlib documentation. Since the legends were covering datapoints I moved them, and the locations for legends can be found here. If there's another way to make a legend, I wasn't able to find it after a few quick searches in the docs.