散点图标记尺寸计算

Question

I have a very specific problem. I have two numpy arrays and the corresponding element of each array represent a 2d point.

a = [1,2,1,6,1]

b = [5,0,3,1,5]

I want to plot a scatter plot where the size of the marker is based on how many times than point occurs.

That is :

1,5 : 2

2,0 : 1

1,3 : 1

6,1 : 1

So the size array must be size = [2,1,1,1] and other two arrays can be

a = [1,2,1,6] and b = [5,0,3,1]

So I must be able to call plt.scatter as follows:

plt.scatter(a,b,s=size)

Answer 1

Since the question is tagged with numpy, we might use numpy. numpy.unique allows to calculate the counts of unique values of an array.

import numpy as np

a = [1,2,1,6,1]
b = [5,0,3,1,5]

u, c = np.unique(np.c_[a,b], return_counts=True, axis=0)

then

# u=
[[1 3]
 [1 5]
 [2 0]
 [6 1]]
# c= 
[1 2 1 1]

This can be plotted like so, where an additional function may be used to normalize the counts to some point sizes for plotting

import matplotlib.pyplot as plt
s = lambda x : (((x-x.min())/float(x.max()-x.min())+1)*8)**2

plt.scatter(u[:,0],u[:,1],s=s(c))

plt.show()

Answer 2

This will do what you want:

from collections import Counter

a = [1, 2, 1, 6, 1]
b = [5, 0, 3, 1, 5]

counts = Counter([(x, y) for x, y in zip(a, b)])

size = [counts[(x, y)] for x, y in zip(a, b)]

counter will keep track of how many times each point appears in your arrays. Then size gets that number from counter .

Note that you actually want size = [2, 1, 1, 1, 2] because you need s to be the same size as your input arrays. This won't matter though; you'll just plot the same point twice.

If you really do want to remove the duplicates, you could do the same thing, but add an extra step, where you create a set of points.

from collections import Counter

a = [1, 2, 1, 6, 1]
b = [5, 0, 3, 1, 5]

counts = Counter([(x, y) for x, y in zip(a, b)])

points = set([(x, y) for x, y in zip(a, b)])
a = list()
b = list()
for x, y in points:
    a.append(x)
    b.append(y)

size = [counts[(x, y)] for x, y in zip(a, b)]