这是一个自动化整个事情的例子。它创建一个临时字典，其中包含 venn 所需的 id 作为键以及该 id 的所有参与集的交集。
如果您不希望对标签进行排序，请删除倒数第二行中的 sorted() 调用。
import math
from matplotlib import pyplot as plt
from matplotlib_venn import venn2, venn3
import numpy as np
# Convert number to indices into binary
# e.g. 5 -> '101' > [2, 0]
def bits2indices(b):
l = []
if b == 0:
return l
for i in reversed(range(0, int(math.log(b, 2)) + 1)):
if b & (1 << i):
l.append(i)
return l
# Make dictionary containing venn id's and set intersections
# e.g. d = {'100': {'c', 'b', 'a'}, '010': {'c', 'd', 'e'}, ... }
def set2dict(s):
d = {}
for i in range(1, 2**len(s)):
# Make venn id strings
key = bin(i)[2:].zfill(len(s))
key = key[::-1]
ind = bits2indices(i)
# Get the participating sets for this id
participating_sets = [s[x] for x in ind]
# Get the intersections of those sets
inter = set.intersection(*participating_sets)
d[key] = inter
return d
# Define some sets
a = set(['a', 'b', 'c'])
b = set(['c', 'd', 'e'])
c = set(['e', 'f', 'a'])
s = [a, b, c]
# Create dictionary from sets
d = set2dict(s)
# Plot it
h = venn3(s, ('A', 'B', 'C'))
for k, v in d.items():
l = h.get_label_by_id(k)
if l:
l.set_text('\n'.join(sorted(v)))
plt.show()
/edit 对不起，我刚刚发现上面的代码没有删除重复的标签，因此是错误的。venn 显示的元素数量和标签数量不同。这是一个新版本，它从其他交叉路口删除了错误的重复项。我想有一种更聪明、更实用的方法来做到这一点，而不是对所有交叉点进行两次迭代......
import math, itertools
from matplotlib import pyplot as plt
from matplotlib_venn import venn2, venn3
import numpy as np
# Generate list index for itertools combinations
def gen_index(n):
x = -1
while True:
while True:
x = x + 1
if bin(x).count('1') == n:
break
yield x
# Generate all combinations of intersections
def make_intersections(sets):
l = [None] * 2**len(sets)
for i in range(1, len(sets) + 1):
ind = gen_index(i)
for subset in itertools.combinations(sets, i):
inter = set.intersection(*subset)
l[next(ind)] = inter
return l
# Get weird reversed binary string id for venn
def number2venn_id(x, n_fill):
id = bin(x)[2:].zfill(n_fill)
id = id[::-1]
return id
# Iterate over all combinations and remove duplicates from intersections with
# more sets
def sets2dict(sets):
l = make_intersections(sets)
d = {}
for i in range(1, len(l)):
d[number2venn_id(i, len(sets))] = l[i]
for j in range(1, len(l)):
if bin(j).count('1') < bin(i).count('1'):
l[j] = l[j] - l[i]
d[number2venn_id(j, len(sets))] = l[j] - l[i]
return d
# Define some sets
a = set(['a', 'b', 'c', 'f'])
b = set(['c', 'd', 'e'])
c = set(['e', 'f', 'a'])
sets = [a, b, c]
d = sets2dict(sets)
# Plot it
h = venn3(sets, ('A', 'B', 'C'))
for k, v in d.items():
l = h.get_label_by_id(k)
if l:
l.set_fontsize(12)
l.set_text('\n'.join(sorted(v)))
# Original for comparison
f = plt.figure(2)
venn3(sets, ('A', 'B', 'C'))
plt.show()
