根据相互数量对网络中用户之间的连接进行分组
Grouping connections between users in a network based on number of mutuals
我有一个元组列表。每个元组代表社交网络中的一个人。第一项是他们的 ID 或“名称”。第二个是字典;每个键都是网络中与他们有相互联系的另一个人,其值是他们有多少共同点。
network = [
(6, {3: 3, 4: 3, 7: 2, 1: 3, 11: 2}),
(1, {7: 3, 11: 4, 6: 3, 4: 3}),
(4, {3: 2, 6: 3, 1: 3, 11: 2, 12: 3}),
(2, {9: 4, 8: 2, 10: 2, 5: 2}),
(12, {3: 2, 4: 3}),
(3, {5: 2, 8: 2, 12: 2, 4: 2, 7: 2, 6: 3}),
(10, {2: 2, 9: 3, 8: 3, 5: 2}),
(5, {3: 2, 8: 3, 9: 4, 10: 2, 2: 2}),
(13, {}),
(8, {2: 2, 9: 3, 10: 3, 3: 2, 5: 3}),
(7, {3: 2, 6: 2, 1: 3}),
(11, {1: 4, 6: 2, 4: 2}),
(9, {2: 4, 8: 3, 10: 3, 5: 4}),
]
如果两个人有 1、2 或 3 个共同点,他们 可能 认识对方。如果他们有 4 个互惠生,他们 可能 彼此认识。我想处理这个列表,这样我就可以确定谁 might/probably 知道谁,结果输出如下:
Name: 1
Might know: 4, 6, 7
Probably knows: 11
Name: 2
Might know: 5, 8, 10
Probably knows: 9
Name: 3
Might know: 4, 5, 6, 7, 8, 12
Probably knows:
Name: 4
Might know: 1, 3, 6, 11, 12
Probably knows:
Name: 5
Might know: 2, 3, 8, 10
Probably knows: 9
Name: 6
Might know: 1, 3, 4, 7, 11
Probably knows:
Name: 7
Might know: 1, 3, 6
Probably knows:
Name: 8
Might know: 2, 3, 5, 9, 10
Probably knows:
Name: 9
Might know: 8, 10
Probably knows: 2, 5
Name: 10
Might know: 2, 5, 8, 9
Probably knows:
Name: 11
Might know: 4, 6
Probably knows: 1
Name: 12
Might know: 3, 4
Probably knows:
这是我目前用来处理它的代码:
might = []
probably = []
for person in network:
name = person[0]
connections = person[1]
for other_name, mutuals in connections.items():
if mutuals > 3:
probably.append(str(other_name))
else:
might.append(str(other_name))
但我只得到了两个列表:
['3', '4', '7', '1', '11', '7', '6', '4', '3', '6', '1', '11', '12', '8', '10',
'5', '3', '4', '5', '8', '12', '4', '7', '6', '2', '9', '8', '5', '3', '8',
'10', '2', '2', '9', '10', '3', '5', '3', '6', '1', '6', '4', '8', '10']
['11', '9', '9', '1', '2', '5']
如何将这些与专有名称相关联?
你想要的输出本质上是一个字典,所以这样构建它是有意义的。每个键都是一个名字;每个值将是另一个字典,键为 'might' 和 'probably'。 (它的值都是列表。)
output = {}
for name, connections in network:
# If we've not added this name yet, create a blank entry:
if name not in output:
output[name] = {'probably': [], 'might': []}
# Now loop through the connected people and add to the correct list:
for other_name, mutuals in connections.items():
if mutuals > 3:
output[name]['probably'].append(other_name)
else:
output[name]['might'].append(other_name)
此时,我们可以使用 Python 的 pprint
函数来检查我们是否走在正确的轨道上。 (对于像这样的嵌套结构,它比 print
更具可读性。)
from pprint import pprint
pprint(output)
输出
{1: {'might': [7, 6, 4], 'probably': [11]},
2: {'might': [8, 10, 5], 'probably': [9]},
3: {'might': [5, 8, 12, 4, 7, 6], 'probably': []},
4: {'might': [3, 6, 1, 11, 12], 'probably': []},
5: {'might': [3, 8, 10, 2], 'probably': [9]},
6: {'might': [3, 4, 7, 1, 11], 'probably': []},
7: {'might': [3, 6, 1], 'probably': []},
8: {'might': [2, 9, 10, 3, 5], 'probably': []},
9: {'might': [8, 10], 'probably': [2, 5]},
10: {'might': [2, 9, 8, 5], 'probably': []},
11: {'might': [6, 4], 'probably': [1]},
12: {'might': [3, 4], 'probably': []},
13: {'might': [], 'probably': []}}
(请注意,pprint
会自动对要显示的键进行排序:它们实际上并非按该顺序排列。)
现在我们需要做的就是将其格式化以供显示,我们可以随心所欲。到目前为止,我将名称保留为整数,以便我们可以正确地对它们进行排序(而不是让 11
在 2
之前结束,就像在对字符串进行排序时那样)。如果这些赋值看起来很复杂,请查看 str.join
and list comprehensions. And you may or may not know about f-strings,它们也非常方便(并且不需要变量甚至是字符串!)
for name, contents in sorted(output.items()):
print(f'Name: {name}')
might = ', '.join([str(i) for i in sorted(contents['might'])])
print(f'\tMight know: {might}')
probably = ', '.join([str(i) for i in sorted(contents['probably'])])
print(f'\tProbably knows: {probably}')
输出:
Name: 1
Might know: 4, 6, 7
Probably knows: 11
Name: 2
Might know: 5, 8, 10
Probably knows: 9
Name: 3
Might know: 4, 5, 6, 7, 8, 12
Probably knows:
Name: 4
Might know: 1, 3, 6, 11, 12
Probably knows:
Name: 5
Might know: 2, 3, 8, 10
Probably knows: 9
Name: 6
Might know: 1, 3, 4, 7, 11
Probably knows:
Name: 7
Might know: 1, 3, 6
Probably knows:
Name: 8
Might know: 2, 3, 5, 9, 10
Probably knows:
Name: 9
Might know: 8, 10
Probably knows: 2, 5
Name: 10
Might know: 2, 5, 8, 9
Probably knows:
Name: 11
Might know: 4, 6
Probably knows: 1
Name: 12
Might know: 3, 4
Probably knows:
Name: 13
Might know:
Probably knows:
我有一个元组列表。每个元组代表社交网络中的一个人。第一项是他们的 ID 或“名称”。第二个是字典;每个键都是网络中与他们有相互联系的另一个人,其值是他们有多少共同点。
network = [
(6, {3: 3, 4: 3, 7: 2, 1: 3, 11: 2}),
(1, {7: 3, 11: 4, 6: 3, 4: 3}),
(4, {3: 2, 6: 3, 1: 3, 11: 2, 12: 3}),
(2, {9: 4, 8: 2, 10: 2, 5: 2}),
(12, {3: 2, 4: 3}),
(3, {5: 2, 8: 2, 12: 2, 4: 2, 7: 2, 6: 3}),
(10, {2: 2, 9: 3, 8: 3, 5: 2}),
(5, {3: 2, 8: 3, 9: 4, 10: 2, 2: 2}),
(13, {}),
(8, {2: 2, 9: 3, 10: 3, 3: 2, 5: 3}),
(7, {3: 2, 6: 2, 1: 3}),
(11, {1: 4, 6: 2, 4: 2}),
(9, {2: 4, 8: 3, 10: 3, 5: 4}),
]
如果两个人有 1、2 或 3 个共同点,他们 可能 认识对方。如果他们有 4 个互惠生,他们 可能 彼此认识。我想处理这个列表,这样我就可以确定谁 might/probably 知道谁,结果输出如下:
Name: 1
Might know: 4, 6, 7
Probably knows: 11
Name: 2
Might know: 5, 8, 10
Probably knows: 9
Name: 3
Might know: 4, 5, 6, 7, 8, 12
Probably knows:
Name: 4
Might know: 1, 3, 6, 11, 12
Probably knows:
Name: 5
Might know: 2, 3, 8, 10
Probably knows: 9
Name: 6
Might know: 1, 3, 4, 7, 11
Probably knows:
Name: 7
Might know: 1, 3, 6
Probably knows:
Name: 8
Might know: 2, 3, 5, 9, 10
Probably knows:
Name: 9
Might know: 8, 10
Probably knows: 2, 5
Name: 10
Might know: 2, 5, 8, 9
Probably knows:
Name: 11
Might know: 4, 6
Probably knows: 1
Name: 12
Might know: 3, 4
Probably knows:
这是我目前用来处理它的代码:
might = []
probably = []
for person in network:
name = person[0]
connections = person[1]
for other_name, mutuals in connections.items():
if mutuals > 3:
probably.append(str(other_name))
else:
might.append(str(other_name))
但我只得到了两个列表:
['3', '4', '7', '1', '11', '7', '6', '4', '3', '6', '1', '11', '12', '8', '10',
'5', '3', '4', '5', '8', '12', '4', '7', '6', '2', '9', '8', '5', '3', '8',
'10', '2', '2', '9', '10', '3', '5', '3', '6', '1', '6', '4', '8', '10']
['11', '9', '9', '1', '2', '5']
如何将这些与专有名称相关联?
你想要的输出本质上是一个字典,所以这样构建它是有意义的。每个键都是一个名字;每个值将是另一个字典,键为 'might' 和 'probably'。 (它的值都是列表。)
output = {}
for name, connections in network:
# If we've not added this name yet, create a blank entry:
if name not in output:
output[name] = {'probably': [], 'might': []}
# Now loop through the connected people and add to the correct list:
for other_name, mutuals in connections.items():
if mutuals > 3:
output[name]['probably'].append(other_name)
else:
output[name]['might'].append(other_name)
此时,我们可以使用 Python 的 pprint
函数来检查我们是否走在正确的轨道上。 (对于像这样的嵌套结构,它比 print
更具可读性。)
from pprint import pprint
pprint(output)
输出
{1: {'might': [7, 6, 4], 'probably': [11]},
2: {'might': [8, 10, 5], 'probably': [9]},
3: {'might': [5, 8, 12, 4, 7, 6], 'probably': []},
4: {'might': [3, 6, 1, 11, 12], 'probably': []},
5: {'might': [3, 8, 10, 2], 'probably': [9]},
6: {'might': [3, 4, 7, 1, 11], 'probably': []},
7: {'might': [3, 6, 1], 'probably': []},
8: {'might': [2, 9, 10, 3, 5], 'probably': []},
9: {'might': [8, 10], 'probably': [2, 5]},
10: {'might': [2, 9, 8, 5], 'probably': []},
11: {'might': [6, 4], 'probably': [1]},
12: {'might': [3, 4], 'probably': []},
13: {'might': [], 'probably': []}}
(请注意,pprint
会自动对要显示的键进行排序:它们实际上并非按该顺序排列。)
现在我们需要做的就是将其格式化以供显示,我们可以随心所欲。到目前为止,我将名称保留为整数,以便我们可以正确地对它们进行排序(而不是让 11
在 2
之前结束,就像在对字符串进行排序时那样)。如果这些赋值看起来很复杂,请查看 str.join
and list comprehensions. And you may or may not know about f-strings,它们也非常方便(并且不需要变量甚至是字符串!)
for name, contents in sorted(output.items()):
print(f'Name: {name}')
might = ', '.join([str(i) for i in sorted(contents['might'])])
print(f'\tMight know: {might}')
probably = ', '.join([str(i) for i in sorted(contents['probably'])])
print(f'\tProbably knows: {probably}')
输出:
Name: 1
Might know: 4, 6, 7
Probably knows: 11
Name: 2
Might know: 5, 8, 10
Probably knows: 9
Name: 3
Might know: 4, 5, 6, 7, 8, 12
Probably knows:
Name: 4
Might know: 1, 3, 6, 11, 12
Probably knows:
Name: 5
Might know: 2, 3, 8, 10
Probably knows: 9
Name: 6
Might know: 1, 3, 4, 7, 11
Probably knows:
Name: 7
Might know: 1, 3, 6
Probably knows:
Name: 8
Might know: 2, 3, 5, 9, 10
Probably knows:
Name: 9
Might know: 8, 10
Probably knows: 2, 5
Name: 10
Might know: 2, 5, 8, 9
Probably knows:
Name: 11
Might know: 4, 6
Probably knows: 1
Name: 12
Might know: 3, 4
Probably knows:
Name: 13
Might know:
Probably knows: