我有一个设备字典，每个设备都有一个可以与之协作的其他设备的列表。将它们视为与之兼容的设备。
devices = {0: [], 1: [9, 10, 14], 2: [9, 11, 14], 3: [], 4: [], 5: [9, 14, 15], 6: [], 7: [], 8: [], 9: [1, 2, 5, 11, 14, 15], 10: [1], 11: [2, 9, 14], 12: [], 13: [], 14: [1, 2, 5, 9, 11, 16], 15: [5, 9], 16: [14], 17: [], 18: [], 19: []}
我希望能够将它们群集到一个列表列表中，该列表包含可以一起协作的所有可能的群集。群集只能包含彼此兼容的设备。换句话说，如果集群[1,9,2]存在，则1必须同时与9和2协作，并且它们也必须彼此协作。在这种情况下，最终结果应如下所示：
[ [1, 9, 14], [2, 9, 14], [5,9,14], [5,15,9] [9,11,14,2], [10,1], [14,16] ]
我可能在手动计算时犯了一个错误，但是我相信这是所有可能的物品簇，同时满足它们的兼容性要求。
但是，在将其转换为代码时遇到一些困难。任何帮助将不胜感激。
答案 0 :(得分：1)
也许会有更干净的方法来获得结果，但是此代码有效。
请注意，您的结果包括[2，9，14]，它是[9,11,14,2]的子集。此代码中的子集已删除。
items = {0: [], 1: [9, 10, 14], 2: [9, 11, 14], 3: [], 4: [], 5: [9, 14, 15], 6: [], 7: [], 8: [],
9: [1, 2, 5, 11, 14, 15], 10: [1], 11: [2, 9, 14], 12: [], 13: [], 14: [1, 2, 5, 9, 11, 16],
15: [5, 9], 16: [14], 17: [], 18: [], 19: []}
grps = []
# create 1-1 groups
for g in items:
for c in items[g]:
if g in items[c]:
grps.append([g,c])
# add other elements to each group
chg = True
while chg: # until no more elements added
chg = False
for g in items: # each single element
for g2 in grps: # check each existing group
if g in g2: continue # if not already in group
ok = True
for c in g2: # check each group member
if not (g in items[c] and c in items[g]): # can we collaborate?
ok = False # no
if ok:
g2.append(g) # add element to group
chg = True
# check for subsets
for i,x in enumerate(grps):
for j,y in enumerate(grps):
if i==j: continue # same group
if set(x) & set(y) == set(x): # if subset
x.clear() # remove elements
grps = [g for g in grps if len(g)] # remove empty groups
print(grps)
输出
[[10, 1], [14, 5, 9], [14, 9, 1], [14, 11, 2, 9], [15, 9, 5], [16, 14]]
答案 1 :(得分：0)
您可以这样做的一种方式：
from itertools import combinations
colaborate2 = lambda args, dic : all(x in dic[y] and y in dic[x] for x, y in combinations(args,2))
def group (x, dic, n=None):
if n is None:
n = len(x)
for i in combinations(x, n):
if colaborate2(i, dic):
return tuple(sorted(i))
if n > 1:
return group(x, dic, n - 1)
def groupings(dic):
m = set([group(val + [key], dic) for key, val in dic.items()])
return [i for i in m if len(i)>1]
groupings(items)
[(5, 9, 15), (14, 16), (5, 9, 14), (1, 9, 14), (1, 10), (2, 9, 11, 14)]
答案 2 :(得分：0)
尽管花了我几个小时，但我对干净整洁的解决方案感到满意。 90％的时间里，在编码时了解需求。经验教训-首先使用铅笔和纸。
代码：
devices = {0: [], 1: [9, 10, 14], 2: [9, 11, 14], 3: [], 4: [], 5: [9, 14, 15], 6: [], 7: [], 8: [], 9: [1, 2, 5, 11, 14, 15], 10: [1], 11: [2, 9, 14], 12: [], 13: [], 14: [1, 2, 5, 9, 11, 16], 15: [5, 9], 16: [14], 17: [], 18: [], 19: []}
# manually calculated expected result [ [1, 9, 14], [2, 9, 14], [5,9,14], [5,15,9] [9,11,14,2], [10,1], [14,16] ]
# true result [[1, 9, 14], [2, 9, 11, 14], [5, 9, 14], [10, 1], [15, 5, 9], [16, 14]]
def is_compatible(k,v):
if v in devices[k]:
return True
return False
clusters = []
for k,v_list in devices.items():
if v_list:
cluster = []
cluster.append(k)
cluster.append(v_list[0])
for v_ele in v_list[1:]:
v_ele_further_check = True
for cluster_member in cluster[1:]: # check v_ele compatible with existing cluster members
if not is_compatible(v_ele, cluster_member):
v_ele_further_check = False
break
if v_ele_further_check:
cluster.append(v_ele)
clusters.append(cluster)
print(clusters)
unique_clusters = []
for cluster in clusters: # eliminate_duplicates
if not sorted(cluster) in unique_clusters:
unique_clusters.append(cluster)
print(unique_clusters)
输出：
[[1, 9, 14], [2, 9, 11, 14], [5, 9, 14], [9, 1, 14], [10, 1], [11, 2, 9, 14], [14, 1, 9], [15, 5, 9], [16, 14]]
[[1, 9, 14], [2, 9, 11, 14], [5, 9, 14], [10, 1], [15, 5, 9], [16, 14]]
评论：
您可以看到我可以轻松摆脱is_compatible函数，但是它增加了可读性。 如果重要的话，可以正确维护群集中群集的顺序。