CODEWITHSUNDEEP
pythonpython-3.xlistdictionarygrouping
iby.helmy
iby.helmy

Reputation: 190

Creating grouped lists based on shared dictionary values in Python3

I have a dictionary of devices where each device has a list of other devices that it can collaborate with. Think of these as the devices that it is compatible to.

 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: []}

I would like to be able to cluster these into a list of lists that contains all possible clusters that can collaborate together. A cluster can only contain devices that are all compatible with each other. In other words, if a cluster [1,9,2] exists then 1 must collaborate with both 9 and 2 and they must collaborate with each other as well. In this scenario the final result should look something like this:

[ [1, 9, 14], [2, 9, 14], [5,9,14], [5,15,9] [9,11,14,2], [10,1], [14,16]  ]

I may have made an error manually computing this, but I believe this to be all possible clusters of the items while satisfying their compatibility requirements.

However, I am having some difficulties translating this into code. Any help would be tremendously appreciated.

Upvotes: 1

Views: 60

Answers (3)

Aaj Kaal
Aaj Kaal

Reputation: 1284

Though it took me few hours I am happy with the neat and clean solution. Understanding the requirement while coding took me 90% of the time. Lesson learnt - use pencil and paper first.

Code:

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)

Output:

[[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]]

Comments:

You can see I can easily get rid of function is_compatible but it adds to the readability. The order of cluster in clusters is maintained correctly if that matters.

Upvotes: 0

Onyambu
Onyambu

Reputation: 79228

One way you could do this:

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)]

Upvotes: 0

Mike67
Mike67

Reputation: 11342

There may be a cleaner way to get the result, but this code works.

Note your result includes [2, 9, 14] which is a subset of [9,11,14,2]. The subset is removed in this code.

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)

Output

[[10, 1], [14, 5, 9], [14, 9, 1], [14, 11, 2, 9], [15, 9, 5], [16, 14]]

Upvotes: 1

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