[Tutor] A dictionary question

[Martin A. Brown]

Hello,

> Just a quick one while I stop for afternoon tea.
> 
> Using the example provide by Mats, if I print(pairs) I get:
> 
> Counter({(4, 6): 2, (3, 4): 1, (3, 6): 1 etc.
> 
> How would I print the pairs so they look like this:
> 
> (4,6) 2
> (3,4) 1
> (3,6) 1

The collections.Counter behaves like a dictionary.  Same methods 
should work.  So, try something like these basic loops that take 
advantage of the items() method.

  c = Counter({(4, 6): 2, (3, 4): 1, (3, 6): 1})
  
  # -- loop through all of the key, value items in the dict-like object
  for k, v in c.items(): 
      print( k, v)

  # -- same but only print where the counter for that entry is 2
  for k, v in c.items(): 
      if v == 2: 
          print( k, v) 

Good luck, enjoy the tea (for a change of pace) and leave your 
dictionaries cleaner than when you arrived.  Only you can stop code 
rot!

-Martin


> My feeble attempt is under the example code. I have looked at a couple of
> collections tutorials but I cannot relate what I'm seeing to the example
> below.
> 
> I'm also wondering how I might do something like this:
> 
> if count == 2:
> 
>     do this
> 
> row = [{7,3},{5},{4,6,8},{7,8},{1},{9,3},{7,9},{4,6,3},{2}]
> #row = [{5},{6},{3},{4,7,5},{1,2,4,5},{1,7,5},{1,2,4,7,9},{8},{1,2,4,7,9}]
> 
> set_list = [{1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {1, 7}, {1, 8}, {1, 9},
>             {2, 3}, {2, 4}, {2, 5}, {2, 6}, {2, 7}, {2, 8}, {2, 9},
>             {3, 4}, {3, 5}, {3, 6}, {3, 7}, {3, 8}, {3, 9},
>             {4, 5}, {4, 6}, {4, 7}, {4, 8}, {4,9},
>             {5, 6}, {5, 7}, {5, 8}, {5, 9},
>             {6, 7}, {6, 8}, {6, 9},
>             {7, 8}, {7, 9},
>             {8, 9}
>             ]
> 
> from collections import Counter
> 
> pairs = Counter()
> 
> for s in set_list:
>     for r in row:
>         if s.issubset(r):
>             pairs[tuple(sorted(s))] += 1
> 
> print(pairs)
> 
> for i in range(9):
>     print(pairs[i])
> 
> 

(Repeated/resent from the correct email address.)

-- 
Martin A. Brown
http://linux-ip.net/