String formatting with permutations of dictionary values

Question

Here is what I currently have:

from collections import defaultdict

output = [{'MPID': 'A', 'CLIENTNAME': 'AAA'},
          {'MPID': 'C', 'CLIENTNAME': 'BBB'},
          {'MPID': 'C1', 'CLIENTNAME': 'CCC'},
          {'MPID': 'C2', 'CLIENTNAME': 'CCC'},
          {'MPID': 'C3', 'CLIENTNAME': 'CCC'}]

d = defaultdict(list)

for item in output:
    d[item['CLIENTNAME']].append(item['MPID'])

final = [{'CLIENTNAME': k, 'MPID': v} for k, v in d.items()]
print final

This output merges the MPID values of matching CLIENTNAMES.

Output:

[{'MPID': ['A'], 'CLIENTNAME': 'AAA'}, 
 {'MPID': ['C'], 'CLIENTNAME': 'BBB'}, 
 {'MPID': ['C1', 'C2', 'C3'], 'CLIENTNAME': 'CCC'}]

What I am trying to do now is format a string with all permutations of each MPID, but ONLY if the dictionary contains more than 1 MPID. (in this example, only CCC has more than 1 MPID).

Here is the query I am formatting:

query = '''x = '{}' and y = '{}' union 
           x = '{}' and y = '{}';'''

This query needs to compare all MPIDS against one another. The desired output would be:

'''x = 'C1' and y = 'C2' union 
   x = 'C2' and y = 'C1';'''

'''x = 'C2' and y = 'C3' union 
   x = 'C3' and y = 'C2';'''

'''x = 'C1' and y = 'C3' union 
   x = 'C3' and y = 'C1';'''

As you can see, X and Y values swap places in the second line of the string.

What would be an efficient way of going about this part?

Thanks.

Zev · Accepted Answer · 2018-06-14 17:39:33Z

1

You can use combinations from itertools

from itertools import combinations

combos = list(combinations(final[2]['MPID'], 2))
combos.sort(key=lambda x: x[1])
for combo in combos:
    a, b = combo
    print '''x = '{}' and y = '{}' union 
               x = '{}' and y = '{}';'''.format(a, b, b, a)

This prints:

x = 'C1' and y = 'C2' union 
               x = 'C2' and y = 'C1';
x = 'C1' and y = 'C3' union 
               x = 'C3' and y = 'C1';
x = 'C2' and y = 'C3' union 
               x = 'C3' and y = 'C2';

You may need to adjust the sort key if that order matters to you.

answered Jun 14, 2018 at 17:39

Zev

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5 Comments

Harrison Over a year ago

Hmm, this isn't doing anything. If I print out combos after you define it, it's an empty list.

Zev Over a year ago

is final[2]['MPID'] the correct target for how you have it?

Harrison Over a year ago

That was just an example output. It can be different. The 3rd one in the output isn't always the one that has multiple values. It should be dynamic.

Zev Over a year ago

Right, I mean you could loop over it and do this but I just wanted to show you how you would generally go about it.

Zev Over a year ago

If you are having trouble getting my example to run, you can see it running at: repl.it/repls/VastMistyProcess. I added a loop there. Nothing will print if there are no combinations to work with.

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