CODEWITHSUNDEEP
pythoncombinationspython-itertools
ecik
ecik

Reputation: 829

Pairing people in a contest

Assume there is a some kind of a contest in which there are five boys and five girls:

boys = ["Boy1", "Boy2", "Boy3", "Boy4", "Boy5"]
girls = ["Girl1", "Girl2", "Girl3", "Girl4", "Girl5"]

Every boy is drawn against one girl and they play a game. What is the Pythonic way (probably involving itertools?) to yield all combinations of all pairs?

For example a combination could be:

combination1 = [("Boy1", "Girl1"), ("Boy2", "Girl2"), 
                ("Boy3", "Girl3"), ("Boy4", "Girl4"), ("Boy5", "Girl5")]

So Boy1 plays against Girl1 etc. If Boy1 is drawn against Girl1 then no other girl can play against him.

Upvotes: 0

Views: 250

Answers (3)

inspectorG4dget
inspectorG4dget

Reputation: 113945

This should do the trick:

The shuffle gives some form of randomness to which boy and girl are paired with each other

In [115]: boys = ["Boy1", "Boy2", "Boy3", "Boy4", "Boy5"]

In [116]: girls = ["Girl1", "Girl2", "Girl3", "Girl4", "Girl5"]

In [117]: random.shuffle(girls)

In [118]: girls
Out[118]: ['Girl5', 'Girl4', 'Girl3', 'Girl1', 'Girl2']

In [119]: for i in itertools.izip(boys, girls):
   .....:     print i
   .....:     
('Boy1', 'Girl5')
('Boy2', 'Girl4')
('Boy3', 'Girl3')
('Boy4', 'Girl1')
('Boy5', 'Girl2')

EDIT: If you want every possible pairing, check this out:

In [126]: boys
Out[126]: ['Boy1', 'Boy2', 'Boy3', 'Boy4', 'Boy5']

In [127]: girls
Out[127]: ['Girl1', 'Girl2', 'Girl3', 'Girl4', 'Girl5']

In [128]: [girls[i:]+girls[:i] for i in xrange(len(girls))]
Out[128]: 
[['Girl1', 'Girl2', 'Girl3', 'Girl4', 'Girl5'],
 ['Girl2', 'Girl3', 'Girl4', 'Girl5', 'Girl1'],
 ['Girl3', 'Girl4', 'Girl5', 'Girl1', 'Girl2'],
 ['Girl4', 'Girl5', 'Girl1', 'Girl2', 'Girl3'],
 ['Girl5', 'Girl1', 'Girl2', 'Girl3', 'Girl4']]

In [129]: for combo in (itertools.izip(boys, g) for g in ( girls[i:]+girls[:i] for i in xrange(len(girls)) )):
   .....:    for pair in combo:
   .....:        print pair,
   .....:    print ''
   .....:     
('Boy1', 'Girl1') ('Boy2', 'Girl2') ('Boy3', 'Girl3') ('Boy4', 'Girl4') ('Boy5', 'Girl5') 
('Boy1', 'Girl2') ('Boy2', 'Girl3') ('Boy3', 'Girl4') ('Boy4', 'Girl5') ('Boy5', 'Girl1') 
('Boy1', 'Girl3') ('Boy2', 'Girl4') ('Boy3', 'Girl5') ('Boy4', 'Girl1') ('Boy5', 'Girl2') 
('Boy1', 'Girl4') ('Boy2', 'Girl5') ('Boy3', 'Girl1') ('Boy4', 'Girl2') ('Boy5', 'Girl3') 
('Boy1', 'Girl5') ('Boy2', 'Girl1') ('Boy3', 'Girl2') ('Boy4', 'Girl3') ('Boy5', 'Girl4')

EDIT 2 (fixes EDIT 1):

>>> perms = itertools.permutations(girls)
>>> len([tuple(p) for p in (itertools.product(boys, g) for g in perms)])
120
>>> perms = itertools.permutations(girls)
>>> len(set(tuple(p) for p in (itertools.product(boys, g) for g in perms)))
120

I had to go with len because there are 120 possible pairings and I didn't want to clutter the post. This is why there is len(...) and len(set(...))

Upvotes: 2

Edwin Dalorzo
Edwin Dalorzo

Reputation: 78579

from itertools import product

l1 = ["boy1","boy2","boy3"]
l2 = ["girl1","girl2","girl3"]

print list(product(l1,l2))

Output is:

[('boy1', 'girl1'), ('boy1', 'girl2'), ('boy1', 'girl3'), 
 ('boy2', 'girl1'), ('boy2', 'girl2'), ('boy2', 'girl3'), 
 ('boy3', 'girl1'), ('boy3', 'girl2'), ('boy3', 'girl3')]

Upvotes: 3

Tim Pietzcker
Tim Pietzcker

Reputation: 336148

Another idea:

>>> zip(boys, random.sample(girls, len(girls)))
[('Boy1', 'Girl3'), ('Boy2', 'Girl2'), ('Boy3', 'Girl4'), ('Boy4', 'Girl1'), 
 ('Boy5', 'Girl5')]

Upvotes: 2

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