overlapping sets
Gerard Flanagan
grflanagan at yahoo.co.uk
Fri Mar 24 15:08:59 CET 2006
kpp9c wrote:
> I have a question... and ... whew ... i am gonna be honest, i haven't
> the slightest clue how to even start ... i am not sure if i used up all
> my good will here or can take a mulligan.. i love to try to at least
> post some lame broken code of my own at first... but like i said, not
> being a math person i am not even sure how to start or if it is even
> possible.
>
> here's the deal... i have a dictionary that defines some collections..
> like so:
>
> sets = { ('one') : [0, 4, 7, 9 ],
> ('two') : [0, 3, 7, 9 ],
> ('three') : [0, 4, 7, 11],
> ('four') : [0, 3, 7, 10 ],
> ('five') : [0, 4, 7, 10 ],
> ('six') : [0, 4, 8, 10 ],
> ('seven') : [0, 3, 6, 10],
> ('eight') : [0, 3, 6, 9 ],
> ('nine') : [0, 3, 7, 11 ],
> ('ten') : [0, 5, 7, 10 ] }
>
> I every time i call this function i would like like it to return a
> collection at random, any collection, so long as it has all but one
> element that is the same. So if i grab [0, 4, 7, 9 ] as my first set
> my next set could be: [0, 3, 7, 9 ], or [0, 4, 8, 9 ], or [0, 4, 7,
> 10], or [1, 4, 7, 9 ], since all these sets contain 3 elements in
> common with the first, and only one that is new or different... but if
> my first run give me: [0, 4, 7, 9 ] i would not get [0, 5, 7, 10 ],
> since this is set has 2 elements that are unique. The goal it to move
> from set to set to set to set always with a maximum of overlap & common
> elements.
>
probably not very efficient but I think it roughly does what you want.
(maybe add a boolean 'sort' parameter, to select sorted output or not):
# k is the length of the required output lists, 0<k<n
# n is the number of lists to output
import random
alphabet = [ 0, 3, 4, 5, 6, 7, 8, 9, 10, 11 ]
def randomiser( a_list, n, k ):
d = len( a_list )
choice = range(d)
vector = []
for i in range(k):
vector.append( choice.pop(random.randint(0,d-i-1)) )
yield [ a_list[s] for s in vector ]
#yield sorted( a_list[s] for s in vector )
for _ in range(n):
rand_vector_idx = random.randint(0,k-1)
rand_choice_idx = random.randint(0,d-k-1)
rand_vector_val = vector[rand_vector_idx] #remember old value
vector[rand_vector_idx] = choice.pop( rand_choice_idx )
choice.append( rand_vector_val ) #add old value back to choice
yield [ a_list[t] for t in vector ]
#yield sorted( a_list[t] for t in vector )
print
for item in randomiser(alphabet, 10, 4):
print item
[3, 6, 9, 8]
[3, 6, 9, 0]
[3, 11, 9, 0]
[3, 11, 10, 0]
[9, 11, 10, 0]
[9, 11, 7, 0]
[9, 4, 7, 0]
[9, 3, 7, 0]
[9, 11, 7, 0]
[9, 11, 7, 8]
[9, 11, 10, 8]
Gerard
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