A question about making a sort-of-counter.
Justin Park
a = list()
And suppose allowed digits as the element are 1,2,3,4,5.
What can I do in order to iterate over all possible values for each element?
For instance, the sequence of the list I want to have would be
[1,1,1,1,1]
[1,1,1,1,2]
[1,1,1,1,3]
....
[5,5,5,5,4]
[5,5,5,5,5]
How can I make it happen?
Thanks,
Justin.
Robert Kern
def counter(n=5, length=5):
for i in xrange(n ** length):
x = []
for j in range(length):
i, r = divmod(i, n)
x.append(r)
yield [k+1 for k in x[::-1]]
--
Robert Kern
"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco
anon
http://docs.python.org/library/itertools.html#itertools.product
from itertools import product
for item in product([1, 2], repeat=2):
print item
Justin Park
It works!
Justin.
Chris Rebert wrote:
> On Tue, Mar 30, 2010 at 2:31 PM, Justin Park <h...@rice.edu> wrote:
>
>> Suppose I have a list.
>> a = list()
>> And suppose allowed digits as the element are 1,2,3,4,5.
>>
>> What can I do in order to iterate over all possible values for each element?
>> For instance, the sequence of the list I want to have would be
>> [1,1,1,1,1]
>> [1,1,1,1,2]
>> [1,1,1,1,3]
>>
>> ....
>>
>> [5,5,5,5,4]
>> [5,5,5,5,5]
>>
>> How can I make it happen?
>>
>
> allowed = range(1,6)
> length = 5
> for counter_tuple in product(allowed, repeat=length):
> counter_list = list(counter_tuple) # if you really need a list
> #do whatever with the counter value
>
> See the docs for itertools.product(); an example virtually identical
> to your situation is given:
> http://docs.python.org/library/itertools.html#itertools.product
>
> Cheers,
> Chris
> --
> http://blog.rebertia.com
>
>
