Probability Algorithm

[月忧茗]

Sorry,   missing some conditions

*already_picked_list*   is get from db.

>  Why keep a counter? Rather than an iterated loop

so ,  if use a iterated loop:

for i in range(43):
    item = choice( ALIST )
    ALIST.remove( item )

    if item in already_picked_list:
        continue

    slot.append( item )


For example, if  we picked item from ALIST 43 times, but all picked items
are in already_picked_list
What's the slot?   It's a empty list,  not contains item from ALIST.

*This is wrong*



> Do you really want to remove an item from the source list?

*Yes*,   i*tems in slot must be unique*

( This is also a condition, which  I have missing... )

> and if you don't want duplicates from within a source list, you should
remove them when building the source list
*
As I mentioned , The source list are all unique elements*



Your last code.
if B ,C ,D are all empty,  result's elements are all from A,  A's
probability is 100% ?

I Know ,  this problem should treated as two situation:

If the four list has not enough elements,   There are no way to ensure the
probability requirements.
If they has enough elements,  Your solution is a good way.
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