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Re: Probability Algorithm

Started byDennis Lee Bieber <wlfraed@ix.netcom.com>
First post2012-08-25 16:47 -0400
Last post2012-08-25 16:47 -0400
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  Re: Probability Algorithm Dennis Lee Bieber <wlfraed@ix.netcom.com> - 2012-08-25 16:47 -0400

#27886 — Re: Probability Algorithm

FromDennis Lee Bieber <wlfraed@ix.netcom.com>
Date2012-08-25 16:47 -0400
SubjectRe: Probability Algorithm
Message-ID<mailman.3814.1345927660.4697.python-list@python.org>
On Sun, 26 Aug 2012 00:03:07 +0800, ??? <yueyoum@gmail.com> declaimed
the following in gmane.comp.python.general:

> Hi,  All,
> 
> I have a  problem of probability algorithm
>
	Smells like homework -- but since you did provide something that
could be turned in as your work, I'm going to go out on a limb
 
> 
> The goal is obtain a list which contains three items.   as the *FinalList*
> 
> There has Four source lists. *
> ALIST, BLIST, CLIST, DLIST
> 
> There are all  Unknown length. They contains unique elements*
> ( In fact,  there are all empty at the program beginning,  when running,
> there growing  )
> 

theLists = {	"A" : [],
			"B" : [],
			"C" : [],
			"D" : []	}

# do whatever is needed to populate the lists 
# I presume reading some file(s) and doing
		...
		theLists[listName].append(listValue)


> Choose items form this source lists. pick up random items to generate the
> FinalList
> Ensure  The Following Requirements
> 
> In the FinalList,
> probability of ALIST's item appeared  is  43%
> probability of BLIST's item appeared  is  37%
> probability of CLIST's item appeared  is  19%
> probability of DLIST's item appeared  is  1%
> 
> 
	<snip>

theProbabilities = {	"A" : 43, 
					"B" : 37+43,
					"C" : 19+37+43,
					"D" : 1+19+37+43	}


> while a_picked_times < 43:

	Why keep a counter? Rather than an iterated loop

for i in range(43):

>     item = choice(ALIST)
>     ALIST.remove(item)

	Do you really want to remove an item from the source list?
Technically, a requirement that an item appears from ALIST 43% of the
time does NOT prohibit it being the SAME ITEM. {This also answers you
problem about only working if the lists are long... as long as the list
contains at least ONE item, you can pick that item to meet the
probability}

> 
>     if item in already_picked_list:
>         continue
>
	This is meaningless with regards to the previous comments: if you
had removed the item from the source list, it will never appear again
(if it did, it was from a different position in the source list, or from
a different source list overall -- and if you don't want duplicates from
within a source list, you should remove them when building the source
list)
 
>     slot.append(item)
>     a_picked_times += 1
> 

	<snip>

	Given that "theLists" contains the four lists:

result = []
while len(result) < 3:
	#using this test handles the case of sublist being empty
	x = random.randint(0, 99)
	if	x < theProbabilities["A"]:
		clist = theLists["A"]
	elif	x < theProbabilities["B"]:
		clist = theLists["B"]
	elif	x < theProbabilities["C"]:
		clist = theLists["C"]
	else:
		clist = theLists["D"]

	if len(clist):
		#is sublist is empty, nothing gets appended, loop goes on
		result.append(random.choice(clist))

-- 
	Wulfraed                 Dennis Lee Bieber         AF6VN
        wlfraed@ix.netcom.com    HTTP://wlfraed.home.netcom.com/

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