list comprehention

[Mathijs]

Hi,

Python beginner here and very much enjoying it. I'm looking for a 
pythonic way to find how many listmembers are also present in a reference 
list. Don't count duplicates (eg. if you already found a matching member 
in the ref list, you can't use the ref member anymore).

Example1: 
ref=[2, 2, 4, 1, 1]
list=[2, 3, 4, 5, 3]
solution: 2

Example2:
ref=[2, 2, 4, 1, 1]
list=[2, 2, 5, 2, 4]
solution: 3 (note that only the first two 2's count, the third 2 in the 
list should not be counted)

Any suggestions or comments?

Thanks.
M.


#my failing effort:
sum([min(r.count(n)-l[:i].count(n),l.count(n)) for i,n in enumerate(l)])

#test lists
import random
#reference list
r=[random.randint(1,5) for n in range(5)]
#list
l=[random.randint(1,5) for n in range(5)]

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Re: list comprehention

[Mathijs]

Op 19 jan 2006 vond "[EMAIL PROTECTED]" :

> another approach:
> 
> ref = [2,2,4,1,1]
> lis = [2,2,5,2,4]
> 
> len([ref.pop(ref.index(x)) for x in lis if x in ref])
> 

This is the type of solution I was hoping to see: one-liners, with no use 
of local variables. As Tim Chase already wrote, it has only one less 
elegant side: it alters the original ref list.

Thanks for your suggestion.
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Re: list comprehention

[Mathijs]

Op 19 jan 2006 vond Peter Otten <[EMAIL PROTECTED]> :

> sum(min(list.count(n), ref.count(n)) for n in set(ref))
> 
> Is that it?

Seems like this is it! Thanks.
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Re: list comprehention

[Mathijs]

Op 19 jan 2006 vond "Paddy" <[EMAIL PROTECTED]>: 
 answer = [ val for val in set(ref) for x in
 range(min(lst.count(val), ref.count(val)))] answer
> [2, 2, 4]

I don't think it's correct. Your algoritm with the ref and lst below gives 
3 as answer. The answer should have been 2 (1,3).

ref=[3, 3, 1, 1, 3]
lst=[5, 1, 4, 5, 3]

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Re: list comprehention

[Mathijs]

Op 20 jan 2006 vond Duncan Booth <[EMAIL PROTECTED]>: 
> Or in other words, define a function to return a dictionary containing
> a count of the number of occurrences of each element in the list (this
> assumes that the list elements are hashable). Then you just add up the
> values in the test list making sure each count is limited to no higher
> than the reference count.

Thanks. Though I don't know much about python (yet), this is more or less 
the way I'de do it the language I'm more comfortable with (object pascal), 
and I wouldn't label this as a pythonic solution. I could be wrong, 
though:)
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Re: list comprehention

[Mathijs]

23 jan 2006 ta (Steven D'Aprano) shuo le:

>> This is the type of solution I was hoping to see: one-liners, with no
>> use of local variables.
> 
> Because you like unreadable, incomprehensible, unmaintainable code?

For practical use: no! But I'm just learning python and to understand 
sets/lists/dicts and it's functions, this seemed like good practise:) and 
besides that: it's fun! A solution with local vaiables, if-statemenst and a 
for loop wouldn't be much different from what I know with my current 
programming skills.

> *wink*

Yeah, yeah, I know, but I felt a response was on its place. :)
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