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/
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
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=
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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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
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,
