list comprehention

[Bryan Olson]

Duncan Booth wrote:
> Here's the way I would do it:
> 
>>>>def occurrences(it):
> 
>     res = {}
>     for item in it:
>         if item in res:
>             res[item] += 1
>         else:
>             res[item] = 1
>     return res

I slightly prefer:

def occurrences(it):
     res = {}
     res[item] = res.get(item, 0) + 1
     return res


[...]
> 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.

Resulting in a linear-time average case, where the posted
list-comprehension-based solutions are quadratic. The title
of the thread is unfortunate.

The generalized problem is multiset (AKA "bag") intersection:

    http://en.wikipedia.org/wiki/Bag_(mathematics)


-- 
--Bryan