dict comprehension question.

Peter Otten __peter__ at web.de
Sat Dec 29 21:32:29 CET 2012


Quint Rankid wrote:

> Newbie question.  I've googled a little and haven't found the answer.
> 
> Given a list like:
> w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
> I would like to be able to do the following as a dict comprehension.
> a = {}
> for x in w:
>     a[x] = a.get(x,0) + 1
> results in a having the value:
> {1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}
> 
> I've tried a few things
> eg
> a1 = {x:self.get(x,0)+1 for x in w}
> results in error messages.
> 
> And
> a2 = {x:a2.get(x,0)+1 for x in w}
> also results in error messages.
> 
> Trying to set a variable to a dict before doing the comprehension
> a3 = {}
> a3 = {x:a3.get(x,0)+1 for x in w}
> gets this result, which isn't what I wanted.
> {1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1}
> 
> I'm not sure that it's possible to do this, and if not, perhaps the
> most obvious question is what instance does the get method bind to?

The name a3 will not be rebound until after the right side is evaluate. To 
spell it with a loop:

a3 = {}
_internal = {} # You have no access to this var. 
               # Even if you can beat a particular 
               # Python implementation -- you shouldn't

for x in w:
    _internal[x] = a3.get(x, 0) + 1

a3 = _internal

That should make it clear that x will be looked up in the "old" empty a3 
dict.

The closest you can get to a self-updating dict is probably

>>> w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
>>> a1 = {}
>>> a1.update((x, a1.get(x, 0)+1) for x in w)
>>> a1
{1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1}

but that it works doesn't mean it is a good idea. 
If your Python version supports it the obvious choice is

>>> from collections import Counter
>>> w = [1, 2, 3, 1, 2, 4, 4, 5, 6, 1]
>>> Counter(w)
Counter({1: 3, 2: 2, 4: 2, 3: 1, 5: 1, 6: 1})





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