incrementing a dictionary value

>
> Is there a more elegant way to increment the value of a dictionary entry
> than what I have done in "this line" below?
>
> elements := #( 1 1 2 2 2 2 3 3 3 3 3 3) asOrderedCollection.
> elementCounter := Dictionary new.
> elements do: [ :key | elementCounter at: key put: (elementCounter at: key
> ifAbsent: [ 0 ]) + 1 ].  "this line"
> elementCounter inspect. "--> a Dictionary(1->2 2->4 3->6 )"
>

> On 5 January 2013 06:06, Ben Coman <[hidden email]> wrote:
>  
>> Is there a more elegant way to increment the value of a dictionary entry
>> than what I have done in "this line" below?
>>
>> elements := #( 1 1 2 2 2 2 3 3 3 3 3 3) asOrderedCollection.
>> elementCounter := Dictionary new.
>> elements do: [ :key | elementCounter at: key put: (elementCounter at: key
>> ifAbsent: [ 0 ]) + 1 ].  "this line"
>> elementCounter inspect. "--> a Dictionary(1->2 2->4 3->6 )"
>>
>>    
> why you don't use bag?
>
> #( 1 1 2 2 2 2 3 3 3 3 3 3) asBag sortedCounts
> {6->3. 4->2. 2->1}
>
>  

> Igor Stasenko wrote:
>>
>> On 5 January 2013 06:06, Ben Coman <[hidden email]> wrote:
>>
>>>
>>> Is there a more elegant way to increment the value of a dictionary entry
>>> than what I have done in "this line" below?
>>>
>>> elements := #( 1 1 2 2 2 2 3 3 3 3 3 3) asOrderedCollection.
>>> elementCounter := Dictionary new.
>>> elements do: [ :key | elementCounter at: key put: (elementCounter at: key
>>> ifAbsent: [ 0 ]) + 1 ].  "this line"
>>> elementCounter inspect. "--> a Dictionary(1->2 2->4 3->6 )"
>>>
>>>
>>
>> why you don't use bag?
>>
>> #( 1 1 2 2 2 2 3 3 3 3 3 3) asBag sortedCounts
>> {6->3. 4->2. 2->1}
>>
>>
>
> Thanks Igor.  That is useful.  So I ended up with the following to gather
> elements that only occur once (probably should have included that in the
> original scope)
> (#( 7 8 1 1 2 2 2 2 3 3 3 3 3 3 ) asBag sortedElements select: [ :asn | asn
> value = 1 ] ) collect: [ :asn | asn key ].
>