Convert list to another form but providing same information

[Mark Lawrence]

On 21/03/2016 18:26, Maurice wrote:
> Hello, hope everything is okay. I think someone might have dealt with a similar issue I'm having.
>
> Basically I wanna do the following:
>
> I have a list such [6,19,19,21,21,21] (FYI this is the item of a certain key in the dictionary)
>
> And I need to convert it to a list of 32 elements (meaning days of the month however first element ie index 0 or day zero has no meaning - keeping like that for simplicity's sake).
> Therefore the resulting list should be:
> [0,0,0,0,0,0,1,0,0,0...,2,0,3,0...0]
>
> So the list index should tell how many occurrences of a certain day in the original list.
>
> My attempt:
>
> weirdList = [[0]*32]*len(dict_days) #list's length should be how many keys in the dict.
>
> counter = 0
> k = 0
> for key in dict_days.keys():
>      for i in range(1,32):
>          if i in dict_days[key]:
>              counter = dict_days[key].count(i)
>              weirdList[k][i] = counter
>      dict_days[key] = weirdList[k]
>      k+=1
>
> However it does not work. weirdList seems to be always the same?
>
> Thanks in advance.
>

 >>> from collections import Counter
 >>> counts = Counter([6,19,19,21,21,21])
 >>> counts
Counter({21: 3, 19: 2, 6: 1})
 >>> weird_list = [0]*32
 >>> weird_list
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0]
 >>> for index, count in counts.items():
...     weird_list[index] = count
...
 >>> weird_list
[0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 3, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0]

-- 
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.

Mark Lawrence