Convert list to another form but providing same information

[Peter Otten]

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.

Rewrite the above as

inner = [0] * 32
weirdList = [inner] * len(dict_days)

Now it should be clearer that weirdList is

[inner, inner, inner, ...]

i. e. len(dict_days) times the same list of 32 zeros. When you modify 
weirdList[0] you also modify weirdList[1] etc. because these are the 
actually the same list.

For the inner list this is not a problem because 0 is immutable, so you can 
only replace it. If the inner list contained a mutable object you'd see the 
same "weird" behaviour you see for the outer list.

The fix is to use multiple inner lists:

not_so_weirdList = [[0]*32 for _ in range(len(dict_days))]

> 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?

Instead of the list you could use a counter and get basically the same 
interface as long as you ensure that the second index satisfies 0<=index<32:

import collections

for key, values in dict_days.items():
    dict_days[key] = collections.Counter(values)