[Tutor] A shorter way to initialize a list?

[Emile van Sebille]

On 12/13/2011 3:26 PM Kaixi Luo said...
<snip>
> Uh, sorry. I made some very bad typos there. It should be:
>
> listA = [[] for i in range(9)]
>
> # some code here...
>
> listB = [[] for i in range(3)]
> count = 0
> for i in range(3):
>      for j in range(3):
>          listB[i].append(listA[count])
>          count+=1
>


This yields listB as [[[], [], []], [[], [], []], [[], [], []]]

So, the trick will be that each list element be unique.  If you're not 
careful, you'll get the same list elements.  For example,

listB = [[[]]*3]*3

appears to yield the same result until you append to one of the elements 
and discover one of the common early gotcha's:

 >>> listB[1][1].append(1)
 >>> print listB
[[[1], [1], [1]], [[1], [1], [1]], [[1], [1], [1]]]

To avoid this, each list container needs to be added separately, which 
means that the two loops will be required.

You can express it as a single statement using list comprehensions:


 >>> listB = [ [ [] for ii in range(3) ] for jj in range(3) ]
 >>> listB
[[[], [], []], [[], [], []], [[], [], []]]
 >>> listB[1][1].append(1)
 >>> listB
[[[], [], []], [[], [1], []], [[], [], []]]


HTH,

Emile