> | def __str__ (self):
>
> string appending is an O(n**2) operations. The usual idiom, applied here,
> would be slist = ['('], slist.append(str(self.value)), .... return
> ''.join(slist). In other words, collect list of pieces and join at end.
I did some timing of operations involved. Doing this I found that
the operation below to get a string representation for trees was
in fact not quadratic.
The final nail in the coffin of this comment is now at:
http://kasterma.wordpress.com/2008/06/30/strings-versus-lists/
(I put it there because the main part is a graph of timing data)
Appending for lists is slower than appending the strings.
This means that the operation using strings is faster.
Again, thanks for all the comments, I enjoyed working this out. Even
better would be to point out any mistakes in my arguments or code.
Best,
Bart
>
> | string = "("
> | if not self.value == None:
> | string += str (self.value)
> |
> | if not (self.left == None and self.right == None):
> | if self.left != None:
> | string += str (self.left)
> | else:
> | string += "()"
> |
> | if self.right != None:
> | string += str (self.right)
> | else:
> | string += "()"
> |
> | string += ")"
> |
> | return string
> |
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