Is there a way to make an automatically growing list in Python? What I mean is to make a list that would grow when an index that does not yet exist is referenced. Basically the behaviour of Ruby arrays.
Thanks in advance!
3 Answers
Sure it's possible, you just have to use a subclass of list to do it.
class GrowingList(list):
def __setitem__(self, index, value):
if index >= len(self):
self.extend([None]*(index + 1 - len(self)))
list.__setitem__(self, index, value)
Usage:
>>> grow = GrowingList()
>>> grow[10] = 4
>>> len(grow)
11
>>> grow
[None, None, None, None, None, None, None, None, None, None, 4]
13 Comments
dan_waterworth
Just note that this is only really useful for dense arrays, if you need a structure for sparse arrays then you are better off with a dictionary based solution.
Ant
add a 'fillvalue' argument in the function...so you can do grow.__setitem__(10, 4, True) for example
dan_waterworth
@Ant,
__setitem__ is a special method, it's not often called via it's name, but like in the example grow[10] = 4.Ant
i know, but if i really want that behaviour, i should be able to do that without rewriting the func. and doing def __setitem__(self, index, value, fillvalue=None) won't affect the normal usage ;)
dan_waterworth
I don't think it is good python to add parameters (even default parameters) to special functions, what if the specification changes and it adds another parameter. I think if you need a fillvalue then you should add it to the
__init__ function, but that wouldn't play well with being inherited from list, you'd need to change the __new__ method, but I'm just trying to write a simple example. |
Lists are dynamic in python. It will automatically grow always (up until you hit sys.maxsize) in your program.
l = list()
l.append(item)
Keep doing that.
3 Comments
Anonymous
No, I mean that it will grow when an out-of-range index is referenced. For example, a=[1,2,3] will grow when a[100]=101 is called.
Tim Pietzcker
What should the values of
a[3] to a[99] be after this?
Patrick Brinich-Langlois
In Ruby, intermediate elements are set to
nil; in Python, None would be the logical choice.No, but you could use a dictionary (hash table) to achieve the same thing.
4 Comments
Anonymous
Thanks, but I need it to be ordered.
Keith
Then you might try the OrderedDict object. ;-)
Keith
They have keys which would act like indices. Not much difference, but saves memory on sparse arrays. The extended list (alternative) approach would also allocate a bunch of unused objects, and also add run time cost to list traversal.
Anonymous
I see what you mean. However, the reason I wanted this is for implementing a mathematical algorithm where the values of indeces are significant.
Noneor some other default value?a[1000]if it doesn't exist yet?IndexErroror should the list grow to this size, too?