Python copying or cloning a defaultdict variable

Question

How can I make a duplicate copy (not just assigning a new pointer to the same location in memory) of Python's defaultdict object?

from collections import defaultdict
itemsChosen = defaultdict(list)    
itemsChosen[1].append(1)
dupChosen = itemsChosen
itemsChosen[2].append(1)
print dupChosen

What the code above does is a shallow copy and returns

defaultdict(<type 'list'>, {1: [1], 2: [1]})

whereas what I'm looking for it to return

defaultdict(<type 'list'>, {1: [1]})

Thanks.

That's not even a shallow copy, that's just two references to the same object. A shallow copy would be `itemsChosen.copy()`. — roippi, Mar 13 '14 at 20:22
also as an aside... `defaultdict`, like many built-ins, has a copy function: `dupChosen = itemsChosen.copy()`... — Corley Brigman, Mar 13 '14 at 20:24
Thanks for the clarifications- I misunderstood shallow/deep copying. I also couldn't a page of functions for defaultdict, but that makes sense since it inherits from dict. — so13eit, Mar 13 '14 at 20:27

score 14 · Accepted Answer · answered Mar 13 '14 at 20:21

14

Use copy:

from copy import copy

dupChosen = copy(itemsChosen)

In case of multiple nestings there is also deepcopy.

answered Mar 13 '14 at 20:21

BoshWash

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How Pythonic! I'll accept the answer once it allows me to, thanks. – so13eit Mar 13 '14 at 20:22
See https://stackoverflow.com/questions/17873384/how-to-deep-copy-a-list on the gotchas of deep copy – heretoinfinity Feb 07 '23 at 10:42

Python copying or cloning a defaultdict variable

1 Answers1