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Python - how to pass a dictionary into defaultdict as value and not as a reference

So say that I have a dictionary with a default value of another dictionary

attributes = { 'first_name': None, 'last_name': None, 'calls': 0 }
accounts = defaultdict(lambda: attributes)

The problem is that the default dictionary that I pass into defaultdict (attributes) is passed as a reference. How can I pass it as a value? So that changing the values in one key doesn't change the values in other keys

For example -

accounts[1]['calls'] = accounts[1]['calls'] + 1
accounts[2]['calls'] = accounts[2]['calls'] + 1
print accounts[1]['calls'] # prints 2
print accounts[2]['calls'] # prints 2

I want each of them to print 1, since I only incremented their respective values for 'calls' once.

like image 440
satnam Avatar asked Feb 07 '17 19:02

satnam


1 Answers

Try:

accounts = defaultdict(attributes.copy)

Since Python 3.3 listss also have copy method so you can use it the same way as above with defaultdicts when you need a dict with a list as a default value.

like image 73
warvariuc Avatar answered Sep 20 '22 23:09

warvariuc