Index dictionaries inside a list

Question

I have a list of dictionaries:

dicty = [defaultdict(list)] * (2)
dicty

[defaultdict(list, {}), defaultdict(list, {})]

Now I want to know how to index into these dictionaries?

I have tried:

dicty[0][0].append('abc')
dicty

[defaultdict(list, {0: ['abc']}), defaultdict(list, {0: ['abc']})]

But as you can see that it appends in both the dictionaries. I want to learn how to append into an individual dictionary.

See the third answer here: https://stackoverflow.com/questions/3459098/create-list-of-single-item-repeated-n-times-in-python — cs95, Jan 19 '18 at 11:01
You have *two references to the same dictionary*, not two different dictionaries. — juanpa.arrivillaga, Jan 19 '18 at 11:06

ettanany · Answer 1 · 2018-01-19T11:04:27.297

2

You can use:

dicty = [defaultdict(list) for _ in range(2)]

When using [defaultdict(list)] * (2), all your dicts point to the same object in memory.

edited Jan 19 '18 at 11:04

answered Jan 19 '18 at 10:58

ettanany

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what is this defaultdict? – Hariom Singh Jan 19 '18 at 11:00
1

@HariomSingh it's from `collections ` module, you can get more details here: https://docs.python.org/3/library/collections.html#collections.defaultdict – ettanany Jan 19 '18 at 11:01

score 1 · Accepted Answer · answered Jan 19 '18 at 11:01

from collections import defaultdict

""" Here, you are saying that you wish to add twice "defaultdict(list)".
    These are the same objects. You are just duplicating the same one with *
"""
dicty = [defaultdict(list)] * (2)
dicty[0][0].append('abc')
print dicty
print dicty[0] is dicty[1], '\n'

""" However, here : you are creating two distinct dictionnaries.
"""
dicty = []
for _ in range(2):
    dicty.append(defaultdict(list))
dicty[0][0].append('abc')
print dicty
print dicty[0] is dicty[1]

# outputs :
#[defaultdict(<type 'list'>, {0: ['abc']}), defaultdict(<type 'list'>, {0: ['abc']})]
#True 

#[defaultdict(<type 'list'>, {0: ['abc']}), defaultdict(<type 'list'>, {})]
#False

Index dictionaries inside a list

2 Answers2