Update value in nested dictionary - Python

Question

I created a dictionary as follows:

gP = dict.fromkeys(range(6), {'a': None, 'b': None, 'c': None, 'd': None})

Now, when I try to modify a value doing:

gP[0]['a'] = 1

for some reason, all the values of a (regardless of the key they belong to) change to 1 as shown below:

{0: {'a': 1, 'b': None, 'c': None, 'd': None},
 1: {'a': 1, 'b': None, 'c': None, 'd': None},
 2: {'a': 1, 'b': None, 'c': None, 'd': None},
 3: {'a': 1, 'b': None, 'c': None, 'd': None},
 4: {'a': 1, 'b': None, 'c': None, 'd': None},
 5: {'a': 1, 'b': None, 'c': None, 'd': None}}

What I am doing wrong? What's the proper assignment statement?

You're only creating one dict. Python doesn't make copies of it for you. — deceze
– deceze ♦, Commented Apr 20, 2017 at 8:13
Could you elaborate? I don't understand how to fix it from your answer, @deceze. — YamiOmar88
– YamiOmar88, Commented Apr 20, 2017 at 8:15

Community · Accepted Answer · 2017-05-23 12:25:45Z

5

As @deceze said, Python doesn't make copies. You are referencing the same dict in all the value parts of the key-value pairs.

An alternative would be:

gP = {x: {'a': None, 'b': None, 'c': None, 'd': None} for x in range(6)}

Update: There is a much cleaner version of this answer by @Chris_Rands:

{x: dict.fromkeys('abcd') for x in range(6)}

edited May 23, 2017 at 12:25

CommunityBot

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answered Apr 20, 2017 at 8:15

UltraInstinct

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4 Comments

Chris_Rands Over a year ago

Shorter would be {x: dict.fromkeys('abcd') for x in range(6)}; or probably faster to take the dict.fromkeys outside the loop

UltraInstinct Over a year ago

@Chris_Rands: You should probably put that out as a separate answer. I like it better than mine. :)

Chris_Rands Over a year ago

You can add it to your answer :)

Rahul K P Over a year ago

Please post someone, It's a nice attempt.

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