Python: Why does a function return a dictionary which has all values filled by assignment?

Question

I have a function that takes in a list and outputs a dictionary. I am creating a dictionary inside the loop and assigning the dictionary values inside the function. A simplified version of what I want to ask is below:

import numpy as np

def get_myDict(tmp_list):
    my_dict = dict.fromkeys(tmp_list, 
                            dict.fromkeys([i**2 for i in tmp_list],0))
    for i in tmp_list:
        my_list = [j for j in np.arange(i)]
        my_dict[i][i**2] = sum(my_list[:]) + i
    return my_dict

Output from this looks like:

[In 1]: get_myDict([1,2,3,4])
[Out 1]: {1: {1: 1, 4: 3, 9: 6, 16: 10},
          2: {1: 1, 4: 3, 9: 6, 16: 10},
          3: {1: 1, 4: 3, 9: 6, 16: 10},
          4: {1: 1, 4: 3, 9: 6, 16: 10}}

Whereas I would expect the output to be

{1: {1: 1, 4: 0, 9: 0, 16: 0},
 2: {1: 0, 4: 3, 9: 0, 16: 0},
 3: {1: 0, 4: 0, 9: 6, 16: 0},
 4: {1: 0, 4: 0, 9: 0, 16: 10}}

I get the desired output if I define the my_dict inside the function with the following:

my_dict = {1: {16: 0, 1: 0, 4: 0, 9: 0}, 
           2: {16: 0, 1: 0, 4: 0, 9: 0}, 
           3: {16: 0, 1: 0, 4: 0, 9: 0}, 
           4: {16: 0, 1: 0, 4: 0, 9: 0}}

I believe that this has to do with shared references in some way. Can someone explain what am I doing wrong here?

All the keys in `my_dict` have the same dictionary as their value – the one you passed to `fromkeys`. — molbdnilo, Oct 14 '18 at 00:07
I'm using Python 2.x . However, I also tried on Python 3.x, with same result. — Vakratund, Oct 14 '18 at 00:08
This question is very similar: https://stackoverflow.com/questions/15516413/dict-fromkeys-all-point-to-same-list — Wolph, Oct 14 '18 at 00:09

score 3 · Accepted Answer · answered Oct 14 '18 at 00:07

3

You are right, you are actually creating a single dictionary with quadratic keys and reference it multiple times.

Use a dict comprehension (since Python 2.7):

my_dict = {k: {k2: 0 for k2 in [i**2 for i in tmp_list]} for k in tmp_list}

answered Oct 14 '18 at 00:07

Michael Butscher

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score 1 · Answer 2 · answered Oct 14 '18 at 00:09

The reason is that you're effectively passing a single value to the dict.fromkeys() method. To illustrate what you're doing:

import numpy as np

def get_myDict(tmp_list):
    value = dict.fromkeys([i**2 for i in tmp_list], 0)
    my_dict = dict.fromkeys(tmp_list, value)
    ...

So all of the dictionaries will reference the same value dictionary.

Perhaps a collection.defaultdict would be more convenient here?

Python: Why does a function return a dictionary which has all values filled by assignment?

2 Answers2