Python dictionary initialization? [duplicate]

Question

I am not sure if this is a bug or a feature. I have a dictionary to be initialized with empty lists.

Lets say

keys =['one','two','three']
sets = dict.fromkeys(keys,[])

What I observed is if you append any item to any of the lists all the lists are modified.

sets = dict.fromkeys(['one','two','three'],[])
sets['one'].append(1)

sets

{'three': [1],'two': [1], 'one': [1]}

But when I do it manually using loop,

for key in keys:
      sets[key] = []
sets['one'].append(1)

sets

{'three': [], 'two': [], 'one': [1]}

I would think the second behavior should be the default.

It's likely you would have been best served here by collections.defaultdict. Declare it like this: sets = collections.defaultdict(list). — Steven Rumbalski
– Steven Rumbalski, Commented May 1, 2012 at 17:16
why are you using lists instead of sets for a variable named sets? — Doboy
– Doboy, Commented May 1, 2012 at 17:57

NPE · Accepted Answer · 2012-05-01 17:02:38Z

10

This is how things work in Python.

When you use fromkeys() in this manner, you end with three references to the same list. When you change one list, all three appear to change.

The same behaviour can also be seen here:

In [2]: l = [[]] * 3

In [3]: l
Out[3]: [[], [], []]

In [4]: l[0].append('one')

In [5]: l
Out[5]: [['one'], ['one'], ['one']]

Again, the three lists are in fact three references to the same list:

In [6]: map(id, l)
Out[6]: [18459824, 18459824, 18459824]

(notice how they have the same id)

edited May 1, 2012 at 17:02

answered May 1, 2012 at 16:54

NPE

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5 Comments

Gareth Latty Over a year ago

The easiest way to explain this, I find, is it's exactly the same as doing mylist = [] followed by sets = dict.fromkeys(keys, mylist) - just as in that case, where mylist is the value for every key, when you pass an empty list [] it's the same empty list every time. The reason for this is that in Python, variables are always references.

ovgolovin Over a year ago

That is why defaultdict uses a factory of default values, not the value itself. It's strange that dict.fromkeys doens't follow the same technique.

MikeyB Over a year ago

Not so strange, the help states: dict.fromkeys(S[,v]) -> New dict with keys from S and values equal to v.

ovgolovin Over a year ago

@MikeyB I mean it would be better if they implemented dict.fromkeys the same way as defaultdict. With the current implementation of dict.fromkeys coders have to use some roundabout ways to achive the necessary functionality.

vijayvammi Over a year ago

Thanks for the comments. I just found it to be a weird behavior from python.

MikeyB · Accepted Answer · 2012-05-01 17:13:17Z

5

Other answers covered the 'Why', so here's the how.

You should use a comprehension to create your desired dictionary:

>>> keys = ['one','two','three']
>>> sets = { x: [] for x in keys }
>>> sets['one'].append(1)
>>> sets
{'three': [], 'two': [], 'one': [1]}

For Python 2.6 and below, the dictionary comprehension can be replaced with:

>>> sets = dict( ((x,[]) for x in keys) )

edited May 1, 2012 at 17:13

answered May 1, 2012 at 17:01

MikeyB

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2 Comments

moooeeeep Over a year ago

otherwise the following might help: dict(zip(keys, ([] for x in keys)))

Paul Seeb Over a year ago

@MikeyB Sorry my mistake. Also dict([(x,[]) for x in keys]) Not sure which is faster, generator or list comp

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