Weird Python behavior

Question

I wrote a code in python which looks like:

maplist=[{}]*11
mylist=[0]*11
maplist[0]['this']=1
print maplist

When I print maplist the output is :

[{'this': 1}, {'this': 1}, {'this': 1}, {'this': 1}, {'this': 1}, {'this': 1}, {'this': 1}, {'this': 1}, {'this': 1}, {'this': 1}, {'this': 1}]

Expected is:

[{'this': 1}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}]

rather than only the first element of the list should have this key in the map. What is causing this problem?

It's amazing how often we get a subtle variation on the same question... in addition to marking things as duplicates, I'm starting to feel like we need to have a way to create a per-tag FAQ. — Karl Knechtel
– Karl Knechtel, Commented Jan 26, 2012 at 10:17

NPE · Accepted Answer · 2012-01-26 08:28:56Z

12

When you do the following:

maplist=[{}]*11

you end up with eleven references to the same dictionary. This means that when you modify one dictionary, they all appear to change.

To fix, replace that line with:

maplist=[{} for in xrange(11)]

Note that, since 0 is a scalar, the next line is fine as it is:

mylist=[0]*11

answered Jan 26, 2012 at 8:23

NPE

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2 Comments

Ohh came across this for the first time. Is there any specific reason why python do so , is it useful someehere?

@Piyush It is just a consequence of the way the * operator works on lists. See this answer for a more in-depth explanation.

Rob Wouters · Accepted Answer · 2012-01-26 08:24:15Z

3

The problem is when you type

maplist=[{}]*11

You're creating a list of 11 references to the same dict. Only one dictionary gets created.

To create 11 separate dictionaries you can do something like this:

>>> maplist = [{} for i in range(11)]
>>> maplist[0]['this'] = 1
>>> maplist
[{'this': 1}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}]

answered Jan 26, 2012 at 8:24

Rob Wouters

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