using zip() and dictionaries

Chris Rebert clp2 at rebertia.com
Thu Apr 30 20:09:15 CEST 2009


> On Apr 30, 12:45 pm, Sneaky Wombat <> wrote:
>> I'm really confused by what is happening here.  If I use zip(), I
>> can't update individual dictionary elements like I usually do.  It
>> updates all of the dictionary elements.  It's hard to explain, so here
>> is some output from an interactive session:
>>
>> In [52]: header=['a','b','c','d']
>> In [53]: columnMap={}
>> In [54]: for (k,v) in zip(header,[[]]*len(header)):
>>    ....:     #print "%s,%s"%(k,v)
>>    ....:     columnMap[k] = v
>>    ....:
>> In [55]: columnMap
>> Out[55]: {'a': [], 'b': [], 'c': [], 'd': []}
>> In [56]: columnMap['a'].append('something')
>> In [57]: columnMap
>> Out[57]:
>> {'a': ['something'],
>>  'b': ['something'],
>>  'c': ['something'],
>>  'd': ['something']}
>>
>> Why does ['something'] get attached to all columnMap elements instead
>> of just element 'a'?
>>
>> In [58]: columnMap={'a': [], 'b': [], 'c': [], 'd': []}
>> In [59]: columnMap['a'].append('something')
>> In [60]: columnMap
>> Out[60]: {'a': ['something'], 'b': [], 'c': [], 'd': []}
>>
>> creating the dictionary without using zip, it works as normal.

On Thu, Apr 30, 2009 at 11:00 AM, Sneaky Wombat <joe.hrbek at gmail.com> wrote:
> quick update,
>
> #change this line:
> for (k,v) in zip(header,[[]]*len(header)):
> #to this line:
> for (k,v) in zip(header,[[],[],[],[]]):
>
> and it works as expected.  Something about the [[]]*len(header) is
> causing the weird behavior.  I'm probably using it wrong, but if
> anyone can explain why that would happen, i'd appreciate it.  My guess
> is that it's iterating through the the whole dictionary because of the
> value on the right in zip().

Read http://www.python.org/doc/faq/programming/#how-do-i-create-a-multidimensional-list
Basically, the multiplication doesn't create new sub-lists, it just
copies references to the one original empty sublist.

Cheers,
Chris
-- 
http://blog.rebertia.com



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