updating dictionaries from/to dictionaries

Edwin.Madari at VerizonWireless.com Edwin.Madari at VerizonWireless.com
Fri Aug 15 03:45:33 CEST 2008


by the way, iterating over bar will throw KeyError if that key does not exist in foo. to see that in action, simply set another key in bar after copy.deepcopy stmt in this example..
bar['xtra'] = 0
and re-run....

fun learning with python...
Edwin

-----Original Message-----
From: Madari, Edwin 
Sent: Thursday, August 14, 2008 9:24 PM
To: 'Brandon'; python-list at python.org
Subject: RE: updating dictionaries from/to dictionaries


if the values for any of the keys are None, both do not work as can be seen below!!. 

since both loops are iterating over keys(), 'get' method which would return a '0' for non existing key does not encounter any non-existing keys.
============================================================
import copy, sys

foo = { 'one' : 1, 'two' : 2, 'three' : None  }
bar = copy.deepcopy(foo)

try:
   for key in foo:
      foo[key] += bar.get(key, 0)
except:
   print 'foo',  sys.exc_info()[:2]

try:
   for key in bar:
      foo[key] += bar[key]
except:
   print 'bar', sys.exc_info()[:2]
========================================
foo (<type 'exceptions.TypeError'>, TypeError("unsupported operand type(s) for +=: 'NoneType' and 'NoneType'",))
bar (<type 'exceptions.TypeError'>, TypeError("unsupported operand type(s) for +=: 'NoneType' and 'NoneType'",))

hope that helps to clarify both point of views.
thanks Edwin

-----Original Message-----
From: python-list-bounces+edwin.madari=verizonwireless.com at python.org
[mailto:python-list-bounces+edwin.madari=verizonwireless.com at python.org]
On Behalf Of Brandon
Sent: Thursday, August 14, 2008 8:35 PM
To: python-list at python.org
Subject: Re: updating dictionaries from/to dictionaries



> (1) iterating over foo:
> for key in foo:
>    foo[key] += bar.get(key, 0)
>
> (2) iterating over bar:
> for key in bar:
>    foo[key] += bar[key]
>
> I (again) challenge you to say *why* you feel that the "iterating over
> bar" solution will not work.


Well if you're going to be clever enough to iterate over bar and then
send the results to another dictionary altogether, I obviously cannot
put up a good argument on this matter!

Thanks for the input, I appreciate it.
--
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