On 09/10/2010 10.25, Alan Gauld wrote:
"Francesco Loffredo" <f...@libero.it> wrote
> On the next iteration you overwrite those two dictionaries
> with new values then append them to the list again.
> So you wind up with 2 copies of the updated dictionaries.
> ...
This is difficult for me too: why does this happen? Or, more correctly,
why should this happen?
It happens because you can't store two objects in one.
There is only one dictionary. If you change its value you
change its value.
How can you save the current contents of a dictionary in a list,
making sure that the saved values won't change if you update the dict?
You need to save a copy of the dictionary. ie Create a new dictionary.
If you put a box of white eggs in your shopping basket you cannot put
brown eggs into that box and expect to still have the white ones as well.
You need to get two boxes!
I obviously haven't been clear in my question, sure I know that if you
change one object you lose the previous contents, but (hope this makes
my question clear) why should the append method of the list store a
pointer to the dictionary, rather then a copy of that dictionary? If it
did, Roelof's code would work perfectly, and you could store in a list
all the subsequent changes of a dictionary without calling them with
different names. For example (and this should make my previous example a
bit fuller), if you had a dictionary containing the current status of a
system, and you wanted to keep an history of that status (well, I'd use
a file, but let's imagine you had to use a list), you could simply add
the (newer version of the) dictionary to the list:
myDictList.append(UpdateIt(myDict))
This would be much more elegant and readable than creating a different
dictionary (meaning a different name) for every state of the said
system, and making sure you never repeat a name twice, or you'll lose
the state you saved some time ago...
I understand that if .append() stored a copy of the dict in the list,
you will end up with lots of copies and a huge amount of memory used by
your list, but that's exactly what will happen if you make those copies
yourself. But you wouldn't have to devise a way to generate a new name
for the dictionary every time you need to update it.
Thank you for your help
Francesco
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