How to create an array which can be used also as a dictionary
Hans Müller
I have a lot of items having a name and a given sequence.
To access these items fast in a sequence order they should be used as
a list, but to be fetched fast by name they also should be in a
dictionary.
Code could be something like this.
class item
def __init__(name, value1, value2, value3):
self.name = name
self.value1 = value1
self.value2 = value2
a = []
a.append(item("foo", "bar", "text1"))
a.append(item("xyz", "basd", "tsddsfxt1"))
a.append(item("aax", "hello", "dont care"))
in a, i have my objects in given order, fast accessible by index, e.g.
a[2] to get the third one. Fine.
Now I'd like to have a dict with references to thes objects like this:
d = {}
for x in a:
d[a.name] = a # do I get a copy of a here or a new reference ?!
In d i now have a dict, fast accessible by name.
But what happens if i modify
a[1].value1 = 1000
is
d["aax"].value1 now 1000 or still "hello" as in this example ?
Any ideas to get access to the SAME object by index (0..n-1) AND by key ?
Emphasis here is on speed.
Thanks a lot,
Hans
Diez B. Roggisch
> Hello,
>
> I have a lot of items having a name and a given sequence.
>
> To access these items fast in a sequence order they should be used as
> a list, but to be fetched fast by name they also should be in a
> dictionary.
>
> Code could be something like this.
>
> class item
> def __init__(name, value1, value2, value3):
> self.name = name
> self.value1 = value1
> self.value2 = value2
>
> a = []
> a.append(item("foo", "bar", "text1"))
> a.append(item("xyz", "basd", "tsddsfxt1"))
> a.append(item("aax", "hello", "dont care"))
>
> in a, i have my objects in given order, fast accessible by index, e.g.
> a[2] to get the third one. Fine.
>
> Now I'd like to have a dict with references to thes objects like this:
>
> d = {}
> for x in a:
> d[a.name] = a # do I get a copy of a here or a new reference ?!
Only a reference.
>
> In d i now have a dict, fast accessible by name.
> But what happens if i modify
> a[1].value1 = 1000
> is
> d["aax"].value1 now 1000 or still "hello" as in this example ?
It's changed. Didn't you try that?
Diez
Hans Müller
> Hans Müller wrote:
>
>> Hello,
>>
>> I have a lot of items having a name and a given sequence.
>>
>> To access these items fast in a sequence order they should be used as
>> a list, but to be fetched fast by name they also should be in a
>> dictionary.
>>
>> Code could be something like this.
>>
>> class item
>> def __init__(name, value1, value2, value3):
>> self.name = name
>> self.value1 = value1
>> self.value2 = value2
>>
>> a = []
>> a.append(item("foo", "bar", "text1"))
>> a.append(item("xyz", "basd", "tsddsfxt1"))
>> a.append(item("aax", "hello", "dont care"))
>>
>> in a, i have my objects in given order, fast accessible by index, e.g.
>> a[2] to get the third one. Fine.
>>
>> Now I'd like to have a dict with references to thes objects like this:
>>
>> d = {}
>> for x in a:
>> d[a.name] = a # do I get a copy of a here or a new reference ?!
>
> Only a reference.
>
>> In d i now have a dict, fast accessible by name.
>> But what happens if i modify
>> a[1].value1 = 1000
>> is
>> d["aax"].value1 now 1000 or still "hello" as in this example ?
>
> It's changed. Didn't you try that?
>
> Diez
btw. how can I get a copy when I need it ?
Thanks a lot, Hans
Chris Rebert
> Diez B. Roggisch wrote:
>> Hans Müller wrote:
>>> But what happens if i modify
>>> a[1].value1 = 1000
>>> is
>>> d["aax"].value1 now 1000 or still "hello" as in this example ?
>>
>> It's changed. Didn't you try that?
>>
>
> btw. how can I get a copy when I need it ?
The `copy` module:
http://docs.python.org/library/copy.html
Cheers,
Chris
--
http://blog.rebertia.com
Terry Reedy
> Hello,
>
> I have a lot of items having a name and a given sequence.
>
> To access these items fast in a sequence order they should be used as a
> list, but to be fetched fast by name they also should be in a dictionary.
Have you looked at namedtuple or OrderedDict in collections module?