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dictionary containing a list

P: n/a
Ben
Hello...

I have set up a dictionary into whose values I am putting a list. I
loop around and around filling my list each time with new values, then
dumping this list into the dictionary. Or so I thought...

It would appear that what I am dumping into the dictionary value is
only a pointer to the original list, so after all my iterations all I
have is a dictionary whose every value is equal to that of the list the
final time I looped around :-(

Is there a way to acheive what I was attempting ? I have done something
almost identical with classes in a list before, and in that case a new
instance was created for each list entry...
I hope this makes some sense, and doesn't seem to head bangingly
simple...
Cheers,

Ben

Oct 6 '06 #1
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10 Replies


P: n/a

Ben wrote:
Hello...

I have set up a dictionary into whose values I am putting a list. I
loop around and around filling my list each time with new values, then
dumping this list into the dictionary. Or so I thought...

It would appear that what I am dumping into the dictionary value is
only a pointer to the original list, so after all my iterations all I
have is a dictionary whose every value is equal to that of the list the
final time I looped around :-(

Is there a way to acheive what I was attempting ? I have done something
almost identical with classes in a list before, and in that case a new
instance was created for each list entry...
I hope this makes some sense, and doesn't seem to head bangingly
simple...
Do you consult your physician over a video link while wearing a ninja
costume down an unlit coal mine at midnight?

Please consider the possibility that your description of what you think
your code might be doing is not enough for diagnosis.

You may need to supply:
(1) a listing of your code
(2) a small amount of input data
e.g. [(1, 'foo'), (42, 'bar'), (1, 'zot')]
(3) the output you expect from that input:
e.g. {1: ['foo', 'zot'], 42: ['bar']}

Cheers,
John

Oct 6 '06 #2

P: n/a
"Ben" <Be*************@gmail.comwrites:
I have set up a dictionary into whose values I am putting a list. I
loop around and around filling my list each time with new values,
then dumping this list into the dictionary. Or so I thought...
Our crystal balls are notoriously unreliable for viewing program code
that hasn't been posted.

--
\ "For man, as for flower and beast and bird, the supreme triumph |
`\ is to be most vividly, most perfectly alive" -- D.H. Lawrence. |
_o__) |
Ben Finney

Oct 6 '06 #3

P: n/a
Ben wrote:
I have set up a dictionary into whose values I am putting a list. I
loop around and around filling my list each time with new values, then
dumping this list into the dictionary. Or so I thought...

It would appear that what I am dumping into the dictionary value is
only a pointer to the original list, so after all my iterations all I
have is a dictionary whose every value is equal to that of the list the
final time I looped around :-(

Is there a way to acheive what I was attempting ?
Where you "loop around ... filling [your] list", use a new
list every time. You can create a new empty list with "[]".

--
--Bryan
Oct 7 '06 #4

P: n/a
On 6 Oct 2006 14:37:59 -0700, Ben <Be*************@gmail.comwrote:
Is there a way to acheive what I was attempting ? I have done something
almost identical with classes in a list before, and in that case a new
instance was created for each list entry...
Not sure what you're trying to pull off, but you may wish to copy the
items in question. (Questions indeed!). Dictionarys have their own
shallow copy method, surprisingly named copy, and there is also a copy
module that does shallow and deep copy (copy and deepcopy, resp.)

HTH,
Theerasak
Oct 7 '06 #5

P: n/a
I think what you mean is that if you change your list, it is changed
somewhere in your dicrionary to. Lists are always copied as pointers,
except explicitly told other wise. So a = b = [] makes a and be the
same list, and a.append(1) makes b - [1].
So do something like mydict[mykey] = mylist[:] (Slicing gives a copy
of the list, not the pointer).
Hope this helps.

Moi
Dolf

On 6 Oct 2006 14:37:59 -0700, "Ben" <Be*************@gmail.comwrote:
>Hello...

I have set up a dictionary into whose values I am putting a list. I
loop around and around filling my list each time with new values, then
dumping this list into the dictionary. Or so I thought...

It would appear that what I am dumping into the dictionary value is
only a pointer to the original list, so after all my iterations all I
have is a dictionary whose every value is equal to that of the list the
final time I looped around :-(

Is there a way to acheive what I was attempting ? I have done something
almost identical with classes in a list before, and in that case a new
instance was created for each list entry...
I hope this makes some sense, and doesn't seem to head bangingly
simple...
Cheers,

Ben
Oct 7 '06 #6

P: n/a
John Machin wrote:
Ben wrote:
>>Hello...

I have set up a dictionary into whose values I am putting a list. I
loop around and around filling my list each time with new values, then
dumping this list into the dictionary. Or so I thought...

It would appear that what I am dumping into the dictionary value is
only a pointer to the original list, so after all my iterations all I
have is a dictionary whose every value is equal to that of the list the
final time I looped around :-(

Is there a way to acheive what I was attempting ? I have done something
almost identical with classes in a list before, and in that case a new
instance was created for each list entry...
I hope this makes some sense, and doesn't seem to head bangingly
simple...


Do you consult your physician over a video link while wearing a ninja
costume down an unlit coal mine at midnight?

Please consider the possibility that your description of what you think
your code might be doing is not enough for diagnosis.

You may need to supply:
(1) a listing of your code
(2) a small amount of input data
e.g. [(1, 'foo'), (42, 'bar'), (1, 'zot')]
(3) the output you expect from that input:
e.g. {1: ['foo', 'zot'], 42: ['bar']}
One of the fascinating things about c.l.py is that sometimes a questin
will be posted that makes almost no sense to me, and somebody else will
casually read the OP's mind, home in on the issue and provide a useful
and relevant answer.

In this case it seems transparent to me, though probably not to you,
that Ben's problem is rootd in the following behaviour, well-known in
python but frequently confusing to noobs:
>>a = [1, 2, 3]
firstlist = a
a.append('another element')
firstlist
[1, 2, 3, 'another element']
>>>
Ben probably needs to look at creating copies using the list() type:
>>a = [1, 2, 3]
firstlist = list(a)
a.append('another element')
firstlist
[1, 2, 3]
>>>
or perhaps, in omore complex circumstances, using the copy module.

regards
Steve
--
Steve Holden +44 150 684 7255 +1 800 494 3119
Holden Web LLC/Ltd http://www.holdenweb.com
Skype: holdenweb http://holdenweb.blogspot.com
Recent Ramblings http://del.icio.us/steve.holden

Oct 7 '06 #7

P: n/a
Steve Holden wrote:
One of the fascinating things about c.l.py is that sometimes a questin
will be posted that makes almost no sense to me, and somebody else will
casually read the OP's mind, home in on the issue and provide a useful
and relevant answer.
if the assertions made by some about the documentation's unsuit-
ability for some are in fact true, that's probably some kind of
natural selection in action.

</F>

Oct 7 '06 #8

P: n/a

Steve Holden wrote:
John Machin wrote:
Ben wrote:
>Hello...

I have set up a dictionary into whose values I am putting a list. I
loop around and around filling my list each time with new values, then
dumping this list into the dictionary. Or so I thought...

It would appear that what I am dumping into the dictionary value is
only a pointer to the original list, so after all my iterations all I
have is a dictionary whose every value is equal to that of the list the
final time I looped around :-(

Is there a way to acheive what I was attempting ? I have done something
almost identical with classes in a list before, and in that case a new
instance was created for each list entry...
I hope this makes some sense, and doesn't seem to head bangingly
simple...

Do you consult your physician over a video link while wearing a ninja
costume down an unlit coal mine at midnight?

Please consider the possibility that your description of what you think
your code might be doing is not enough for diagnosis.

You may need to supply:
(1) a listing of your code
(2) a small amount of input data
e.g. [(1, 'foo'), (42, 'bar'), (1, 'zot')]
(3) the output you expect from that input:
e.g. {1: ['foo', 'zot'], 42: ['bar']}
One of the fascinating things about c.l.py is that sometimes a questin
will be posted that makes almost no sense to me, and somebody else will
casually read the OP's mind, home in on the issue and provide a useful
and relevant answer.

In this case it seems transparent to me, though probably not to you,
that Ben's problem is rootd in the following behaviour, well-known in
python but frequently confusing to noobs:
>>a = [1, 2, 3]
>>firstlist = a
>>a.append('another element')
>>firstlist
[1, 2, 3, 'another element']
>>>
It's quite transparent to me that his symptom is caused by the one list
being used throughout the exercise, instead of one per different dict
key. What you have described is one possibility.

Here's another possibility: Making the charitable assumption that he
has an outer loop and an inner loop, maybe (as I think another poster
has already suggested) all he needs to do is move "mylist = []" inside
the outer loop. Note that he doesn't say explicitly whether the one
list that he gets is the *correct* list for the last key, or whether
it's the catenation of all the correct lists, or something else.

Yet another: Noobs do all sorts of funny things. He could be operating
on a "clean the bucket out after each use instead making a new one"
paradigm:

| >>d= {}
| >>L = []
| >>L.append(1)
| >>L.append(2)
| >>d['a'] = L
| >>d
| {'a': [1, 2]}
| >>del L[:]
| >>d
| {'a': []}
| >>L.append(3)
| >>L.append(4)
| >>d['b'] = L
| >>d
| {'a': [3, 4], 'b': [3, 4]}

Cheers,
John

Oct 7 '06 #9

P: n/a
Ben
I think what you mean is that if you change your list, it is changed
somewhere in your dicrionary to. Lists are always copied as pointers,
except explicitly told other wise. So a = b = [] makes a and be the
same list, and a.append(1) makes b - [1].
So do something like mydict[mykey] = mylist[:] (Slicing gives a copy
of the list, not the pointer).
Hope this helps.

Moi
Dolf

Ah -this is exactly what I was doing wrong -thaks very much! Aologies
also for not posting sooner, I have been away for a few days.

Thanks for all of your help,

Ben

On 6 Oc
John Machin wrote:
Steve Holden wrote:
John Machin wrote:
Ben wrote:
>
>>Hello...
>>
>>I have set up a dictionary into whose values I am putting a list. I
>>loop around and around filling my list each time with new values, then
>>dumping this list into the dictionary. Or so I thought...
>>
>>It would appear that what I am dumping into the dictionary value is
>>only a pointer to the original list, so after all my iterations all I
>>have is a dictionary whose every value is equal to that of the list the
>>final time I looped around :-(
>>
>>Is there a way to acheive what I was attempting ? I have done something
>>almost identical with classes in a list before, and in that case a new
>>instance was created for each list entry...
>>
>>
>>I hope this makes some sense, and doesn't seem to head bangingly
>>simple...
>>
>
>
Do you consult your physician over a video link while wearing a ninja
costume down an unlit coal mine at midnight?
>
Please consider the possibility that your description of what you think
your code might be doing is not enough for diagnosis.
>
You may need to supply:
(1) a listing of your code
(2) a small amount of input data
e.g. [(1, 'foo'), (42, 'bar'), (1, 'zot')]
(3) the output you expect from that input:
e.g. {1: ['foo', 'zot'], 42: ['bar']}
>
One of the fascinating things about c.l.py is that sometimes a questin
will be posted that makes almost no sense to me, and somebody else will
casually read the OP's mind, home in on the issue and provide a useful
and relevant answer.

In this case it seems transparent to me, though probably not to you,
that Ben's problem is rootd in the following behaviour, well-known in
python but frequently confusing to noobs:
>>a = [1, 2, 3]
>>firstlist = a
>>a.append('another element')
>>firstlist
[1, 2, 3, 'another element']
>>>

It's quite transparent to me that his symptom is caused by the one list
being used throughout the exercise, instead of one per different dict
key. What you have described is one possibility.

Here's another possibility: Making the charitable assumption that he
has an outer loop and an inner loop, maybe (as I think another poster
has already suggested) all he needs to do is move "mylist = []" inside
the outer loop. Note that he doesn't say explicitly whether the one
list that he gets is the *correct* list for the last key, or whether
it's the catenation of all the correct lists, or something else.

Yet another: Noobs do all sorts of funny things. He could be operating
on a "clean the bucket out after each use instead making a new one"
paradigm:

| >>d= {}
| >>L = []
| >>L.append(1)
| >>L.append(2)
| >>d['a'] = L
| >>d
| {'a': [1, 2]}
| >>del L[:]
| >>d
| {'a': []}
| >>L.append(3)
| >>L.append(4)
| >>d['b'] = L
| >>d
| {'a': [3, 4], 'b': [3, 4]}

Cheers,
John
Oct 8 '06 #10

P: n/a
"Fredrik Lundh" <fr*****@pythonware.comwrote:

Steve Holden wrote:
One of the fascinating things about c.l.py is that sometimes a questin
will be posted that makes almost no sense to me, and somebody else will
casually read the OP's mind, home in on the issue and provide a useful
and relevant answer.

if the assertions made by some about the documentation's unsuit-
ability for some are in fact true, that's probably some kind of
natural selection in action.

</F>
LOL - Whining about documentation is what programmers do - its driven by the
"fact" that the docs and the implementation are "never" in sync, except on
something that is either trivial, or as old as the hills...

And it does not matter if the software is free and open source, or bought at
great expense - there are always these differences - sometimes niggly, and often
major - it sometimes looks as if the docs were a statement of intent, with the
implementation taking a left turn at the first crossroads.

- Hendrik
- Hendrik

Oct 9 '06 #11

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