Can python create a dictionary from a list comprehension?
Hi,
I'm trying to turn o list of objects into a dictionary using a list
comprehension.
Something like
entries = {}
[entries[int(d.date.strftime('%m'))] = d.id] for d in links]
I keep getting errors when I try to do it. Is it possible? Do
dictionary objects have a method equivalent to [].append? Maybe a
lambda?
Thanks for your help!
Erik
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Re: Can python create a dictionary from a list comprehension?
On May 27, 1:55 pm, erikcw <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I'm trying to turn o list of objects into a dictionary using a list
> comprehension.
>
> Something like
>
> entries = {}
> [entries[int(d.date.strftime('%m'))] = d.id] for d in links]
>
> I keep getting errors when I try to do it. Is it possible? Do
> dictionary objects have a method equivalent to [].append? Maybe a
> lambda?
>
> Thanks for your help!
> Erik
try...
[entries.__setitem__(int(d.date.strftime('%m'))], d.id) for d in
links]
btw...I was curious of this too. I used 'dir(dict)' and looked for a
method that might do what we wanted and bingo!
~Sean
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Re: Can python create a dictionary from a list comprehension?
erikcw schrieb:
> Hi,
>
> I'm trying to turn o list of objects into a dictionary using a list
> comprehension.
>
> Something like
>
> entries = {}
> [entries[int(d.date.strftime('%m'))] = d.id] for d in links]
>
> I keep getting errors when I try to do it. Is it possible? Do
> dictionary objects have a method equivalent to [].append? Maybe a
> lambda?
>
> Thanks for your help!
> Erik
>
>
normally a dict(whatEver) will do ;-)
Example:
a = [1,2,3,4,5,6,7,8,9,10]
aDict = dict([(x,x+1) for x in a if x%2==0])
print aDict
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Re: Can python create a dictionary from a list comprehension?
On 27 mai, 22:55, erikcw <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I'm trying to turn o list of objects into a dictionary using a list
> comprehension.
>
> Something like
>
> entries = {}
> [entries[int(d.date.strftime('%m'))] = d.id] for d in links]
>
> I keep getting errors when I try to do it. Is it possible? Do
> dictionary objects have a method equivalent to [].append? Maybe a
> lambda?
>
> Thanks for your help!
> Erik
entries = dict([ (int(d.date.strftime('%m')),d.id) for d in links] )
With Python2.4 and above you can use a "generator expression"
entries = dict( (int(d.date.strftime('%m')),d.id) for d in links )
Regards,
Pierre
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Re: Can python create a dictionary from a list comprehension?
In <[EMAIL PROTECTED]>, half.italian
wrote:
> [entries.__setitem__(int(d.date.strftime('%m'))], d.id) for d in
> links]
>
> btw...I was curious of this too. I used 'dir(dict)' and looked for a
> method that might do what we wanted and bingo!
This is really ugly. Except `__init__()` it's always a code smell if you
call a "magic" method directly instead of using the corresponding
syntactic sugar or built in function. And using a list comprehension just
for side effects is misleading because the reader expects a (useful) list
to be build when stumbling over a list comp and it's wasteful because an
unnecessary list of `None`\s is build and thrown away for no reason other
than to have a one liner. This is not Perl! ;-)
Ciao,
Marc 'BlackJack' Rintsch
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Re: Can python create a dictionary from a list comprehension?
> Example:
>
> a = [1,2,3,4,5,6,7,8,9,10]
>
> aDict = dict([(x,x+1) for x in a if x%2==0])
>
> print aDict
>
When I run this program I get:
{8: 9, 2: 3, 4: 5, 10: 11, 6: 7}
why this output isn't ordered, giving:
{2: 3, 4: 5, 6: 7, 8: 9, 10: 11 }
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Re: Can python create a dictionary from a list comprehension?
Pierre Quentel a écrit :
> On 27 mai, 22:55, erikcw <[EMAIL PROTECTED]> wrote:
>> Hi,
>>
>> I'm trying to turn o list of objects into a dictionary using a list
>> comprehension.
...
>
> entries = dict([ (int(d.date.strftime('%m')),d.id) for d in links] )
>
> With Python2.4 and above you can use a "generator expression"
>
> entries = dict( (int(d.date.strftime('%m')),d.id) for d in links )
>
You can also create dictionaries knowing only the keys the same way (ie.
a two-dimensional array) :
In [77]: dict.fromkeys((a, b) for a in range(4) for b in range(2))
Out[78]:
{(0, 0): None,
(0, 1): None,
(1, 0): None,
(1, 1): None,
(2, 0): None,
(2, 1): None,
(3, 0): None,
(3, 1): None}
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Re: Can python create a dictionary from a list comprehension?
En Mon, 28 May 2007 05:20:16 -0300, Wim Vogelaar
<[EMAIL PROTECTED]> escribió:
>> Example:
>>
>> a = [1,2,3,4,5,6,7,8,9,10]
>>
>> aDict = dict([(x,x+1) for x in a if x%2==0])
>>
>> print aDict
>>
>
> When I run this program I get:
> {8: 9, 2: 3, 4: 5, 10: 11, 6: 7}
>
> why this output isn't ordered, giving:
> {2: 3, 4: 5, 6: 7, 8: 9, 10: 11 }
A dictionary is not ordered, no matter how you create it. If you want to
process the keys in order:
for key in sorted(aDict):
print key, '=', aDict[key]
(Note that sorted(aDict) returns a *list*, not a dictionary!)
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Re: Can python create a dictionary from a list comprehension?
>
> why this output isn't ordered, giving:
> {2: 3, 4: 5, 6: 7, 8: 9, 10: 11 }
>
>
I made the original list two elements longer: a =
[1,2,3,4,5,6,7,8,9,10,11,12]
and to my surprise the output is now ordered, giving: {2: 3, 4: 5, 6: 7, 8:
9, 10: 11, 12: 13}
I am running ActiveState ActivePython 2.5
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Re: Can python create a dictionary from a list comprehension?
On May 28, 12:25 am, Marc 'BlackJack' Rintsch <[EMAIL PROTECTED]> wrote:
> In <[EMAIL PROTECTED]>, half.italian
> wrote:
>
> > [entries.__setitem__(int(d.date.strftime('%m'))], d.id) for d in
> > links]
>
> > btw...I was curious of this too. I used 'dir(dict)' and looked for a
> > method that might do what we wanted and bingo!
>
> This is really ugly. Except `__init__()` it's always a code smell if you
> call a "magic" method directly instead of using the corresponding
> syntactic sugar or built in function. And using a list comprehension just
> for side effects is misleading because the reader expects a (useful) list
> to be build when stumbling over a list comp and it's wasteful because an
> unnecessary list of `None`\s is build and thrown away for no reason other
> than to have a one liner. This is not Perl! ;-)
>
> Ciao,
> Marc 'BlackJack' Rintsch
It's ugly I agree, but it was the first solution I found. I need you
guys for the _right_ solutions :) I have stumbled over the same
situation myself. I don't see that the list comprehension itself is
misleading. If nothing is catching the empty list that is returned,
it signals that the returned list is unimportant, and if wrapped by a
call to dict() its obvious also.
Do you think we just shouldn't use list comprehensions to build
dictinaries at all? Or is Stefan's solution acceptable (and pythonic)?
~Sean
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Re: Can python create a dictionary from a list comprehension?
[EMAIL PROTECTED] wrote: > Do you think we just shouldn't use list comprehensions to build > dictinaries at all? Or is Stefan's solution acceptable (and pythonic)? Use list comprehensions where you need the resulting list; if you want nothing but the side effects, use a for loop. [Stefan Sonnenberg-Carstens] > a = [1,2,3,4,5,6,7,8,9,10] > aDict = dict([(x,x+1) for x in a if x%2==0]) Stefan's example meets the above criterion, so yes, it's acceptable. In Python 2.5 you would use a generator expression, though: aDict = dict((x, x+1) for x in a if x % 2 ==0) Peter -- http://mail.python.org/mailman/listinfo/python-list
Re: Can python create a dictionary from a list comprehension?
En Mon, 28 May 2007 05:37:12 -0300, Wim Vogelaar
<[EMAIL PROTECTED]> escribió:
> I made the original list two elements longer: a =
> [1,2,3,4,5,6,7,8,9,10,11,12]
>
> and to my surprise the output is now ordered, giving: {2: 3, 4: 5, 6: 7,
> 8:
> 9, 10: 11, 12: 13}
>
> I am running ActiveState ActivePython 2.5
Keys in a dictionary are listed in an arbitrary order; the *only* thing
about the ordering you can say is that, given a FIXED dictionary (already
constructed, and without any other intervening operation that could alter
its content), when you iterate over its keys (using .keys(), .iterkeys()),
its values (.values(), .itervalues()) or items (.items(), .iteritems())
you will always get the same things in the same order over and over.
If you create the dictionary using a different sequence of insertions and
deletions, you may get different results. If you insert and delete things
afterwards, you may get different results. If you exit the program and run
it again, you may get different results. The *only* guaranteed fact is
that you will get the same results provided you don't modify the
dictionary at all.
See note (3) in http://docs.python.org/lib/typesmapping.html
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Re: Can python create a dictionary from a list comprehension?
"Wim Vogelaar" wrote:
>
>>
>> why this output isn't ordered, giving:
>> {2: 3, 4: 5, 6: 7, 8: 9, 10: 11 }
>>
>>
>
> I made the original list two elements longer: a =
> [1,2,3,4,5,6,7,8,9,10,11,12]
>
> and to my surprise the output is now ordered, giving: {2: 3, 4: 5, 6:
> 7, 8: 9, 10: 11, 12: 13}
>
... and it will become unordered again when you get up to storing 32 in the
dictionary, and ordered again when you get up to 44.
It is an unfortunate side-effect of the way that Python dictionaries work
that it often appears when you store small integers as though they are
being stored sorted. In fact the order depends on the order in which you
inserted the values into the dictionary, the current size allocated to the
dictionary, and whether any slots have previously been occupied by other
values.
In your original output:
{8: 9, 2: 3, 4: 5, 10: 11, 6: 7}
the keys 2, 4, and 6 are inserted into the 2nd, 4th, 6th slots of the
dictionary (counting from 0 and there are initially 8 slots), 8 goes into
the 0th slot, 10 would go into the 2nd slot except that is filled so it
ends up in the 5th slot.
When you insert 12, the initial dictionary is too full (it never fills all
the slots), so it is resized to have 32 slots. Again the 32 wraps round to
the 0th slot and higher values all collide with filled slots so they end up
in less obvious positions:
>>> for i in range(12,42,2):
print dict.fromkeys(range(2,i,2))
{8: None, 2: None, 4: None, 10: None, 6: None}
{2: None, 4: None, 6: None, 8: None, 10: None, 12: None}
{2: None, 4: None, 6: None, 8: None, 10: None, 12: None, 14: None}
{2: None, 4: None, 6: None, 8: None, 10: None, 12: None, 14: None, 16:
None}
{2: None, 4: None, 6: None, 8: None, 10: None, 12: None, 14: None, 16:
None, 18: None}
{2: None, 4: None, 6: None, 8: None, 10: None, 12: None, 14: None, 16:
None, 18: None, 20: None}
{2: None, 4: None, 6: None, 8: None, 10: None, 12: None, 14: None, 16:
None, 18: None, 20: None, 22: None}
{2: None, 4: None, 6: None, 8: None, 10: None, 12: None, 14: None, 16:
None, 18: None, 20: None, 22: None, 24: None}
{2: None, 4: None, 6: None, 8: None, 10: None, 12: None, 14: None, 16:
None, 18: None, 20: None, 22: None, 24: None, 26: None}
{2: None, 4: None, 6: None, 8: None, 10: None, 12: None, 14: None, 16:
None, 18: None, 20: None, 22: None, 24: None, 26: None, 28: None}
{2: None, 4: None, 6: None, 8: None, 10: None, 12: None, 14: None, 16:
None, 18: None, 20: None, 22: None, 24: None, 26: None, 28: None, 30: None}
{32: None, 2: None, 4: None, 6: None, 8: None, 10: None, 12: None, 14:
None, 16: None, 18: None, 20: None, 22: None, 24: None, 26: None, 28: None,
30: None}
{32: None, 2: None, 4: None, 6: None, 8: None, 10: None, 12: None, 34:
None, 14: None, 16: None, 18: None, 20: None, 22: None, 24: None, 26: None,
28: None, 30: None}
{32: None, 2: None, 4: None, 6: None, 8: None, 10: None, 12: None, 34:
None, 14: None, 16: None, 18: None, 20: None, 22: None, 24: None, 36: None,
26: None, 28: None, 30: None}
{32: None, 2: None, 4: None, 38: None, 6: None, 8: None, 10: None, 12:
None, 34: None, 14: None, 16: None, 18: None, 20: None, 22: None, 24: None,
36: None, 26: None, 28: None, 30: None}
The fact that integers hash to themselves may be unlikely to change, but
the sizes of the hash tables, or the conflict resolution strategy when an
insertion hits an already used slot could all vary in other versions of
Python.
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