Can python create a dictionary from a list comprehension?

[erikcw]

Hi,

I'm trying to turn o list of objects into a dictionary using a list
comprehension.

Something like

entries = {}
[entries[int(d.date.strf time('%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

[half.italian]

On May 27, 1:55 pm, erikcw <erikwickst...@ gmail.comwrote:

Hi,

I'm trying to turn o list of objects into a dictionary using a list
comprehension.

Something like

entries = {}
[entries[int(d.date.strf time('%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.__setit em__(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

[Stefan Sonnenberg-Carstens]

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.strf time('%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

[Pierre Quentel]

On 27 mai, 22:55, erikcw <erikwickst...@ gmail.comwrote:

Hi,

I'm trying to turn o list of objects into a dictionary using a list
comprehension.

Something like

entries = {}
[entries[int(d.date.strf time('%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.str ftime('%m')),d. id) for d in links] )

With Python2.4 and above you can use a "generator expression"

entries = dict( (int(d.date.str ftime('%m')),d. id) for d in links )
Regards,
Pierre

[Marc 'BlackJack' Rintsch]

In <11************ *********@o11g2 000prd.googlegr oups.com>, half.italian
wrote:

[entries.__setit em__(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

[Wim Vogelaar]

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 }

[Maric Michaud]

Pierre Quentel a écrit :

On 27 mai, 22:55, erikcw <erikwickst...@ gmail.comwrote:

>Hi,

I'm trying to turn o list of objects into a dictionary using a list
comprehensio n.

....

>
entries = dict([ (int(d.date.str ftime('%m')),d. id) for d in links] )

With Python2.4 and above you can use a "generator expression"

entries = dict( (int(d.date.str ftime('%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}

[Tim Churches]

http://shaheeilyas.com/flags/

Scroll to the bottom to see why this is not entirely off-topic. Are
there other public examples in which Python has been used to harvest and
represent public information in useful and/or interesting ways? Ideas
for some more?

Tim C

[Wim Vogelaar]

>
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

[Gabriel Genellina]

En Mon, 28 May 2007 05:20:16 -0300, Wim Vogelaar
<wi************ **************@ bag.python.orge scribió:

>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!)

--
Gabriel Genellina