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@gmail.com]

Re: Can python create a dictionary from a list comprehension?

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]

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.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]

Re: Can python create a dictionary from a list comprehension?

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]

Re: Can python create a dictionary from a list comprehension?

In <1180299814.129 770.93220@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]

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 }

[Maric Michaud]

Re: Can python create a dictionary from a list comprehension?

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]

Flags of the world

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]

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

[Gabriel Genellina]

Re: Can python create a dictionary from a list comprehension?

En Mon, 28 May 2007 05:20:16 -0300, Wim Vogelaar
<wim.vogelaarat mc2worlddotorg@ 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

[half.italian@gmail.com]

Re: Can python create a dictionary from a list comprehension?

On May 28, 12:25 am, Marc 'BlackJack' Rintsch <bj_...@gmx.net wrote:

In <1180299814.129 770.93...@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

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

[Peter Otten]

Re: Can python create a dictionary from a list comprehension?

half.italian@gm ail.com 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

[Gabriel Genellina]

Re: Can python create a dictionary from a list comprehension?

En Mon, 28 May 2007 05:37:12 -0300, Wim Vogelaar
<wim.vogelaarat mc2worlddotorg@ bag.python.orge scribió:

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

--
Gabriel Genellina

[Duncan Booth]

Re: Can python create a dictionary from a list comprehension?

"Wim Vogelaar" <wim.vogelaar at mc2world dot orgwrote:

>

>>
>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(r ange(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.

[Ben Finney]

Re: Flags of the world

Tim Churches <tchur@optushom e.com.auwrites:

http://shaheeilyas.com/flags/

>
Scroll to the bottom to see why this is not entirely off-topic.

I fail to see what it has to do with the thread you're replyiing to,
which is a discussion of creating a dictionary from a list
comprehension.

If you want to start a new thread, compose a new message, not a reply
to an existing thread.

--
\ "If you ever catch on fire, try to avoid seeing yourself in the |
`\ mirror, because I bet that's what REALLY throws you into a |
_o__) panic." -- Jack Handey |
Ben Finney