Is there a function to initialize a dictionary?
Right now I'm using:
d = {x+1:[] for x in range(50)}
Is there any better solution?
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On 2009-12-22 15:33 PM, mattia wrote:
Is there a function to initialize a dictionary?
Right now I'm using:
d = {x+1:[] for x in range(50)}
Is there any better solution?
For things like this? No. If you find yourself writing this pattern frequently,
though, you can wrap it up in a function and
On 12/22/2009 1:33 PM mattia said...
Is there a function to initialize a dictionary?
Right now I'm using:
d = {x+1:[] for x in range(50)}
Is there any better solution?
I tend to use setdefault and fill in as I go, but if you need to have a
complete 50-element dict from the get go, I'd probably
On 22-12-2009 22:33, mattia wrote:
Is there a function to initialize a dictionary?
Right now I'm using:
d = {x+1:[] for x in range(50)}
Is there any better solution?
Maybe you can use:
dict.fromkeys(xrange(1,51))
but this will initialize all values to None instead of an empty list...
-
"mattia" wrote in message
news:4b313b3a$0$1135$4fafb...@reader1.news.tin.it...
Is there a function to initialize a dictionary?
Right now I'm using:
d = {x+1:[] for x in range(50)}
Is there any better solution?
Depending on your use case, a defaultdict might suite you:
from collections impo
mattia wrote:
> Is there a function to initialize a dictionary?
> Right now I'm using:
> d = {x+1:[] for x in range(50)}
> Is there any better solution?
There is a dictionary variant that you don't have to initialize:
from collections import defaultdict
d = defaultdict(list)
Peter
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http://ma
Il Tue, 22 Dec 2009 23:09:04 +0100, Peter Otten ha scritto:
> mattia wrote:
>
>> Is there a function to initialize a dictionary? Right now I'm using:
>> d = {x+1:[] for x in range(50)}
>> Is there any better solution?
>
> There is a dictionary variant that you don't have to initialize:
>
> from
Il Tue, 22 Dec 2009 23:09:04 +0100, Peter Otten ha scritto:
> mattia wrote:
>
>> Is there a function to initialize a dictionary? Right now I'm using:
>> d = {x+1:[] for x in range(50)}
>> Is there any better solution?
>
> There is a dictionary variant that you don't have to initialize:
>
> from
On Dec 22, 11:51 pm, mattia wrote:
> Il Tue, 22 Dec 2009 23:09:04 +0100, Peter Otten ha scritto:
>
> > mattia wrote:
>
> >> Is there a function to initialize a dictionary? Right now I'm using:
> >> d = {x+1:[] for x in range(50)}
> >> Is there any better solution?
>
> > There is a dictionary varia
mattia wrote:
Il Tue, 22 Dec 2009 23:09:04 +0100, Peter Otten ha scritto:
mattia wrote:
Is there a function to initialize a dictionary? Right now I'm using:
d = {x+1:[] for x in range(50)}
Is there any better solution?
There is a dictionary variant that you don't have to initialize:
from co
I'd like to run this idea by the community to see if it's PEP worthy and
hasn't been already rejected.
Background
Just as keyword arguments enhance code readability and diminish the risk of
bugs, so too would named
return values. Currently, we can write
val1, val2, val3 = myfunc()
but we must
On 6/21/2016 3:34 AM, Ari Freund via Python-list wrote:
I'd like to run this idea by the community to see if it's PEP worthy and
hasn't been already rejected.
Background
Just as keyword arguments enhance code readability and diminish the risk of
bugs, so too would named
return values. Currently
On 2016-06-21 18:12, Terry Reedy wrote:
On 6/21/2016 3:34 AM, Ari Freund via Python-list wrote:
I'd like to run this idea by the community to see if it's PEP worthy and
hasn't been already rejected.
Background
Just as keyword arguments enhance code readability and diminish the risk of
bugs, so
On Tue, Jun 21, 2016, 10:14 AM Terry Reedy wrote:
> On 6/21/2016 3:34 AM, Ari Freund via Python-list wrote:
> > I'd like to run this idea by the community to see if it's PEP worthy and
> > hasn't been already rejected.
> >
> There was a recent (last couple of months?) discussion on python-ideas
>
On Tue, 21 Jun 2016 05:34 pm, Ari Freund wrote:
> I'd like to run this idea by the community to see if it's PEP worthy and
> hasn't been already rejected.
>
> Background
> Just as keyword arguments enhance code readability and diminish the risk
> of bugs, so too would named
> return values.
I do
Thanks everybody. There seems to be a lot of resistance to dict unpacking, in
addition to the problem with my proposed shorthand dict() initialization syntax
pointed out by Steven D'Aprano, so I won't be pursuing this.
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On 26 June 2016 at 18:28, Ari Freund via Python-list
wrote:
> Thanks everybody. There seems to be a lot of resistance to dict unpacking,
> in addition to the problem with my proposed shorthand dict() initialization
> syntax pointed out by Steven D'Aprano, so I won'
On Tue, Jun 21, 2016 at 9:41 PM Steven D'Aprano wrote:
> On Tue, 21 Jun 2016 05:34 pm, Ari Freund wrote:
> > var3, var1, var2 = **d
>
> But I don't want to use the key names your function uses. I want to
> use names which makes sense for my application
>
Note that my dict unpacking synt
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