Hi,
I posted this (by accident) off the list:
On 2012-11-14, at 23:43 , Chris Withers wrote:
On 14/11/2012 22:37, Chris Withers wrote:
On 14/11/2012 10:11, mar...@v.loewis.de wrote:
def xdict(**kwds):
return kwds
Hah, good call, this trumps both of the other options:
100 -r 5 -v
mar...@v.loewis.de wrote:
It's faster than calling dict() because the dict code will
create a second dictionary, and discard the keywords dictionary.
Perhaps in the case where dict() is called with keyword
args only, it could just return the passed-in keyword
dictionary instead of creating
Greg Ewing, 15.11.2012 11:48:
mar...@v.loewis.de wrote:
It's faster than calling dict() because the dict code will
create a second dictionary, and discard the keywords dictionary.
Perhaps in the case where dict() is called with keyword
args only, it could just return the passed-in keyword
On 11/15/2012 9:58 AM, Stefan Behnel wrote:
Greg Ewing, 15.11.2012 11:48:
mar...@v.loewis.de wrote:
It's faster than calling dict() because the dict code will
create a second dictionary, and discard the keywords dictionary.
Perhaps in the case where dict() is called with keyword
args only,
On 15/11/2012 4:21pm, Terry Reedy wrote:
I was thinking that CPython could check the ref count of the input
keyword dict to determine whether it is newly created and can be
returned or is pre-existing and must be copied. But it seems not so.
def d(**x): return sys.getrefcount(x)
import sys
Stefan Behnel wrote:
This should work as long as this still creates a copy of d at some point:
d = {...}
dict(**d)
It will -- the implementation of the function call opcode always
creates a new keyword dict for passing to the called function.
--
Greg
On Nov 14, 2012, at 5:37 PM, Chris Withers wrote:
On 14/11/2012 10:11, mar...@v.loewis.de wrote:
Zitat von Chris Withers ch...@simplistix.co.uk:
a_dict = dict(
x = 1,
y = 2,
z = 3,
...
)
What can we do to speed up the former case?
It should be possible to
On 14/11/2012 10:11, mar...@v.loewis.de wrote:
Zitat von Chris Withers ch...@simplistix.co.uk:
a_dict = dict(
x = 1,
y = 2,
z = 3,
...
)
What can we do to speed up the former case?
It should be possible to special-case it. Rather than creating
a new dictionary from
On 14/11/2012 22:37, Chris Withers wrote:
On 14/11/2012 10:11, mar...@v.loewis.de wrote:
def xdict(**kwds):
return kwds
Hah, good call, this trumps both of the other options:
$ python2.7 -m timeit -n 100 -r 5 -v
{'a':1,'b':2,'c':3,'d':4,'e':5,'f':6,'g':7}
raw times: 1.45 1.45 1.44
On 2012-11-14, at 23:43 , Chris Withers wrote:
On 14/11/2012 22:37, Chris Withers wrote:
On 14/11/2012 10:11, mar...@v.loewis.de wrote:
def xdict(**kwds):
return kwds
Hah, good call, this trumps both of the other options:
$ python2.7 -m timeit -n 100 -r 5 -v
Zitat von Chris Withers ch...@simplistix.co.uk:
$ python2.7 -m timeit -n 100 -r 5 -v
{'a':1,'b':2,'c':3,'d':4,'e':5,'f':6,'g':7}
raw times: 1.49 1.49 1.5 1.49 1.48
100 loops, best of 5: 1.48 usec per loop
$ python2.7 -m timeit -n 100 -r 5 -v 'dict(a=1,b=2,c=3,d=4,e=5,f=6,g=7)'
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