Awesome! thats what i was looking for - pure python one-liner!
On 09/09/2010 00:29, Steve Dower wrote:
Since we're all contributing here:
d = { key: (v1, v2) for key, v1, v2 in zip(seq[::3], seq[1::3], seq[2::3]) }
Or
d = dict((key, (v1, v2)) for key, v1, v2 in zip(seq[::3], seq[1::3], seq[2::3])
using Python 2.6. The first uses a dictionary comprehension, which is
new in 2.7/3 (but not IronPython 2.7 it would seem...)
On Thu, Sep 9, 2010 at 04:08, Iivari Mokelainen<iiv...@mokelainen.com> wrote:
I thought you guys would suggest something like this
def parseOffsets(str):
seq = str.split()
dict = {}
while seq != []:
name, x, y = seq[:3]
dict[name] = (int(x),int(y))
del seq[:3]
return dict
which seems to me as something in more of python-spirit? :D
well, anyways, thanks - now i know i dont have to bend backwards with
restricting myself from using simple old paradigms (like for loop) in my
python code (:
Cheers!
On 08/09/2010 17:13, Vernon Cole wrote:
If you bring up Python in interactive mode, and type "import this", some
sage advice will appear:
v v v v v
C:\Users\vernon>ipy
C:\Users\vernon>"c:\program files\IronPython 2.7\ipy.exe"
IronPython 2.7 Alpha 1 (2.7.0.1) on .NET 4.0.30319.1
import this
Explicit is better than implicit.
Simple is better than complex.
Readability counts.
^ ^ ^ ^
In that spirit, may I suggest the following simple, readable code?
<code>
ml = ['name1','x1','y1','name2','x2','y2']
i = 0
d = {}
while True:
try:
d[ml[i]] = (ml[i+1],ml[i+2])
except IndexError:
break
i += 3
print repr(d)
</code>
On Wed, Sep 8, 2010 at 5:29 AM, Iivari Mokelainen<iiv...@mokelainen.com>
wrote:
I need to unpack a list of
[name, x, y, name, x, y]
to a dictionary
{name : (x,y), name:(x,y)}
how would one do that?
Also, is it ok to ask such newbie questions here, or is there a better
place for that?
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