creating and appending to a dictionary of a list of lists | | |
Hey,
I started with this:
factByClass = {}
def update(key, x0, x1, x2, x3):
x = factByClass.setdefault(key, [ [], [], [], [] ])
x[0].append(x0)
x[1].append(x1)
x[2].append(x2)
x[3].append(x3)
update('one', 1, 2, 3, 4)
update('one', 5, 6, 7, 8)
update('two', 9, 10, 11, 12)
print factByClass
{'two': [[9], [10], [11], [12]], 'one': [[1, 5], [2, 6], [3, 7], [4,
8]]}
I then 'upgraded' to this:
def update(key, *args):
x = factByClass.setdefault(key, [[], [], [], [] ])
for i, v in enumerate(args):
x[i].append(v)
Is there a better way?
Cheers!
| | | | re: creating and appending to a dictionary of a list of lists
On Aug 15, 3:30 am, pyscottish...@hotmail.com wrote: Quote:
Hey,
>
I started with this:
>
factByClass = {}
>
.... Quote:
def update(key, *args):
x = factByClass.setdefault(key, [[], [], [], [] ])
for i, v in enumerate(args):
x[i].append(v)
>
Is there a better way?
Well, the following is perhaps neater: Quote: Quote: Quote:
>>factByClass = defaultdict(lambda: [[],[],[],[]])
>>def update(key, *args):
.... map(list.append, factByClass[key], args)
.... Quote: Quote: Quote:
>>update('one', 1, 2, 3, 4)
>>update('one', 5, 6, 7, 8)
>>update('two', 9, 10, 11, 12)
>>>
>>print factByClass
defaultdict(<function <lambdaat 0x00F73430>, {'two': [[9], [1
0], [11], [12]], 'one': [[1, 5], [2, 6], [3, 7], [4, 8]]})
It abuses the fact that list.append modifies the list in place -
normally you would use map to get a new list object. In this case the
new list returned by map is just a list of None's (since append
returns None - a common idiom for functions that operate by side
effect), and so is not used directly.
--
Ant... http://antroy.blogspot.com/ | | | | re: creating and appending to a dictionary of a list of lists
On Aug 15, 8:08 am, Ant <ant...@gmail.comwrote: Quote:
On Aug 15, 3:30 am, pyscottish...@hotmail.com wrote:
> > Quote:
I started with this:
> >
... Quote:
def update(key, *args):
x = factByClass.setdefault(key, [[], [], [], [] ])
for i, v in enumerate(args):
x[i].append(v)
> Quote:
Is there a better way?
>
Well, the following is perhaps neater:
> Quote: Quote:
>factByClass = defaultdict(lambda: [[],[],[],[]])
>def update(key, *args):
>
... map(list.append, factByClass[key], args)
...>>update('one', 1, 2, 3, 4) Quote: Quote:
>update('one', 5, 6, 7, 8)
>update('two', 9, 10, 11, 12)
> Quote: Quote:
>print factByClass
>
defaultdict(<function <lambdaat 0x00F73430>, {'two': [[9], [1
0], [11], [12]], 'one': [[1, 5], [2, 6], [3, 7], [4, 8]]})
>
It abuses the fact that list.append modifies the list in place -
normally you would use map to get a new list object. In this case the
new list returned by map is just a list of None's (since append
returns None - a common idiom for functions that operate by side
effect), and so is not used directly.
>
--
Ant...
> http://antroy.blogspot.com/ Nice. I like it. Thanks a lot! |  |
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