trictionary?

Adam Tomjack adamtj at adamtj.org
Mon Aug 29 00:55:00 CEST 2005


Oops, I found a bug in my previous code.

If you say

   bin_item = bin.setdefault(x, [1, 0])
   bin_item[0] += 1

then if x wasn't in bin, it'll get initialized to [1, 0], then 
incremented to [2, 0] in the first loop.  The code you asked about would 
have produced [1, 0].  Instead, you can say:

   bin_item = bin.setdefault(x, [0, 0])
   bin_item[0] += 1

That should be equivalent.  My example with the class is similarly broken.

Adam




Adam Tomjack wrote:
> Randy,
> 
> I'd probably use a two element list.
> 
> Instead of using an if/else to check if an element is in your dict and 
> initialize it, you can use the setdefault() function.  The docs for 
> dictionaries explain it pretty well.
> 
>       bin = {}
>       for whatever:
>          for [a, b] in foo:
>   	  x = 42 - a
>   	  bin_item = bin.setdefault(x, [1, 0])
>   	  bin_item[0] += 1
>       for x, (y, z) in bin.iteritems():
>          print x, y, z
> 
> You could also use a class like a C-style struct if you want named items:
> 
> class BinItem:
>    def __init__(self, s=0, t=0):
>      self.s = s
>      self.t = t
> 
> bin = {}
> for a, b in foo:
>    x = 42 - a
>    bin_item = bin.setdefault(x, BinItem(1, 0))
>    bin_item.s += 1
> 
> for x, item in bin.iteritems():
>    print x, item.s, item.t
> 
> 
> Luck in battle,
> 
> Adam
> 
> 
> Randy Bush wrote:
> 
>>i have some code which looks kinda like 
>>
>>    bin = {}
>>    for whatever:
>>       for [a, b] in foo:
>>	  x = 42 - a
>>	  y = 42 - b
>>	  if bin.has_key(x):
>>	     bin[x] += 1
>>	  else:
>>	     bin[x] = 1
>>    for i, j in bin.iteritems():
>>       print i, j
>>
>>now i want to add a second count column, kinda like
>>
>>    bin = {}
>>    for whatever:
>>       for [a, b] in foo:
>>	  x = 42 - a
>>	  if bin.has_key(x):
>>	     bin[x.b] += 1
>>	  else:
>>	     bin[x.b] = 1
>>	     bin[x.not b] = 0
>>    for x, y, z in bin.iteritems():
>>       print x, y, z
>>
>>should the dict value become a two element list, or is
>>there a cleaner way to do this?
>>
>>randy
>>
> 
> 
> 
> 
> 




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