On Thu, May 15, 2008 at 2:40 AM, GTXY20 <gtxy20 <at> gmail.com> wrote:
> Hello all,
>
> I have dictionary like the following:
>
> d={(1,23A):[a,b,c,d],  (1,24A):[b,c,d], (2,23A):[a,b], (2,24A):[a,b,c,d]}
>
> I would like to iterate through the dictionary such that if it finds
> the value 'a' in the values of the key that it would remove the value
> 'b' from the values list. In addition if it finds 'a' in the values
> list I would like it to take the first item from the key tuple k[0]
> and and then look through the dictionary for that k[0] value and then
> remove 'b' from its value list even if 'a' is not present in that
> list. So at the end I would like my dictionary to end with:
>
> d={(1,23A):[a,c,d],  (1,24A):[c,d], (2,23A):[a], (2,24A):[a,c,d]}
>
> Initally I did the following:
>
> for k,v in d.items():
> u=k[0]
> b=k[1]
> if 'a' in v:
> for k,v in d.items():
> if k[0] == u:
> for values in v:
> if values == 'b':
> v.remove(values)
>
> However this will be a very big dictionary and it ended up running for
> a very long time - in fact I had to kill the process.

The main problem is the nested loops. Every time you find an 'a' you
search the whole dict for items starting with matching keys.  There
are some smaller problems as well - you could use d.iteritems()
instead of d.items() to avoid creating intermediate lists, and there
are better ways to remove 'b' from v.

The first thing I would do is reduce this to two passes over the dict
- the first pass can find the keys, the second to delete 'b' from the
list. For example,
toRemove=set(k[0] for k, v in d.iteritems() if 'b' in v)
for k,v in d.iteritems():
   if k[0] in toRemove:
       try:
           v.remove('b')
       except ValueError:
           pass

If this is too slow, consider splitting up your keys - make a
two-level dict so you can find all the keys starting with a particular
value by a dict lookup instead of search.

Kent