No that makes sense Tim, Thanks again for all your help on this one, it'll all prove invaluable I'm sure. I managed to crack all the big troubles last week with my reoccurring tasks, it's just a case of tidying up a few of these loose ends, then that'll be python project no.1 over and done with :-D
Thanks, Rob -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Tim Golden Sent: 30 April 2007 12:54 Cc: python-list@python.org Subject: Re: Dict Copy & Compare Robert Rawlins - Think Blue wrote: > On[e] quick question, how can I order a dict by > the 'values' (not keys) before looping? Is that possible? Depends on what you want to do. You can loop on the sorted values very easily: <code> d1 = dict (a=2, b=1) for value in sorted (d1.values): print value </code> but inferring the equivalent key is, in effect, not possible since more than one key might correspond to that value. Depending on what you're after, the following technique might be of use: <code> import operator d = dict (a=2, b=1, c=-1, d=4) for k, v in sorted ( d.items (), key=operator.itemgetter (1) ): print k, "=>", v </code> It may look a bit hairy, but break it down: d.items () returns a list of 2-tuples, each one corresponding to a key-value pair from the dict. In our case, that'll be: [('a', 2), ('b', 1), ('c', -1), ('d', 4)] Although I've written them out like that, the order they'll come in is undefined. sorted () will return a sorted version of whatever iterable you chuck at it. Under normal Python semantics, sorted() on the list above will return no change since I've listed things out in alphanumeric order. The extra key= parameter tells the sorted routine to call the function you provide against each of the items in the list (in our case that means against each of the 2-tuples) and using the result of that function as the sorting order. The operation.itemgetter (1) bit is a touch complicated unless you're already familiar with partial functions, but it basically returns *another* function which takes the item you give it and returns the -- in this case -- 1st item. Just believe me: it works. So, in summary: + Get a list of key-value pairs + Sort them according to the 1st item (Python-style) which in this case is the value. + Do something with the result Before the key= param was introduced into sort/sorted, people used to do the same thing with what's often called DSU (short for decorate-sort-undecorate), a technique which here would look something like this: items = d.items () sortable_items = [(i[1], i) for i in items] sortable_items.sort () sorted_items = [i[-1] for i in sortable_items] I mention this because (a) you still see it around a fair bit and (b) there are occasions where it's still useful, for example where a simple function call can't really cope. (Did I answer the question, or was I just rambling?) TJG -- http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list
