Brian van den Broek wrote:
<SNIP>
Finally, in the first instance, I was aiming for the OP's stated end. To make this more general and reusable, I think I'd do:
<code> def get_list_dup_dict(a_list, threshold=1):
items_dict, dup_dict = {}, {} # Question below
for i in a_list: items_dict[i] = items_dict.get(i, 0) + 1
for k, v in items_dict.iteritems(): if v >= threshold: dup_dict[k] = v #Question below
return dup_dict
def print_list_dup_report(a_list, threshold=1):
dup_dict = get_list_dup_dict(a_list, threshold) for k, v in sorted(dup_dict.iteritems()): print '%s occurred %s times' %(k, v) </code>
My question (from comment in new code):
Since I've split the task into two functions, one to return a duplication dictionary, the other to print a report based on it, I think the distinct dup_dict is needed. (I do want the get_list_dup_dict function to return a dict for possible use in other contexts.)
You could have a function that gets all of the counts, and then filter on threshold when you print.
<SNIP>
Am I overlooking yet another dict technique that would help, here? Any other improvements?
Thanks Kent. Above the last snip would indeed be one such improvement! (How'd I miss that?)
Best,
Brian vdB
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