Sunil Tech wrote:
> Danny i did it like this
>
> result_dict = {}
> for i in tes:
> if i['a'] in result_dict:
> temp = result_dict[i['a']]
> temp['b'].append(i['b'])
> temp['c'].append(i['c'])
> temp['a'] = i['a']
> result_dict[i['a']] = temp
> else:
> result_dict[i['a']] = {
> 'b': [i['b']],
> 'c': [i['c']],
> 'a': i['a']}
> pprint.pprint(result_dict.values())
>
> result is
>
> [{'a': 1, 'b': ['this', 'is', 'sentence'], 'c': [221, 875, 874]},
> {'a': 2, 'b': ['this', 'another', 'word'], 'c': [215, 754, 745]}]
>
> any can one improve this method in terms of performance, etc..
What you have is a good solution; the most important part performance-wise
is that you collect records with the same `a` value in a dict.
For reference here's my two-pass solution to the problem as originally
specified:
bc = collections.defaultdict(lambda: ([], []))
for rec in tes:
b, c = bc[rec["a"]]
b.append(rec["b"])
c.append(rec["c"])
result = [{"a": a,
"b": ", ".join(b),
"c": ", ".join(map(str, c))}
for a, (b, c) in bc.items()]
If you are flexible with the result data structure you could omit the second
loop and use bc.items() directly.
_______________________________________________
Tutor maillist - Tutor@python.org
To unsubscribe or change subscription options:
https://mail.python.org/mailman/listinfo/tutor
