Short circuting

[Alexander Kapps]

On 22.01.2011 01:10, Alexander Kapps wrote:
> On 22.01.2011 00:33, Ed Connell wrote:
>> Hi,
>>
>> Consider the following please: (re_section, re_name, etc are
>> previously compiled patterns)
>>
>> result1 = re_section.search(line);
>> result2 = re_name.search(line);
>> result3 = re_data1.search(line);
>> result4 = re_data2.search(line);
>>
>> if result1:
>> last_section = result1.group()[18:-5]
>> elif result2:
>> last_name = result2.group(0)[6:-1]
>> elif result3:
>> data[last_section] = {last_name:
>> result3.group()[13:-5]}
>> elif result4:
>> data[last_section] = {last_name:
>> result4.group()[17:-5]}
>>
>> It gets my goat to have to obtain all resultx when I just want the
>> first that is not None. (In theory, the number of results can be
>> much longer.) I can think of alternatives (raising exceptions), but
>> they all use deep indenting.
>>
>> Ideas?
>>
>> Ed
>>
>
>
> Maybe something like this (totally untested and probably wrong, I'm
> already quite tired):
>
>
> for pattern in (re_section, re_name, re_data1, re_data2):
> result = pattern.search(line):
> if result:
> if pattern == re_section:
> last_section = result1.group()[18:-5]
> elif pattern == re_name:
> last_name = result2.group(0)[6:-1]
> elif pattern == re_data1:
> data[last_section] = {last_name: result3.group()[13:-5]}
> elif pattern == re_data2:
> data[last_section] = {last_name: result4.group()[17:-5]}
>
>
> Also, if you have long if/elif ladders, look if you can apply the
> dictionary dispatch pattern.

Correction. Of course you need to break out of the loop as soon as a 
not None result is found:

     if result:
         if pattern == re_section:
             last_section = result1.group()[18:-5]
         ...
         break
...