[Piglit] [PATCH 2/8] framework: Make the set of tests per page always be a set.

[Dylan Baker]

On Wednesday 18 September 2013 15:00:27 Eric Anholt wrote:
> Starting with appending to the dictionary entry as a list, then making
> the entry into a set later is pretty weird.  Just always use a set.
> ---
>  framework/summary.py | 24 ++++++++----------------
>  1 file changed, 8 insertions(+), 16 deletions(-)
> 
> diff --git a/framework/summary.py b/framework/summary.py
> index 3f9b883..26bdda9 100644
> --- a/framework/summary.py
> +++ b/framework/summary.py
> @@ -383,8 +383,8 @@ class Summary:
> 
>          self.status = {}
>          self.fractions = {}
> -        self.tests = {'all': [], 'changes': [], 'problems': [], 'skipped':
> [], -                      'regressions': [], 'fixes': []}
> +        self.tests = {'all': set(), 'changes': set(), 'problems': set(),
> 'skipped': set(), +                      'regressions': set(), 'fixes':
> set()}
> 
>          for each in self.results:
>              # Build a dict of the status output of all of the tests, with
> the @@ -395,7 +395,7 @@ class Summary:
>              self.status.update({each.name: status})
> 
>              # Create a list with all the test names in it
> -            self.tests['all'] = list(set(self.tests['all']) |
> set(each.tests)) +            self.tests['all'] = set(self.tests['all']) |
> set(each.tests)
> 
>      def __generate_lists(self, lists):
>          """
> @@ -444,36 +444,28 @@ class Summary:
>                  # out the list will reduce it's length to 1 if all entries
>                  # are the same, meaning it is not a change
>                  if len(set(status)) > 1:
> -                    self.tests['changes'].append(test)
> +                    self.tests['changes'].add(test)
> 
>              if 'problems' in lists:
>                  # If the result contains a value other than 1 (pass) or 4
>                  # (skip) it is a problem. Skips are not problems becasuse
>                  # they have Their own page.
>                  if [i for e in [2, 3, 5] for i in status if e is i]:
> -                    self.tests['problems'].append(test)
> +                    self.tests['problems'].add(test)
> 
>              if 'skipped' in lists:
>                  # Find all tests with a status of skip
>                  if 4 in status:
> -                    self.tests['skipped'].append(test)
> +                    self.tests['skipped'].add(test)
> 
>              if 'fixes' in lists:
>                  # Find both fixes and regressions, and append them to the
>                  # proper lists
>                  for i in xrange(len(status) - 1):
>                      if status[i] < status[i + 1] and status[i] != 0:
> -                        self.tests['regressions'].append(test)
> +                        self.tests['regressions'].add(test)
>                      if status[i] > 1 and status[i + 1] == 1:
> -                        self.tests['fixes'].append(test)
> -
> -        # Remove duplicate entries from the status lists
> -        # If there are 4+ results can result in mutiple passes or
> regressions -        # and changes in other words: "pass fail pass fail"
> will result in a -        # regression, a fix, and a regression and it will
> be printed twice in -        # the summary. Turning them into sets remove
> duplicates
> -        for (result, value) in self.tests.items():
> -            self.tests[result] = set(value)
> +                        self.tests['fixes'].add(test)
> 
>      def __find_totals(self):
>          """

This of course makes sense.
Reviewed-by: Dylan Baker <baker.dylan.c at gmail.com>
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