[FAF PATCH v2 1/2] bugzilla: pick report with highest count

[Richard Marko]

When creating new ticket make sure we pick
report with highest count so we don't end up
linking to unrelated report if clustering
splits the problem.

Closes #172.

Signed-off-by: Richard Marko 
---
 pyfaf/bugzilla.py        | 2 +-
 pyfaf/storage/problem.py | 8 ++++++++
 2 files changed, 9 insertions(+), 1 deletion(-)

diff --git a/pyfaf/bugzilla.py b/pyfaf/bugzilla.py
index f135655..a8db6ec 100644
--- a/pyfaf/bugzilla.py
+++ b/pyfaf/bugzilla.py
@@ -713,7 +713,7 @@ class Bugzilla(object):
             # pick first and assign this bug to it
             data['component'] = components.pop()

-            report = problem.reports[0]
+            report = problem.sorted_reports[0]
             if not report.backtraces:
                 logging.warning('Refusing to process report with no backtrace.')
                 continue
diff --git a/pyfaf/storage/problem.py b/pyfaf/storage/problem.py
index 7242517..e8ee1a5 100644
--- a/pyfaf/storage/problem.py
+++ b/pyfaf/storage/problem.py
@@ -92,3 +92,11 @@ class Problem(GenericTable):
         sorted by quality.
         '''
         return sorted(self.backtraces, key=lambda bt: bt.quality, reverse=True)
+
+    @property
+    def sorted_reports(self):
+        """
+        Return list of all reports sorted by report count.
+        """
+
+        return sorted(self.reports, key=lambda report: report.count, reverse=True)