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Simplify arch-sorting key lambda

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Stefano Rivera 2010-11-27 09:24:32 +02:00
parent 282aa33614
commit cb64c34c22
1 changed files with 1 additions and 1 deletions
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2
wrap-and-sort
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@ -67,7 +67,7 @@ class Control(object):
if "Architecture" in paragraph: if "Architecture" in paragraph:
archs = set(paragraph["Architecture"].split()) archs = set(paragraph["Architecture"].split())
# Sort, with wildcard entries (such as linux-any) first: # Sort, with wildcard entries (such as linux-any) first:
archs = sorted(archs, key=lambda x: (1 - int("any" in x), x)) archs = sorted(archs, key=lambda x: ("any" not in x, x))
paragraph["Architecture"] = " ".join(archs) paragraph["Architecture"] = " ".join(archs)
if sort_paragraphs: if sort_paragraphs: