On Wed, 11 Dec 2013 23:25:53 -0800, Robert Voigtländer wrote:
> I have a list like this:
>
> a = [(52, 193), ...... (36, 133)]
# iterate over the list of tuples
# creates a dictionary n0:[n1a, n1b, n1c ... ]
# from tuples (n0,n1a), (n0,n1b), (n0,n1c) ...
b = {}
for x in a:
if x[0] in b:
pass
else:
b[x[0]] = []
if x[1] not in b[x[0]]:
b[x[0]].append( x[1] )
# iterate over the dictionary
# create the list of result tuples (n0, n1min, n1max) .....
c = [ (x, min(y), max(y) ) for x,y in b.iteritems() ]
print c
There may be a more efficient test for whether b[x[0]] eists than
trapping keyError.
--
Denis McMahon, denismfmcma...@gmail.com
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