min max from tuples in list

Peter Otten __peter__ at web.de
Thu Dec 12 03:35:57 EST 2013


Robert Voigtländer wrote:

> Hi,
> 
> I have a list like this:
> 
> a = [(52, 193), (52, 193), (52, 192), (51, 193), (51, 191), (51, 190),
> (51, 189), (51, 188), (50, 194), (50, 187), (50, 186), (50, 185), (50,
> 184), (49, 194), (49, 183), (49, 182), (49, 181), (48, 194), (48, 180),
> (48, 179), (48, 178), (48, 177), (47, 194), (47, 176), (47, 175), (47,
> 174), (47, 173), (46, 195), (46, 172), (46, 171), (46, 170), (46, 169),
> (45, 195), (45, 168), (45, 167), (45, 166), (44, 195), (44, 165), (44,
> 164), (44, 163), (44, 162), (43, 195), (43, 161), (43, 160), (43, 159),
> (43, 158), (42, 196), (42, 157), (42, 156), (42, 155), (41, 196), (41,
> 154), (41, 153), (41, 152), (41, 151), (40, 196), (40, 150), (40, 149),
> (40, 148), (40, 147), (39, 196), (39, 146), (39, 145), (39, 144), (39,
> 143), (38, 196), (38, 142), (38, 141), (38, 140), (37, 197), (37, 139),
> (37, 138), (37
>  , 137), (37, 136), (36, 197), (36, 135), (36, 134), (36, 133)]
> 
> 
> I need to find a -performant- way to transform this into a list with
> tuples (a[0],[a[0][1]min],[a[0][1]max]).
> 
> Hard to explaint what I mean .. [0] of the first three tuples is 52. [1]
> is 193,193 and 192. What I need as result for these three tuples is:
> (52,192,193).
> 
> For the next five tuples it is (51,188,193).
> 
> 
> Extra challenges:
> - This list is sorted. For performance reasons I would like to keep it
> unsorted. - There may be tuples where min=max.
> - There my be tupples where [0] only exists once. So mix is automatically
> max
> 
> 
> I hope I was able to explain what I mean.

I have a hunch that sorting the list might be quite efficient. You should at 
least try

import operator
import itertools

a = ...
a.sort()
result = []
for key, group in itertools.groupby(a, key=operator.itemgetter(0)):
    minpair = maxpair = next(group)
    for maxpair in group:
        pass
    result.append((key, minpair[1], maxpair[1]))

for item in result:
    print(item)

to see whether it is good enough.




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