# Is there a more elegant way to do this?

Aahz Maruch aahz at panix.com
Wed Sep 13 20:55:14 CEST 2000

```In article <8poglt\$10fq\$1 at news.rchland.ibm.com>,
Larry Whitley <ldw at us.ibm.com> wrote:
>
>I have a list of counters that will have a wide variety of different values
>in them.  At intervals while the program runs, I will print out the indexes
>of the counters with the five largest counts.  The counters are in a list
>identified below as self.counters.  Here's my inelegant way of doing it.
>
>def runningReport(self): # a method of a larger class
>    temp = [] # to make sure that temp is not just another reference to
>self.counters
>    temp = temp + self.counters # there are 100 individual counts in
>self.counters
>    temp.sort()
>    temp.reverse() # now largest value is first
>    temp2 = [] # for the result
>    for i in range( 5 ):
>        temp2.append( self.counters.index( temp[i] ) # find the index of the
>next (largest) counter and store it in temp2
>    print temp2

This isn't much more elegent, but it's much more efficient:

def runningReport(self, numItems=5): # a method of a larger class
if numItems > len(self.counter):
numItems = len(self.counter)
numItems = -1 * abs(int(numItems)) # make sure numItems is negative int
temp = self.counters[:] #make copy of self.counters
temp.sort()
temp2 = temp[numItems:] #negative numItems pulls from end of list
temp2.reverse() #Change to descending order?
print temp2

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