Sorting a dictionary field belonging to a list

[Jocknerd]

I have a list called teamlist which contain dictionaries of teams. These
are the fields in my team dictionary:

name, won, lost, tied, pf, pa

To print standings I do the following:

def printStandings(teamlist):
for team in teamlist:
print (team['name'], team['won'], team['lost'], team['tied'],
team['pf'], team['pa'])

My question is how would I go about sorting the output based on team['pf']
for instance?

 

[Chris Green]

My question is how would I go about sorting the output based on team['pf']
for instance?

The simplest way would be to use the sort method on the list in the
first place. That method takes an optional comparison function that
you would have to write. See help(lst.sort)

lst = [ {'name': 'z'}, {'name': 'd'}, {'name': 'e'} ]

def mycmp(a,b):
return cmp(a['name'],b['name'])

lst.sort(mycmp)

If the original order is important to you, you will be to use
something like copy.copy() to make a copy of the list.

 

[John Lenton]

I have a list called teamlist which contain dictionaries of teams. These
are the fields in my team dictionary:

name, won, lost, tied, pf, pa

To print standings I do the following:

def printStandings(teamlist):
for team in teamlist:
print (team['name'], team['won'], team['lost'], team['tied'],
team['pf'], team['pa'])

My question is how would I go about sorting the output based on team['pf']
for instance?

replace

for team in teamlist:

using the decorate-sort-undecorate pattern,

sortedlist = [(i['pf'], i) for i in teamlist]
sortedlist.sort()
for pf, team in sortedlist:

(this is far faster than calling a python function on each comparison);
in 2.4 you will be able to do

for team in sorted(teamlist, operator.itemgetter('pf')):

--
John Lenton ([email protected]) -- Random fortune:
Do not drink coffee in early A.M. It will keep you awake until noon.

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