many ways
http://blog.client9.com/2007/11/sorting-python-dict-by-value.html
alist = sorted(adict.iteritems(), key=lambda (k,v): (v,k))
http://coreygoldberg.blogspot.com/2008/06/python-sort-dictionary-by-values.html
sorted(foo.items(), key=lambda(k,v):(v,k))
http://writeonly.wordpress.com/2008/08/30/sorting-dictionaries-by-value-in-python-improved/
def sbv0(adict,reverse=False):
''' proposed at Digital Sanitation Engineering
http://blog.modp.com/2007/11/sorting-python-dict-by-value.html '''
return sorted(adict.iteritems(), key=lambda (k,v): (v,k), reverse=reverse)
def sbv1(d,reverse=False):
''' explicit list expansion '''
L = [(k,v) for (k,v) in d.iteritems()]
return sorted(L, key=lambda x: x[1] , reverse=reverse)
def sbv2(d,reverse=False):
''' generator '''
L = ((k,v) for (k,v) in d.iteritems())
return sorted(L, key=lambda x: x[1] , reverse=reverse)
def sbv3(d,reverse=False):
''' using a lambda to get the key, rather than "double-assignment" '''
return sorted(d.iteritems(), key=lambda x: x[1] , reverse=reverse)
def sbv4(d,reverse=False):
''' using a formal function to get the sorting key, rather than a lambda'''
def sk(x): return x[1]
return sorted(d.iteritems(), key=sk , reverse=reverse)
def sk(x): return x[1]
def sbv5(d,reverse=False):
''' using a formal function, defined in outer scope
to get the sorting key, rather than a lambda
'''
return sorted(d.iteritems(), key=sk , reverse=reverse)
from operator import itemgetter
def sbv6(d,reverse=False):
''' proposed in PEP 265, using the itemgetter '''
return sorted(d.iteritems(), key=itemgetter(1), reverse=True)
D = dict(zip(range(100),range(100)))
from profile import run
run("for ii in xrange(10000): sbv0(D, reverse=True)")
run("for ii in xrange(10000): sbv1(D, reverse=True)")
run("for ii in xrange(10000): sbv2(D, reverse=True)")
run("for ii in xrange(10000): sbv3(D, reverse=True)")
run("for ii in xrange(10000): sbv4(D, reverse=True)")
run("for ii in xrange(10000): sbv5(D, reverse=True)")
run("for ii in xrange(10000): sbv6(D, reverse=True)")
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