# Order by value in dictionary

Abandoned besturk at gmail.com
Wed Oct 17 17:09:50 CEST 2007

```Very very thanks everbody..

These are some method..
Now the fastest method is second..

==== 1 ===
def sortt(d):
items=d.items()
backitems=[ [v[1],v[0]] for v in items]
backitems.sort()
#boyut=len(backitems)
#backitems=backitems[boyut-500:]
a=[ backitems[i][1] for i in range(0,len(backitems))]
a.reverse()
return a

==== 2 =====
import operator
def sortt(d):
backitems=d.items()
boyut=len(backitems)
backitems=backitems[boyut-500:]
backitems=sorted(backitems, key=operator.itemgetter(1))
a=[ backitems[i][0] for i in range(0,len(backitems))]
a.reverse()
return a

==== 3 =====
def sortt(d):
backitems=d.items()
backitems.sort(lambda x,y:cmp(x[1],y[1]))
backitems=sorted(backitems, key=operator.itemgetter(1))
a=[ backitems[i][0] for i in range(0,len(backitems))]
a.reverse()
return a

====== 4 =======
import heapq

def sortt(d):
backitems=d.items()
backitems=heapq.nlargest(1000, backitems, operator.itemgetter(1))
a=[ backitems[i][0] for i in range(0,len(backitems))]
a.reverse()
return a

```