Python - Sorting a Tuple-Keyed Dictionary by Tuple Value -

i have dictionary constituted of tuple keys , integer counts , want sort third value of tuple (key[2]) so

data = {(a, b, c, d): 1, (b, c, b, a): 4, (a, f, l, s): 3, (c, d, j, a): 7} print sorted(data.iteritems(), key = lambda x: data.keys()[2]) 

with desired output

>>> {(b, c, b, a): 4, (a, b, c, d): 1, (c, d, j, a): 7, (a, f, l, s): 3} 

but current code seems nothing. how should done?

edit: appropriate code

sorted(data.iteritems(), key = lambda x: x[0][2]) 

but in context

from collections import ordered dict  data = {('a', 'b', 'c', 'd'): 1, ('b', 'c', 'b', 'a'): 4, ('a', 'f', 'l', 's'): 3, ('c', 'd', 'j', 'a'): 7} xxx = [] yyy = [] zzz = ordereddict() key, value in sorted(data.iteritems(), key = lambda x: x[0][2]):     x = key[2]     y = key[3]     xxx.append(x)     yyy.append(y)     zzz[x + y] = 1 print xxx print yyy print zzz 

zzz unordered. know because dictionaries default unordered , need use ordereddict sort don't know use it. if use checked answer suggests 'tuple index out of range' error.

solution:

from collections import ordereddict  data = {('a', 'b', 'c', 'd'): 1, ('b', 'c', 'b', 'a'): 4, ('a', 'f', 'l', 's'): 3, ('c', 'd', 'j', 'a'): 7} xxx = [] yyy = [] zzz = ordereddict() key, value in sorted(data.iteritems(), key = lambda x: x[0][2]):     x = key[2]     y = key[3]     xxx.append(x)     yyy.append(y)     zzz[x + y] = 1 print xxx print yyy print zzz 

dictionaries unordered in python. can use ordereddict.

you have sort like:

from collections import ordereddict  result = ordereddict(sorted(data.iteritems(),key=lambda x:x[0][2]))

you need use key=lambda x:x[0][2] because elements tuples (key,val) , obtain key, use x[0].

this gives:

>>> data = {('a', 'b', 'c', 'd'): 1, ('b', 'c', 'b', 'a'): 4, ('a', 'f', 'l', 's'): 3, ('c', 'd', 'j', 'a'): 7} >>> collections import ordereddict >>> result = ordereddict(sorted(data.iteritems(),key=lambda x:x[0][2])) >>> result ordereddict([(('b', 'c', 'b', 'a'), 4), (('a', 'b', 'c', 'd'), 1), (('c', 'd', 'j', 'a'), 7), (('a', 'f', 'l', 's'), 3)]) 

edit:

in order make zzz ordered well, can update code to:

data = {('a', 'b', 'c', 'd'): 1, ('b', 'c', 'b', 'a'): 4, ('a', 'f', 'l', 's'): 3, ('c', 'd', 'j', 'a'): 7} xxx = [] yyy = [] zzz = ordereddict() key, value in sorted(data.iteritems(), key = lambda x: x[0][2]):     x = key[2]     y = key[3]     xxx.append(x)     yyy.append(y)     zzz[x + y] = 1 print xxx print yyy print zzz