Python - Sorting a Tuple-Keyed Dictionary by Tuple Value

Question

I have a dictionary constituted of tuple keys and integer counts and I want to sort it by the third value of the tuple (key[2]) as 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 this desired output

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

but my current code seems to do nothing. How should this be done?

Edit: The appropriate code is

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

but in the 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()
for 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 is unordered. I know that this is because dictionaries are by default unordered and that I need to use OrderedDict to sort it but I don't know where to use it. If I use it as the checked answer suggests I get a '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()
for 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

@JonathanConnell: That doesn't mean you can get your desired output. — Martijn Pieters
– Martijn Pieters, Commented Mar 28, 2017 at 13:30
So why are you using data.keys()[2]? The key order is arbitrary, whatever key at index 2 is should not inform your sort. And all your items are sorted on the basis of that one key. Did you mean x[0][2] perhaps (so the first element of the (key, value) pair or key, then the third element in that tuple). — Martijn Pieters
– Martijn Pieters, Commented Mar 28, 2017 at 13:32

willeM_ Van Onsem · Accepted Answer · 2017-03-28 15:10:07Z

3

Dictionaries are unordered in Python. You can however use an OrderedDict.

You then have to sort like:

from collections import OrderedDict

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

You need to use key=lambda x:x[0][2] because the elements are tuples (key,val) and so to obtain the key, you 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}
>>> from 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 to make zzz ordered as well, you can update your 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()
for 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

edited Mar 28, 2017 at 15:10

answered Mar 28, 2017 at 13:29

willeM_ Van Onsem

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12 Comments

Martijn Pieters Over a year ago

The actual problem is the sort key, which you fixed but didn't explain.

willeM_ Van Onsem Over a year ago

@MartijnPieters: that's why they invented something as edit. To first give an answer the op can test, and then explain. So that less time is wasted if the answer does not meet the expectations :). I notice you actually nearly always do the same :)

Martijn Pieters Over a year ago

The revisions of the post show that the key=lambda x: data.keys()[2] version was there in the first 5 minutes.

Sadie LaBounty Over a year ago

There is an unusual caveat though: I am using this to say 'for key, value in sorted_dict: x = key[2], y = key[3] etc. For this, it reverts back to being unsorted even though if I print sorted_dict it is alphabetically sorted.

willeM_ Van Onsem Over a year ago

@JonathanConnell: but as specified in the answer: a dictionary is unoredered. You have to make zzz an OrderedDict as well ito maintain order at that level.

|

Daniel Roseman · Accepted Answer · 2017-03-28 13:31:58Z

3

Your key function is completely broken. It's passed the current value as x, but you ignore that and instead always get the second item from the list of keys.

Use key=lambda x: x[0][2] instead.

edited Mar 28, 2017 at 13:31

answered Mar 28, 2017 at 13:30

Daniel Roseman

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1 Comment

willeM_ Van Onsem Over a year ago

I think you have to use x[0][2] since the iteritems() emit (key,value) tuples.

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