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Dictionary Sorting By Values Other Than Keys

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xm1014

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Dictionary Sorting By Values Other Than Keys

Nov 8th, 2009

Hello everyone,

I'm learning Python and I have a few questions about Dictionaries and sorting them by a value other than the key.

Lets say, for example, I have a Dictionary full of data that needs to be sorted on the 3rd value instead of the key. How would I go about doing this? I have tried just the .sort() and that just sorts by the key.

So if,

Data = [('1023', ['Name', '3.1', 'SomeData']), ('1000', ['SomeName', '4.0', 'MoreData'])

how would I sort "Data" by the value where 3.1 is?

Please forgive my lack of correct terminology if I am missing some somewhere! Any help would be greatly appreciated.

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mn_kthompson mn_kthompson is offline

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Nov 8th, 2009

It doesn't look like you're using a dictionary. It appears that you have a list of tuples, and each tuple consists of a string and a list.

I can't think of a built in way to do this. You might have to look at writing your own version of a bubble sort or something. To do that, you would go through the entire outer list and compare the 2nd element of the inner list. If they are not in the correct order, then switch them around. Repeat until you can go through the whole list without having to make any switches.

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Gribouillis Gribouillis is online now

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Nov 8th, 2009

You can use the 'key' argument to 'sort' or 'sorted'. Here, Data is a list. You can write

 Help with Code Tags 
 python Syntax (Toggle Plain Text) 
def my_key(item):
    return item[1][1] # this returns 3.1 if item is ('1023', ['Name', '3.1', 'SomeData'])
 
Data.sort(key = my_key)
def my_key(item):
    return item[1][1] # this returns 3.1 if item is ('1023', ['Name', '3.1', 'SomeData'])

Data.sort(key = my_key)

Data is easily turned into a dictionary:

 Help with Code Tags 
 Python Syntax (Toggle Plain Text) 
Data = dict(Data)
L = sorted(Data.items(), key = my_key)
Data = dict(Data)
L = sorted(Data.items(), key = my_key)

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vegaseat

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Nov 8th, 2009

Similar to Gribouillis' code, but uses lambda as a helper function ...

 Help with Code Tags 
 python Syntax (Toggle Plain Text) 
# sort a more complex combination of lists/tuples:
mylist = [
(1, ['a', '3.1', 'ad']),
(2, ['b', '4.0', 'bd']),
(3, ['c', '2.5', 'cd']),
]
# sort by item at index [1][1] of each tuple
# using a helper function like lambda
newlist = sorted(mylist, key=lambda tup: tup[1][1])
 
print newlist
 
"""my result (made pretty) -->
[
(3, ['c', '2.5', 'cd']),
(1, ['a', '3.1', 'ad']), 
(2, ['b', '4.0', 'bd'])
]
"""
# sort a more complex combination of lists/tuples:
mylist = [
(1, ['a', '3.1', 'ad']),
(2, ['b', '4.0', 'bd']),
(3, ['c', '2.5', 'cd']),
]
# sort by item at index [1][1] of each tuple
# using a helper function like lambda
newlist = sorted(mylist, key=lambda tup: tup[1][1])

print newlist

"""my result (made pretty) -->
[
(3, ['c', '2.5', 'cd']),
(1, ['a', '3.1', 'ad']), 
(2, ['b', '4.0', 'bd'])
]
"""

May 'the Google' be with you!

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vegaseat

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Nov 8th, 2009

You can also use a somewhat older concept of sorting, the Schwartzian transform algorithm. This is made very easy with Python's list comprehension ...

 Help with Code Tags 
 python Syntax (Toggle Plain Text) 
# sort a more complex combination of lists/tuples:
 
mylist = [
(1, ['a', '3.1', 'ad']),
(2, ['b', '4.0', 'bd']),
(3, ['c', '2.5', 'cd']),
]
 
# use the Schwartzian transform algorithm
# temporarily put a copy of indexed item in front
templist = [(x[1][1], x) for x in mylist]
templist.sort()
# remove temporary front item after sorting
newlist = [val for (temp, val) in templist]
 
print newlist
 
"""my result (made pretty) -->
[
(3, ['c', '2.5', 'cd']),
(1, ['a', '3.1', 'ad']), 
(2, ['b', '4.0', 'bd'])
]
"""
# sort a more complex combination of lists/tuples:
    
mylist = [
(1, ['a', '3.1', 'ad']),
(2, ['b', '4.0', 'bd']),
(3, ['c', '2.5', 'cd']),
]

# use the Schwartzian transform algorithm
# temporarily put a copy of indexed item in front
templist = [(x[1][1], x) for x in mylist]
templist.sort()
# remove temporary front item after sorting
newlist = [val for (temp, val) in templist]

print newlist

"""my result (made pretty) -->
[
(3, ['c', '2.5', 'cd']),
(1, ['a', '3.1', 'ad']), 
(2, ['b', '4.0', 'bd'])
]
"""

Last edited by vegaseat; Nov 8th, 2009 at 8:37 pm.

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xm1014

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Nov 10th, 2009

I've tried all three of these methods, however, I haven't been able to get them to work correctly for my particular set of data. I believe I may have pasted my dictionary incorrectly.

Here is what I have in my dictionary:
data = {'1234': ['Matt', '2.5', 'CS'], '1000': ['John', '4.0', 'Music'], '1023': ['Aaron', '3.1', 'PreMed'], '1001': ['Paul', '3.9', 'Music'], '9000': ['Kris', '3.5', 'Business']}

Would this have affected the methods of sorting that were mentioned?

Last edited by xm1014; Nov 10th, 2009 at 4:38 am.

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vegaseat

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DaniWeb's Hypocrite

Nov 10th, 2009

So you have a dictionary after all!

Dictionaries are indexed by key and the keys are in a hash order for fast lookups. They will always display in this order! If you want to sort by a certain item, you have to convert them to lists. I thought that is what you did.

 Help with Code Tags 
 Python Syntax (Toggle Plain Text) 
data_dict = {
'1234': ['Matt', '2.5', 'CS'], 
'1000': ['John', '4.0', 'Music'], 
'1023': ['Aaron', '3.1', 'PreMed'], 
'1001': ['Paul', '3.9', 'Music'], 
'9000': ['Kris', '3.5', 'Business']
}
 
# convert dictionary to a list of tuples
data_list = [(key, val) for key, val in data_dict.items()]
 
print(data_list)
"""the raw list -->
[
('1234', ['Matt', '2.5', 'CS']), 
('9000', ['Kris', '3.5', 'Business']), 
('1001', ['Paul', '3.9', 'Music']), 
('1023', ['Aaron', '3.1', 'PreMed']), 
('1000', ['John', '4.0', 'Music'])
]
"""
 
# now you can sort the list by item[1][1]
sorted_list = sorted(data_list, key=lambda tup: tup[1][1])
 
print(sorted_list)
"""result sorted by item[1][1] -->
[
('1234', ['Matt', '2.5', 'CS']), 
('1023', ['Aaron', '3.1', 'PreMed']), 
('9000', ['Kris', '3.5', 'Business']), 
('1001', ['Paul', '3.9', 'Music']), 
('1000', ['John', '4.0', 'Music'])]
"""
 
# however if you go back to a dictionary
new_dict = dict(sorted_list)
 
print(new_dict)
"""result is the dictionary order again -->
{
'1234': ['Matt', '2.5', 'CS'], 
'1000': ['John', '4.0', 'Music'], 
'1001': ['Paul', '3.9', 'Music'], 
'1023': ['Aaron', '3.1', 'PreMed'], 
'9000': ['Kris', '3.5', 'Business']}
"""
data_dict = {
'1234': ['Matt', '2.5', 'CS'], 
'1000': ['John', '4.0', 'Music'], 
'1023': ['Aaron', '3.1', 'PreMed'], 
'1001': ['Paul', '3.9', 'Music'], 
'9000': ['Kris', '3.5', 'Business']
}

# convert dictionary to a list of tuples
data_list = [(key, val) for key, val in data_dict.items()]

print(data_list)
"""the raw list -->
[
('1234', ['Matt', '2.5', 'CS']), 
('9000', ['Kris', '3.5', 'Business']), 
('1001', ['Paul', '3.9', 'Music']), 
('1023', ['Aaron', '3.1', 'PreMed']), 
('1000', ['John', '4.0', 'Music'])
]
"""

# now you can sort the list by item[1][1]
sorted_list = sorted(data_list, key=lambda tup: tup[1][1])

print(sorted_list)
"""result sorted by item[1][1] -->
[
('1234', ['Matt', '2.5', 'CS']), 
('1023', ['Aaron', '3.1', 'PreMed']), 
('9000', ['Kris', '3.5', 'Business']), 
('1001', ['Paul', '3.9', 'Music']), 
('1000', ['John', '4.0', 'Music'])]
"""

# however if you go back to a dictionary
new_dict = dict(sorted_list)

print(new_dict)
"""result is the dictionary order again -->
{
'1234': ['Matt', '2.5', 'CS'], 
'1000': ['John', '4.0', 'Music'], 
'1001': ['Paul', '3.9', 'Music'], 
'1023': ['Aaron', '3.1', 'PreMed'], 
'9000': ['Kris', '3.5', 'Business']}
"""

So, if you want to do any processing that needs a sorted order, to have to convert your dictionary to a temporary list.

Last edited by vegaseat; Nov 10th, 2009 at 9:16 am.

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xm1014

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Nov 10th, 2009

Ah ha! That worked! Now, since you're putting it in a list to sort the values, is there also a way to separate those values you are sorting on into a list by themselves?

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Nov 10th, 2009

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Originally Posted by xm1014

Ah ha! That worked! Now, since you're putting it in a list to sort the values, is there also a way to separate those values you are sorting on into a list by themselves?

Is this what you mean?

 Help with Code Tags 
 python Syntax (Toggle Plain Text) 
>>> my_dict = {'a':1, 'b':2, 'c':3}
>>> my_dict.keys()
['a', 'c', 'b']
>>> my_dict.values()
[1, 3, 2]
>>>
>>> my_dict = {'a':1, 'b':2, 'c':3}
>>> my_dict.keys()
['a', 'c', 'b']
>>> my_dict.values()
[1, 3, 2]
>>>

1. Use Code Tags.
2. Homework? Show Effort.
3. Keep discussions on the forum: no PMs

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xm1014

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#10

Nov 10th, 2009

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Originally Posted by jlm699

Is this what you mean?
Help with Code Tags

python Syntax (Toggle Plain Text)
>>> my_dict = {'a':1, 'b':2, 'c':3}
>>> my_dict.keys()
['a', 'c', 'b']
>>> my_dict.values()
[1, 3, 2]
>>>
>>> my_dict = {'a':1, 'b':2, 'c':3} >>> my_dict.keys() ['a', 'c', 'b'] >>> my_dict.values() [1, 3, 2] >>>

Not exactly, I'd like to take the values that are being sorted on and put them into a new list/dict etc.

For example, extracting only these values from the dictionary above:
new_list = ('2.5','3.5','3.9','4.0','3.1')