CODEWITHSUNDEEP
pythonpython-3.x
Faller
Faller

Reputation: 1698

Sorting by value in a dictionary if the value is a list

I know that dictionaries aren't sortable, that being said, I want the representation of the dictionary that is sorted upon that value which is a list. Example:

some_dict= {
    "a": [0, 0, 0, 0, 0],
    "b": [1400, 50, 30, 18, 0],
    "c": [800, 30, 14, 14, 0],
    "d": [5000, 100, 30, 50, .1],
    "for fun": [140000, 1400, 140, 140, .42]
}

I want to sort by the first item in the list of each.

I tried something like:

sorted(some_dict.items(), key = lambda for x in some_dict: some_dict[x][0])

but I get invalid syntax.

Thanks in advance.

Upvotes: 1

Views: 72

Answers (2)

TerryA
TerryA

Reputation: 59974

You're on the right track, but you can't put a for-loop in a lambda (hence the error):

>>> sorted(some_dict.items(), key=lambda x: x[1][0])
[('a', [0, 0, 0, 0, 0]), ('c', [800, 30, 14, 14, 0]), ('b', [1400, 50, 30, 18, 0]), ('d', [5000, 100, 30, 50, 0.1]), ('for fun', [140000, 1400, 140, 140, 0.42])]

If you want to keep this order in a dictionary, you can use collections.OrderedDict:

>>> from collections import OrderedDict
>>> mydict = OrderedDict(sorted(some_dict.items(), key=lambda x: x[1][0]))
>>> print(mydict)
OrderedDict([('a', [0, 0, 0, 0, 0]), ('c', [800, 30, 14, 14, 0]), ('b', [1400, 50, 30, 18, 0]), ('d', [5000, 100, 30, 50, 0.1]), ('for fun', [140000, 1400, 140, 140, 0.42])])
>>> print(mydict['a'])
[0, 0, 0, 0, 0]

Upvotes: 4

Blender
Blender

Reputation: 298166

lambdas are anonymous functions, but they can only execute expressions in Python:

lambda pair: pair[1][0]

You could also write it verbosely:

def sorting_func(pair):
    key, value = pair
    return value[0]

sorted(some_dict.items(), key=sorting_func)

Upvotes: 1

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