How do I sort a python dictionary values and rearrange keys?

Question

How can I sort the following dictionary by its values and rearrange keys?.

{1: [0, 0, 1, 1], 2: [0, 1, 1, 1], 3: [1, 1, 1, 1], 4: [1, 0, 1, 1]}

Expected result :

{1: [0, 0, 1, 1], 2: [0, 1, 1, 1], 3: [1, 0, 1, 1], 4: [1, 1, 1, 1]}

Asked By: Bala

||

Source

Answer 1

You don’t need to use a dictionary.

You can just use a list and it’s indexes since your values are just incrementing numbers.

To convert from a dictionary to a list of it’s values, call:

value_list = list(dictionaryname.values())

Then, simply call sort on the list:

value_list.sort()

Also, python dictionaries are an unordered data type.

Answered By: Craze XD

Answer 2

Python dict doesn’t have a order, it’s implemented by hash and you can not sort it. You should convert it to another type then sort.

Answered By: wityu

Answer 3

The easiest way is to create a new dictionary by combining the keys and the sorted values from the original dict.
Assign that dict to the original variable, and you’re done.

In [1]: orig = {1: [0, 0, 1, 1], 2: [0, 1, 1, 1], 3: [1, 1, 1, 1], 4: [1, 0, 1, 1]}
Out[1]: {1: [0, 0, 1, 1], 2: [0, 1, 1, 1], 3: [1, 1, 1, 1], 4: [1, 0, 1, 1]}

In [2]: k = list(orig.keys())
Out[2]: [1, 2, 3, 4]

In [3]: v = sorted(orig.values())
Out[3]: [[0, 0, 1, 1], [0, 1, 1, 1], [1, 0, 1, 1], [1, 1, 1, 1]]

In [4]: orig = dict(zip(k, v))
Out[4]: {1: [0, 0, 1, 1], 2: [0, 1, 1, 1], 3: [1, 0, 1, 1], 4: [1, 1, 1, 1]}

It can even be done in a single line:

In [1]: orig = {1: [0, 0, 1, 1], 2: [0, 1, 1, 1], 3: [1, 1, 1, 1], 4: [1, 0, 1, 1]};

In [2]: orig = dict(zip(orig.keys(), sorted(orig.values())))
Out[2]: {1: [0, 0, 1, 1], 2: [0, 1, 1, 1], 3: [1, 0, 1, 1], 4: [1, 1, 1, 1]}

Answered By: Roland Smith

Answer 4

If I get you right, you might want do this:

def value_of_list(li):
    sum = 0
    base = 2
    count = 0
    for digit in reversed(li):
        sum += digit * base ** count
        count += 1
    return sum

dic = {
    0: [0, 0, 0, 0],
    6: [0, 1, 1, 0],
    1: [0, 0, 0, 1],
    3: [0, 0, 1, 1],
    4: [0, 1, 0, 0],
    2: [0, 0, 1, 0],
    5: [0, 1, 0, 1],
}

sorted(dic.items(), key=lambda x: value_of_list(x[1]))
print(dic)

It will give a "ordered" output just like the given example.

{0: [0, 0, 0, 0], 1: [0, 0, 0, 1], 2: [0, 0, 1, 0], 3: [0, 0, 1, 1], 4: [0, 1, 0, 0], 5: [0, 1, 0, 1], 6: [0, 1, 1, 0]}

Answered By: BakaFT

How do I sort a python dictionary values and rearrange keys?

Question:

Answers: