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I have a string of integers, such as string = {11, 2, 14, 1, 17, 1, 123, 1} What I need to do is turn it into a list, and then sort them into two separate lists of odd and even elements.

I had a dictionary that looked like {11: 2, 14: 1, 17: 1, 123: 1}, converted it to a string, and removed the colons into commas, so now I have the first string. I did this so I can split up the elements into having the numbers 11, 14, 17, and 123 in one list, and 2, 1, 1, and 1 in the other list.

I think I can convert this string into a list, and then use a loop to append each element into the list I need it to be. The problem is, I can't figure out how to turn this string into a list again. How would I do that? Alternatively, can I split the values the way I want straight from the dictionary and save steps? Thanks for any help

Yes, you can split the values straight from the dictionary! You'd use the .keys() and .values() on the dictionary object.

>>> obj = {11: 2, 14: 1, 17: 1, 123: 1}

>>> list(obj.keys())

[123, 17, 11, 14]

>>> list(obj.values())

[1, 1, 2, 1]

String conversion isn't necessary. In fact, there's no need to create a list of all keys for your problem. You can use collections.defaultdict and iterate your dictionary:

from collections import defaultdict



x = {11: 2, 14: 1, 17: 1, 123: 1}



dd = defaultdict(list)



for key in x:

    dd['odd' if key % 2 else 'even'].append(key)

The result is a dictionary mapping of odd and even keys:

defaultdict(list, {'odd': [11, 17, 123],

                   'even': [14]})

You can then access odd keys via dd['odd'], even keys via dd['even'].

To split a string into ints, use the built-in split method to return a list of each of the items:

new_list = string.split(", ")