How to create a dictionary based on odd and even values from another dictionary?

Question

I have the following dictionary:

{'key1': [value1], 'key2': [value2], 'key3': [value3], 'key4': [value4] }

what I want to do is to use the values to create a new dictionary. The value of the odd elements will be the new keys, and the value of the next element will be the new value of that key. I mean:

{'value1': [value2], 'value3': [value4]}

how can i do it? i would like to clarify that this is a small example, since dictionaries have hundreds of elements, therefore the solution should be scalable to dictionaries with any amount of elements.

Answer 1

Assuming, that 1) no KeyError will occur, ie there are no odd-indexed dictionary values with the same values, and 2) you have even number of entries in your dict, I would do something like this:

vals = my_dict.values()
new_dict = {}
for i in range(0, len(vals), 2):
    new_dict[vals[i]] = vals[i+1]

But, as others pointed out, dictionaries don't actually have order, so it's a "let's say" solution.

Answer 2

Steps to follow:

1. Extract Keys Array from DictObj
2. List them to category Even and Odd Indexed Array

3. Create a dictionary out of these two array.

    dictObj = {'key1': 'value1', 'key2': 'value2', 'key3': 'value3', 'key4': 'value4' }; print (dictObj.keys()); evenArray = dictObj.keys()[1::2]; oddArray = dictObj.keys()[::2]; dictObjNew = dict(zip(evenArray, oddArray)) print dictObjNew