Converting a list or string into a dictionary and returning only specific keys/values from that dictionary

Question

I've been working on a function that can convert a list/string into a dictionary and I'm able to run it on a bunch of different samples successfully. For instance, I took a list of the letters in the alphabet (az) and assigned them a number (1-26):

import string
mylist = list(string.ascii_lowercase)

def list_conversion(items):
    print(dict(enumerate(items, 1)))

so when I call list_conversion(mylist) I get the output that I'm looking for:

{1: 'a', 2: 'b', 3: 'c', 4: 'd', 5: 'e', 6: 'f', 7: 'g', 8: 'h', 9: 'i', 10: 'j', 11: 'k', 12: 'l', 13: 'm', 14: 'n', 15: 'o', 16: 'p', 17: 'q', 18: 'r', 19: 's', 20: 't', 21: 'u', 22: 'v', 23: 'w', 24: 'x', 25: 'y', 26: 'z'}

However, I'm not sure how to run this function and only return specific keys/values. For instance, I want to return on the values for odd number keys. I understand how to do this in an isolated instance:

d = {'x': 1, 'y': 2, 'z': 3} 
for key, value in d.items():
    if value%2 != 0:
        print(key)

But I can't figure out how to build this into my function correctly. How would I build such a for loop into my function that returns only even/odd values and more broadly, how can I return only specific keys/values that I choose from this function? I thought that something like this would work:

def list_conversion(items):
    print(dict(enumerate(items, 1)))

list_conversion(mylist)

for key, value in mylist.items():
    if value%2 != 0:
        print(key)

However, I get AttributeError: 'list' object has no attribute 'items' so I get the sense I'm going down the wrong path here.

Answer 1

To answer the more broad question, you need to provide some mechanism by which you can determine whether or not the key is allowable for your result. I would modify your function like so:

>>> import string
>>> 
>>> def list_conversion(alist, filterfunc = None):
...   # Handle the case where you just want all values
...   if filterfunc is None:
...     return dict(enumerate(alist, 1))
...   # Handle other cases where filterfunc is a callable
...   return {k: v for k, v in enumerate(alist, 1) if filterfunc(k)}
... 
>>> list_conversion(string.ascii_lowercase, lambda x: x % 2) # odds
{1: 'a', 3: 'c', 5: 'e', 7: 'g', 9: 'i', 11: 'k', 13: 'm', 15: 'o', 17: 'q', 19: 's', 21: 'u', 23: 'w', 25: 'y'}
>>> list_conversion(string.ascii_lowercase, lambda x: not x % 2) # evens
{2: 'b', 4: 'd', 6: 'f', 8: 'h', 10: 'j', 12: 'l', 14: 'n', 16: 'p', 18: 'r', 20: 't', 22: 'v', 24: 'x', 26: 'z'}

You can pass a more specific function in as well:

>>> def arbitrary(key):
...   wanted = [1,5,12,13,15]
...   return key in wanted
... 
>>> list_conversion(string.ascii_lowercase, arbitrary)
{1: 'a', 15: 'o', 12: 'l', 5: 'e', 13: 'm'}

Answer 2

You better take the print out of the function.

def list_conversion(items):
    return {i: x for i, x in enumerate(items, 1) if i % 2 == 1}

Answer 3

Various changes to your code. list_conversion should return a dictionary. No need to convert string.ascii_lowercase to a list, as enumerate will iterate over the elements of the string. Finally, I use a list comprehension for efficiency rather than a loop.

import string

def list_conversion(items):
    return dict(enumerate(items, 1))

mylist = list_conversion(string.ascii_lowercase)
[k for k, v in mylist.items() if k%2 != 0]

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