replace None from list with two values

Question

I would like to replace None value with 0 or 1 depending on some external condition. My solution is:

my_list = [3, 4, 5, None, 6, 7, None, 8, None]
my_list = [(1 if external_string == "ONE" else 0) if v is None else v for v in my_list]

Is this a pythonic way to solve the problem? Let's suppose to have more than two possible value of string in order to assign, for example, 0 if external_string is "ZERO" , 1 if external_string if "ONE" , 2 if external_string is "TWO" and so on: in this case the way I write the code above is stylistically acceptable?

Answer 1

Since external_string doesn't change in the loop, so you can calculate the replacement value once:

replacement = 1 if external_string == 'ONE' else 0
my_list = [replacement if v is None else v for v in my_list]

If the external_string test uses data that changes or is more complex, just create a function:

def replace(value):
    if value is not None:
        return value
    return 1 if external_string == 'ONE' else 0

my_list = [replace(v) for v in my_list]

Don't try to cram everything into a list comprehension; readability counts!

For multiple options, consider using a dictionary mapping the external string to replacement values:

external_string_map = {'ONE': 1, 'TWO': 2}  # etc.
replacement = external_string_map.get(external_string, 0)
my_list = [replacement if v is None else v for v in my_list]

Answer 2

If there are a lot of options for external_string you better to use a dictionary that keeps all the possibilities then choose the proper replacing value using external_string :

all_options = {'ZERO':0, 'ONE':1, 'TWO':2, 'THREE': 3}

my_list = [all_options[external_string] if v is None else v for v in my_list]

Note that instead of using a direct indexing you can also use dict.get() method which will return None (by default) if the key doesn't exist in the dictionary or you can pass a custom value in order to be passed in case of lacking the key.