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Exercise 6

Let's try another variation on the even/odd-numbers theme.

We'll return to the simpler one-dimensional version of my_list. In this problem, you should write code that creates a new list with one element for each number in my_list. If the original number is an even, then the corresponding element in the new list should contain the string 'even'; otherwise, the element should contain 'odd'.

my_list = [
    1, 3, 6, 11,
    4, 2, 4, 9,
    17, 16, 0,
]

# pretty-printed for clarity
[
    'odd', 'odd', 'even', 'odd',
    'even', 'even', 'even', 'odd',
    'odd', 'even', 'even'
]

Solution

my_list = [
    1, 3, 6, 11,
    4, 2, 4, 9,
    17, 16, 0,
]

result = []
for number in my_list:
    if number % 2 == 0:
        result.append('even')
    else:
        result.append('odd')

print(result)

Our approach is straightforward: we iterate over all the numbers in the list and check whether each is even. Based on the result, we append either 'even' or 'odd' to the result list.

You may have struggled if you tried to use a list comprehension for this problem. Since comprehensions don't have an else capability, trying to generate 'even' for some values and 'odd' for others is challenging. You can use a ternary expression in the comprehension, but this is a little confusing visually:

my_list = [
    1, 3, 6, 11,
    4, 2, 4, 9,
    17, 16, 0,
]

#highlight
result = [ 'even' if number % 2 == 0 else 'odd'
           for number in my_list ]
#endhighlight
print(result)

On line 7, we've used a ternary expression to choose between the two values. The ternary is equivalent to:

if number % 2 == 0:
    return 'even'
else:
    return 'odd'

A cleaner approach is to use a helper function to determine whether we should add 'even' or 'odd' to the new list:

my_list = [
    1, 3, 6, 11,
    4, 2, 4, 9,
    17, 16, 0,
]

#highlight
def odd_or_even(number):
    return 'even' if number % 2 == 0 else 'odd'

result = [ odd_or_even(number)
#endhighlight
           for number in my_list ]
print(result)

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