Using if, elif and else in list comprehension in python3

Question

I want to convert a loop-based script containing if, elif and else into a list comprehension but I cannot figure out how to do it.

Here is the script I wrote (it prints the numbers from 1 to 100, but for multiples of 3 it prints 'fizz', for multiples of 5 it prints 'buzz' and for multiples of both 3 and 5 it prints 'fizzbuzz'):

for num in range(1, 101):
    if num % 3 == 0 and num % 5 == 0:
        print('fizzbuzz')
    elif num % 3 == 0:
        print('fizz')
    elif num % 5 ==0:
        print('buzz')
    else:
        print(num)

@Sayse I guess they want to re-factor their `for` loop as a comprehension, but it's not clear — Chris_Rands, Jul 17 '19 at 08:58
The question is to reduce the lines of code using list comprehension — Flahmez, Jul 17 '19 at 09:01
You shouldn't (ab)use list comprehensions here, because they're for quick one-off data transform/filter operations. Don't try to fit complex logic inside comprehensions. What you have here is just fine — abdusco, Jul 17 '19 at 09:04
You cannot use elif in comprehension lists but you can use if/else expression : `[(1 if x % 2 == 0 else 0) for x in range(100)]` — Louis Saglio, Jul 17 '19 at 09:08
You should reperate IO (this is what printing is) from logic. So what do you want? You want to have a list with numbers and `"fizzbuzz"` at certain places. Never use list comprehension to do IO. — Elmex80s, Jul 17 '19 at 09:20
Did you tried [`elif` in list comprehension conditionals](https://stackoverflow.com/q/9987483/6194097) question? — Kushan Gunasekera, Jul 17 '19 at 17:32

bonnal-enzo · Accepted Answer · 2022-06-20T10:15:38.423

4

elif is not part of the if-else short-hand, aka if else conditional operator, but you can achieve the same logic by chaining these operators, for example:

if A:
    v = a
elif B:
    v = b
else:
    v = c

turns into

v = a if A else b if B else c

And you can use a conditional operator expression in your list comprehension:

[a if A else b if B else c for something in someiterator]

Note that things can become unreadable quite quickly, for example this may not be recommended in your example:

['fizzbuzz' if num%3 == 0 and num%5 == 0  else 'fizz' if num%3 == 0 else 'buzz' if num%5 == 0 else num for num in range(1, 101)]

edited Jun 20 '22 at 10:15

answered Jul 17 '19 at 09:06

bonnal-enzo

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19

You should NOT use list comprehension unless you plan to use the resulting list. List comprehension is not just a way to do a one line for loop. At bare minimum two line it with a generator. If you want it in one line than just use `;` though the OP should break out of the mentality that one line of code is better than multiple. – Error - Syntactical Remorse Jul 18 '19 at 13:34

score 1 · Answer 2 · answered Jul 19 '19 at 17:51

List comprehension is not the right tool since you want to do something (print) and not to produce a list.

First, you should replace the "switch" by a function:

def foobar(num):
    if num % 3 == 0 and num % 5 == 0:
        return 'fizzbuzz'
    elif num % 3 == 0:
        return 'fizz'
    elif num % 5 ==0:
        return 'buzz'
    else:
        return str(num) # return a string to have a consistent return type

(You can make a one liner of this function as in @EnzoBnl's answer if you want, but that's not a good idea.). Now, you code looks like:

for num in range(1, 101):
    print(foobar(num))

If you want a list comprehension (a generator here), use:

print("\n".join(foobar(num) for num in range(1, 101)))

Using if, elif and else in list comprehension in python3

2 Answers2