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 · 2019-07-17T09:13:10.263

2

You can't use elif in list comprehension because it's not part of the if-else short expression syntax.

But you can get the same logic with chaining:

if b1:
    a
elif b2:
    b
else:
    c

becomes

a if b1 else b if b2 else c

So this do the trick for you :

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

Note: it is totally discouraged to use such unreadable list comprehensions in real life projects !

edited Jul 17 '19 at 09:13

answered Jul 17 '19 at 09:06

bonnal-enzo

1,055
8
17

A pointless list of `None`s vs. a pointless more heavy list of `str`s objects that are just aimed to be printed. About readability: for me list comprehension are maily usefull for one-line readability, because if you have to split it in several lines, you would rather initialize a list of *n* `None`s and fill it with a regular loop because it looks standard and it is faster (slightly but still). – bonnal-enzo Jul 17 '19 at 09:58
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
I agree with you. But I think the author question was in a study/training context. Maybe an exercice to play with python syntax possibilities or something like that. So for me it does have sense to wonder how to convert some regular loops into list comprehension, even if it's not pythonic or fast or good looking in this specific case. – bonnal-enzo Jul 18 '19 at 13:41

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