如何对多个 if elif 和 else 语句使用 python 列表推导式

Question

I am trying to write using list comprehensions for the scenario.

I want fizzbuzz if it is divisible by both 3 and 5; fizz for divisble by 3 and buzz if divisible by 5.

It seems like in my logic I have to use else logic in the end using '' to get the output. if I don't use the last else I am getting an error. I am sure I am missing something silly.

['fizzbuzz' if (i%3==0 and i%5==0) else 'fizz' if i%3==0 else 'buzz'if i%5==0 else ''   for i in range(1,51) ]

Answer 1

Sorry, I didn't read the question completely; if none of the scenarios match then we need to print the number. so the solution I wrote needed an 'i' to print in the default scenarios.

print(['fizzbuzz' if (i%3==0 and i%5==0) else 'fizz' if i%3==0 else 'buzz'if i%5==0 else i   for i in range(1,51) ])

Answer 2

Note: this answer was created before OP changed their question, so you might need to tweak the code you use. The process that you use will be very similar to the write-up below, so I figure I should keep this answer posted for now.

As a disclaimer, this answer is more from a code golf perspective than a practical perspective. Don't try to use this approach an actual project (or even a job interview) because your code will be too confusing to understand.

First, we need something to test our modifications against, because these modifications can be very error-prone. Feel free to ignore this, it's just so we can compare the list comprehensions we develop to the expected output:

list(filter(None, ['fizzbuzz' if (i%3==0 and i%5==0) else 'fizz' if i%3==0 else 'buzz'if i%5==0 else '' for i in range(1,51)]))

---

Now, let's try to make the comprehension shorter :)

You can apply a filter operation when constructing lists by placing an if statement on the right side of the range statement:

['fizzbuzz' if (i%3==0 and i%5==0) else 'fizz' if i%3==0 else 'buzz' for i in range(1,51) if (i%5==0 or i%3 == 0)]

It seems like you are trying to optimize for code length (which can be pretty fun sometimes so I totally get it). If this is the case, we can further reduce your comprehension.

First, Python evaluates 1 as true and 0 as false. This allows us to re-write (i%3==0 and i%5==0) as (not i%3 and not i%5) .

Next, we can apply DeMorgan's Law:

(not i%3 and not i%5) EQUALS not (i%3 or i%5) .

By applying this discovery we can get a decent character reduction in our original statement, which is pretty neat:

['fizzbuzz' if not (i%3 or i%5) else 'fizz' if i%3==0 else 'buzz' for i in range(1,51) if not (i%5 and i%3)]

Testing against the original spec we set up at the top:

list(filter(None, ['fizzbuzz' if (i%3==0 and i%5==0) else 'fizz' if i%3==0 else 'buzz'if i%5==0 else '' for i in range(1,9999)])) == ['fizzbuzz' if not (i%3 or i%5) else 'fizz' if i%3==0 else 'buzz' for i in range(1,9999) if not (i%5 and i%3)]
# evaluates to True :)

---

Another trick we can use to clean things up is the fact that Python allows us to multiply strings by integers (try in in interactive mode and see what happens):

['fizz'*(i%3==0) + 'buzz'*(i%5==0) for i in range(1,51) if not (i%5 and i%3)]

Testing against the original spec we set up at the top:

list(filter(None, ['fizzbuzz' if (i%3==0 and i%5==0) else 'fizz' if i%3==0 else 'buzz'if i%5==0 else '' for i in range(1,9999)])) == ['fizz'*(i%3==0) + 'buzz'*(i%5==0) for i in range(1,9999) if not (i%5 and i%3)]
# evaluates to True :)

---

If anybody can reduce the line count any further (without just shortening variables/whitespace), feel free to edit this answer, since that seems to be the goal for this question.