Python FizzBuzz

问题:

I have been given this question to do in Python:

Take in a list of numbers from the user and run FizzBuzz on that list.

When you loop through the list remember the rules:

1) If the number is divisible by both 3 and 5 print FizzBuzz

2) If it's only divisible by 3 print Fizz

3) If it's only divisible by 5 print Buzz

4) Otherwise just print the number

Also remember elif!

I have the following script created, but it gives me an error at if n%3=True

n=input()
    if n%3=True:
        print("Fizz")
    else if n%5=True:
        print ("Buzz")
    elif print n

Can anyone help? Thank you very much!

回答1:

A few issues with your code here. The first issue is that, for comparison, you should be using ==, not =, which is for assignment.

The second issue is that you want to check that the remainder of the divisions (which is what the modulo operator calculates) is zero, not that it's true, which doesn't really make sense.

You should be using elif for "otherwise if..." and else for "otherwise." And you need to fix the formatting of your else clause.

You want:

n=input()
if n%3 == 0:
    print("Fizz")
elif n%5 == 0:
    print ("Buzz")
else:
    print n

Finally, your code does not meet the spec:

1) If the number is divisible by both 3 and 5 print "FizzBuzz"

The above will not do this. This part I'm going to leave to you because I'm not here to solve the assignment for you :)

回答2:

Based on this

FizzBuzz: For integers up to and including 100, prints FizzBuzz if the integer is divisible by 3 and 5 (15); Fizz if it's divisible by 3 (and not 5); Buzz if it's divisible by 5 (and not 3); and the integer otherwise.

def FizzBuzz():
    for i in range(1,101):
        print {
            3 : "Fizz",
            5 : "Buzz",
            15 : "FizzBuzz"}.get(15*(not i%15) or
                                 5*(not i%5 ) or
                                 3*(not i%3 ), '{}'.format(i))

The .get() method works wonders here.

Operates as follows

For all integers from 1 to 100 (101 is NOT included),
print the value of the dictionary key that we call via get according to these rules.

"Get the first non-False item in the get call, or return the integer as a string."

When checking for a True value, thus a value we can lookup, Python evaluates 0 to False. If i mod 15 = 0, that's False, we would go to the next one.

Therefore we NOT each of the 'mods' (aka remainder), so that if the mod == 0, which == False, we get a True statement. We multiply True by the dictionary key which returns the dictionary key (i.e. 3*True == 3)

When the integer it not divisible by 3, 5 or 15, then we fall to the default clause of printing the int '{}'.format(i) just inserts i into that string - as a string.

Some of the output

Fizz
79
Buzz
Fizz
82
83
Fizz
Buzz
86
Fizz
88
89
FizzBuzz
91
92
Fizz
94
Buzz
Fizz
97
98
Fizz
Buzz

回答3:

n % 3 (or n % any number) does not evaluate to True or False, it's not a Boolean expression. n % 3 == 0 on the other hand, does.

As an aside, what happens when n % 3 == 0 and n % 5 == 0 both evaluate to True?

回答4:

Make it universal for any integer, positive or negative. Also make it easily expandable to other keywords for any integer by creating a dictionary of keywords.

def checkdict(divdict, i):
    out = ""
    for key in divdict:
        if key != 0:
            if i%key==0:
                out+=divdict[key]
        if key == 0 and i == 0:
                out+=divdict[key]
    if out == "":
        out = i
    print(out)

if __name__ == "__main__":
    mydict = {3:"Fizz",5:"Buzz"}
    for i in range(-50,50):
        checkdict(mydict, i)

回答5:

def check(num):
    finalWord = ''
    for k,v in numWordDict.items():
        if num % k == 0:
            finalWord += v

    if not finalWord:
        return num
    else: 
        return finalWord

def FizzLoop(start=0, stop=10, step=1):
    for i in range(start, stop, step):
        print(check(i))


numWordDict = {3:'fizz', 6:'buzz', 5:'fiver'}

FizzLoop(0, 10)
print("----------")
FizzLoop(0, 50, 5)

回答6:

one of the shortest answers i have found is

c=1
while c<101:print((c%3<1)*'Fizz'+(c%5<1)*'Buzz'or c);c+=1

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