Python Zodiac Sign Generator by Year[HELP]

[cr4zyfr0g]

Hello PHC Fam, Ano po kaya idadagdag ko dito na code? Ito po yung isa sa pinapagawa samin.
1.Year must be from included 1980-2014. Should the user inputs outside the range, the program will throw a ValueError.

Year = int(input("Year: "))
if (Year - 2000) % 12 == 0:
   chinesezodiac_sign = 'Dragon'
elif (Year - 2000) % 12 == 1:
    chinesezodiac_sign = 'Snake'
elif (Year - 2000) % 12 == 2:
    chinesezodiac_sign = 'Horse'
elif (Year - 2000) % 12 == 3:
    chinesezodiac_sign = 'sheep'
elif (Year - 2000) % 12 == 4:
    chinesezodiac_sign = 'Monkey'
elif (Year - 2000) % 12 == 5:
    chinesezodiac_sign = 'Rooster'
elif (Year - 2000) % 12 == 6:
    chinesezodiac_sign = 'Dog'
elif (Year - 2000) % 12 == 7:
    chinesezodiac_sign = 'Pig'
elif (Year - 2000) % 12 == 8:
    chinesezodiac_sign = 'Rat'
elif (Year - 2000) % 12 == 9:
    chinesezodiac_sign = 'Ox'
elif (Year - 2000) % 12 == 10:
    chinesezodiac_sign = 'Tiger'
else:
    chinesezodiac_sign = 'Hare'
print("Your Chinese Zodiac is", chinesezodiac_sign)

 

[zackmark29]

Ahm pwede mo siguro iadd yung pag check kung valid ba yung year
and pwede mo rin adopt yung ginawa ko instead of if else statement, dictionary nalang

while True:
    year = int(input('Input Year: '))
    if year in range(1980, 2014):
        break
    print('Year must be 1980-2014 only')

zodiacs = {
    0: 'Dragon',
    1: 'Snake',
    2: 'Horse',
    3: 'Sheep',
    4: 'Monkey',
    5: 'Rooster',
    6: 'Dog',
    7: 'Pig',
    8: 'Rat',
    9: 'Ox',
    10: 'Tiger'
}
zodiac_sign = (year - 2000) % 12
default_zodiac_sign = 'Hare'
chinese_zodiac_sign = zodiacs.get(zodiac_sign, default_zodiac_sign)

print('Zodiac Sign:', zodiac_sign)
print('Your Chinese Zodiac is:', chinese_zodiac_sign)

Output:

Input Year: 123
Year must be 1980-2014 only
Input Year: 1970
Year must be 1980-2014 only
Input Year: 1980
Zodiac Sign: 4
Your Chinese Zodiac is: Monkey

But since you want to throw an error, ganito gawin mo instead of While True

year = int(input('Input Year: '))
if year not in range(1980, 2014):
    raise ValueError(f'Invalid year: {year}. Year must be 1980-2014 only')

Input Year: 123
Traceback (most recent call last):
  File "C:\Users\ADMIN\Desktop\python\zodiac.py", line 4, in <module>
    raise ValueError(f'Invalid year: {year}. Year must be 1980-2014 only')
ValueError: Invalid year: 123. Year must be 1980-2014 only

 

[cr4zyfr0g]

Ahm pwede mo siguro iadd yung pag check kung valid ba yung year
and pwede mo rin adopt yung ginawa ko instead of if else statement, dictionary nalang

while True:
    year = int(input('Input Year: '))
    if year in range(1980, 2014):
        break
    print('Year must be 1980-2014 only')

zodiacs = {
    0: 'Dragon',
    1: 'Snake',
    2: 'Horse',
    3: 'Sheep',
    4: 'Monkey',
    5: 'Rooster',
    6: 'Dog',
    7: 'Pig',
    8: 'Rat',
    9: 'Ox',
    10: 'Tiger'
}
zodiac_sign = (year - 2000) % 12
default_zodiac_sign = 'Hare'
chinese_zodiac_sign = zodiacs.get(zodiac_sign, default_zodiac_sign)

print('Zodiac Sign:', zodiac_sign)
print('Your Chinese Zodiac is:', chinese_zodiac_sign)

Output:

Input Year: 123
Year must be 1980-2014 only
Input Year: 1970
Year must be 1980-2014 only
Input Year: 1980
Zodiac Sign: 4
Your Chinese Zodiac is: Monkey

But since you want to throw an error, ganito gawin mo instead of While True

year = int(input('Input Year: '))
if year not in range(1980, 2014):
    raise ValueError(f'Invalid year: {year}. Year must be 1980-2014 only')

Input Year: 123
Traceback (most recent call last):
  File "C:\Users\ADMIN\Desktop\python\zodiac.py", line 4, in <module>
    raise ValueError(f'Invalid year: {year}. Year must be 1980-2014 only')
ValueError: Invalid year: 123. Year must be 1980-2014 only

Maraming salamat lods

 

[zackmark29]

Maraming salamat lods

Gawin mo yung sa last na code instead of While True yun pala instruction sa inyo =D