有沒有辦法使 if 語句更短？

Question

我讓用戶輸入過敏症狀並告訴用戶該症狀是過敏症狀，如果症狀在元組中，如果不是，那么它會告訴他們“不是過敏症狀。我曾嘗試使用元組並寫了“if user_input ==”tuple_name“：”，但它不起作用。有沒有辦法做到這一點？如果元組不起作用，有沒有辦法讓 if else 語句更短？

這是迄今為止的小程序：

# most common questions asked by patients at pharmacy
# allergy treatment
def symptoms(allergies):
while True:
    user_input = input("enter symptom: ")
    if user_input == "runny nose":
        print("allergy symptom")
    elif user_input == "itchy throat":
        print("allergy symptom")
    elif user_input == "watery eyes":
        print("allergy symptom")
    elif user_input == "itchy nose":
        print("allergy symptom")
    elif user_input == "trouble breathing":
        print("this is a severe allergic reaction. Call 911!")
        break
    elif user_input == "hives":
        print("this is an allergic reaction")
    elif user_input == "rash":
        print("this is an allergic reaction")
    elif user_input == "throat closing":
        print("this is a severe allergic reaction. Call 911!")
        break
    elif user_input == "swelling":
        print("this is an allergic reaction")
    else:
        print("not an allergy symptom.")


symptoms('allergies')


# Rph recommended otc products for mild allergic reactions and for allergies
def allergy_otc():
while True:
    pt_otc_age = input("Is the patient younger than 12 years old? ")
    if pt_otc_age == "yes":
        print("Recommended: Children's Zyrtec, Claritin, Allegra, & Benadryl")
    else:
        print("Recommended: Claritin, Zyrtec, Allegra, & Benadryl")
    pt_pregnancy_status = input("Is the patient pregnant? ")
    if pt_pregnancy_status == "yes":
        print("Recommended allergy medication for pregnant women would be Claritin.")
    else:
        break


allergy_otc()

Answer 1

in運算符檢查序列中是否存在 object。 所以：

allergy_symptoms = (
    "runny nose",
    "itchy throat",
    "watery eyes",
    "itchy nose"
 )
allergic_reactions = (
    "hives",
    "rash",
    "swelling"
)
severe_reactions = (
    "trouble breathing",
    "throat closing"
)

if user_input in allergy_symptoms:
    print("allergy symptom")
elif user_input in allergic_reactions:
    print("this is an allergic reaction")
elif user_input in severe_reactions:
    print("this is a severe allergic reaction. Call 911!")
    break
else:
    print("not an allergy symptom.")

Answer 2

將症狀分組在一個列表中，將反應分組在另一個列表中，然后使用if x in list （如果 x 存在於列表中）

symptoms = ["runny nose", "itchy throat", "watery eyes"]

user_input = input()

if user_input in symptoms:
    print("you are gonna die")
else:
    print("don't worry, be happy")

Answer 3

如果您要查找一個術語並返回一個值，使用字典會更方便。 嘗試這個：

# You should generally avoid "magic values" by declaring them up-front. This
# prevents hard-to-find typing errors
SEVERE_REACTION = "this is a severe allergic reaction. Call 911!"

# define the dictionary outside the get_diagnosis function to avoid 
# re-calculating it every time
symptoms = {
    "runny nose": "allergy symptom",
    "itchy throat": "allergy symptom",
    "watery eyes": "allergy symptom",
    "itchy nose": "allergy symptom",
    "trouble breathing": SEVERE_REACTION,
    "hives": "this is an allergic reaction",
    "rash": "this is an allergic reaction",
    "throat closing": SEVERE_REACTION,
    "swelling": "this is an allergic reaction" }

def get_diagnosis(user_input):
    try:
        return symptoms[user_input]
    # if the user_input isn't in the dictionary, catch the exception and return
    # a message
    except KeyError:
        return "not an allergy symptom."

def symptoms(allergies):
while True:
    user_input = input("enter symptom: ")
    
    diagnosis = get_diagnosis(user_input)
    print(diagnosis)

    if diagnosis == SEVERE_REACTION:
        break