Question

我需要帮助。

a = ["cat","dog","fish","hamster"]

 user = raw_input("choose your fav pet ")

if user == a[0]:

    print a[0]

elif user == a[1]:

    print a[1]

elif user == a[2]:

    print a[2]

elif user == a[3]:

    print a[3]

else:

    print "sorry, the aninimal you type does not exist"

我想要做的是测试移动应用程序，所以我使用动物作为测试。该程序确实有效，但问题是世界上有超过100种动物，我将它们放在一个列表中，我不想创建许多elif语句。

有没有办法让它更短更快？

Answer 1

使用for循环：

for animal in a:
    if user == animal:
        print animal
        break
else:
    print "Sorry, the animal you typed does not exist"

但是我注意到这段代码有点傻。如果您发现匹配用户条目的动物打印时，您只需检查该条目是否在a列表中，如果是print user，则可以：

if user in a:
    print user
else:
    print "Sorry, the animal you typed does not exist"

Answer 2

我会选择：

if user in a:
    print user

这将检查user输入是否在宠物列表中，如果是，则会打印出来。

Answer 3

在运营商中使用：

if user in a:
    print user
else : 
    print "sorry, the aninimal you type does not exist"

Answer 4

 a = ["cat","dog","pig","cat"]
    animal = input("enter animal:")

    if animal in a:
        print (animal , "found")
    else:
        print ("animal you entered not found in the list")

以上代码适合您的要求。

Answer 5

我想添加一种新的方式，我无法猜测每个身体是如何错过的。我们必须使用不区分大小写的标准检查字符串。

a = ["cat","dog","fish","hamster"]

user = raw_input("choose your fav pet ")
// adding for user case where user might have entered DOG, but that is valid match
matching_obj = [data for data in a if data.lower() == user.lower()]

if matching_obj:
    print("Data found ",matching_obj)
else:
    print("Data not found")

写少elif陈述

5 个答案: