当我只需要一个时从我的字典中获取多个输出答案

【问题标题】：Getting multiple outputs from my dictionary when I need only one当我只需要一个时从我的字典中获取多个输出
【发布时间】：2022-01-15 02:08:53
【问题描述】：

我只打印存储在字典中的相应缩写有点问题

user = input("Enter a Abbreviation: ")

dictionary = {"ADSL": "Application Programming Interface",
              "IDE": "Integrated Development Enviroment",
              "SDK": "Software Development Kit",
              "UI": "User Interface",
              "UX": "User eXperience",
              "OPP": "Object Oriented Programming"
              }

for x in dictionary:
    if user in x:
        print(user + ":" + " " + dictionary[x])

    elif x != dictionary:
        print("Abbreviation not found")

这是我的输出

Enter a Abbreviation: UI
Abbreviation not found
Abbreviation not found
Abbreviation not found
UI: User Interface
Abbreviation not found
Abbreviation not found

我只需要输入的缩写键值而不是所有的 Abbreviation not found 输出，我希望这是有道理的

【问题讨论】：

什么你不只是删除elif 部分？也许您也可以尝试输入您想要的输出。

标签： python dictionary for-loop

【解决方案1】：

字典的好处是它是“索引的”，所以你不需要找到关键。您正在遵循数组方法，这是一种错误的方法。

user = input("Enter a Abbreviation: ")

dictionary = {"ADSL": "Application Programming Interface",
              "IDE": "Integrated Development Enviroment",
              "SDK": "Software Development Kit",
              "UI": "User Interface",
              "UX": "User eXperience",
              "OPP": "Object Oriented Programming"
              }

if user in dictionary:
   print(user + ":" + " " + dictionary[user])
else:
   print("Abbreviation not found")

您可以在 google 上找到有关字典 here、here 的更多信息。知道何时使用数组、字典、列表、队列非常重要……

【讨论】：

【解决方案2】：

也许就是这样：

user = input("Enter a Abbreviation: ")

dictionary = {"ADSL": "Application Programming Interface",
              "IDE": "Integrated Development Enviroment",
              "SDK": "Software Development Kit",
              "UI": "User Interface",
              "UX": "User eXperience",
              "OPP": "Object Oriented Programming"
              }

for x in dictionary:
    if user in x:
        print(user + ":" + " " + dictionary[x])
    else:
        continue

【讨论】：

【解决方案3】：

我尝试了这段代码，它可以工作...看起来您正在检查所有选项...

user = input("Enter a Abbreviation: ")

dictionary = {"ADSL": "Application Programming Interface",
              "IDE": "Integrated Development Enviroment",
              "SDK": "Software Development Kit",
              "UI": "User Interface",
              "UX": "User eXperience",
              "OPP": "Object Oriented Programming"
              }

found = 0
for x in dictionary:
    if user in x:
        found = 1
        print(user + ":" + " " + dictionary[x])

if found == 0:
    print("Abbreviation not found")

【讨论】：

【解决方案4】：

看来你需要dict.get。

user = input("Enter a Abbreviation: ")

dictionary = {"ADSL": "Application Programming Interface",
              "IDE": "Integrated Development Enviroment",
              "SDK": "Software Development Kit",
              "UI": "User Interface",
              "UX": "User eXperience",
              "OPP": "Object Oriented Programming"
              }

result = dictionary.get(user.upper())
if result:
    print(user + ":" + " " + result )
else:
    print("Abbreviation not found")

如果您有key，则无需循环

【讨论】：