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python - 通过函数合并来自用户输入的两个字典

问题描述

我是 Python 新手，我正在编写一个函数，它可以合并来自不同用户输入的两个字典。它有效，但在我看来，我的代码不必要又长又麻烦。有没有办法让它更简单和流畅？这里的代码：

key1 = int(input("Give an integer as first key"))
key2 = int(input("Give an integer as second key"))

value1 = input("Give a a first value")
value2 = input("Give a second value")


class_list1 = {}
class_list2 = {}

class_list1[key1] = value1

class_list2[key2] = value2


def merge_dictionaries(x,y):
    z = {**x,**y}
    print("The merged dictionary is : ")
    return z
    
    

print(merge_dictionaries(class_list1, class_list2))

输出：

Give an integer as first key 1
Give an integer as second key 2
Give a a first value value1
Give a second value value2
The merged dictionary is : 
{1: 'value1', 2: 'value2'}

标签： pythonfunctiondictionary

解决方案

您可以将用户输入直接保存在字典中而无需合并：

userStorage = {}
for inputNum in range(2):
    # Temporary variables
    _key, _value = None, None
    
    while not (_key and _value):
        # both variables must have a value!
        # 'validate' at least the key as integer
        try:
            _key = int(input("Give an integer as key#%d:" % inputNum))
        except:
            print("No integer entered!")
            continue
        _value = input("Give a value for key#%d:" % inputNum)
        
        if _key and _value:
            userStorage[_key] = _value
            break
print(userStorage)

输出：

Give an integer as key#0:asd
No integer entered!
Give an integer as key#0:9
Give a value for key#0:foo
Give an integer as key#1:10
Give a value for key#1:bar
{9: 'foo', 10: 'bar'}

请注意，“最短”版本可能是（但现在代码容易出现错误的用户输入）：

userInput = lambda x, y: int(input("Give an integer as key#%d:" % x)) if y == 0 else input("Give a value for key#%d:" % x)
userStorage = {userInput(x, 0): userInput(x, 1) for x in range(2)}
print(userStorage)

输出：

Give an integer as key#0:87
Give a value for key#0:foo
Give an integer as key#1:88
Give a value for key#1:baz
{87: 'foo', 88: 'baz'}

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