通过功能合并用户输入中的两个字典

如何解决通过功能合并用户输入中的两个字典

我是Python的新手，我正在编写一个函数，该函数合并了来自不同用户输入的两个字典。它有效，但是在我看来，我的代码既不必要又冗长。有没有办法使它更简单和流畅？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'}

解决方法

尝试以下

key1 = int(input("Give an integer as first key"))
key2 = int(input("Give an integer as second key"))
x= y= {}
x[key1]=input("Give a a first value")
y[key2]=input("Give a second value")
print({**x,**y})

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'}

您的代码不太长，没有被重构。
看一下代码重构。

创建一个用于获取输入的功能，一个用于创建字典并使用主要功能的功能

         def merge_dictionaries(x,y):
            z = {**x,**y}
            print("The merged dictionary is : {0} ".format(z))
            return z

         def main:
             class_list1,class_list2 = function_input_dictionary() #function that returns dicts
             key1,key2 = function_input_key() #function that returns input of key
             value1,value2 = function_input_value() # function that returns values
             print(merge_dictionaries(class_list1,class_list2))

#look for design patterns and double return of parameters

以下是对原始文档的一些修整（简化功能并在可能的情况下包装input行）：

class_list1 = {}
class_list2 = {}

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

class_list1[key1] = input("Give a a first value")
class_list2[key2] = input("Give a second value")

def merge_dictionaries(x,y):
    return {**x,**y}

print("The merged dictionary is : {}".format(merge_dictionaries(class_list1,class_list2)))

以上是我在保留原始代码的逻辑的同时要保存行的方法。

如果您愿意采用其他方法，则可以选择一个input，然后使用split将其切成创建字典所需的部分。这里的代码更短，并且可以采用可变数量的key：value对，但是input的结构更复杂：

s = input('Enter int:string pairs,separated by commas\n').split(',')
d = {int(p.split(':')[0]) : p.split(':')[-1] for p in s}
print("The merged dictionary is : {}".format(d))

因此input中的1:a,2:b,3:c给出了{1: 'a',2: 'b',3: 'c'}

尝试以下一项，您会发现它很有用，因为在我的示例中，您可以通过修改名为length的变量来简单地合并两个以上的词典

示例

length = 2
key = 0
value = ''
class_list = {}
for i in range(1,length):
    key = int(input(f"Give an integer as {i} key: "))
    value = input(f"Give a a {i} value: ")
    class_list[key] = value


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

尝试像这样创建字典，而不要使用其他变量。并且不会更改预期的输出。

class_list1[int(input("Give an integer as first key"))] = input("Give a a first value")

class_list2[int(input("Give an integer as second key"))] = input("Give a second value")