如何查看字典/列表并获取位置和布尔值 [重复]

Posted 2023-02-23

【中文标题】如何查看字典/列表并获取位置和布尔值 [重复]

【英文标题】：How to Look inside a Dict/List and get the position and a bool [duplicate]

【发布时间】：2021-05-16 19:22:55

【问题描述】：

我想知道当有人进入时是否无论如何，比如说'070718604545'。它查找它，如果它在那里，它会在列表中打印其他人，示例如下

Patt = [
                'Phone': "0718604545", 'Name': "Tom", 'Age': '2007'
                'Phone': "0718123567", 'Name': "Katy", 'Age': '1998'
                'Phone': "0718604578", 'Name': "BillyW", 'Age': '1970'
                 ....
                 ....
                 ....
                 ....
                'Phone': "0714565778", 'Name': "Sony", 'Age': '1973'
            ]

他会输入'0718604545'作为例子。

x = input("Enter Phone")
Search for x in Patt[Phone]:
   name = Patt[Name] where Phone = x
   print(name)

所以答案应该是汤姆。

谢谢，

【问题讨论】：

如果多人拥有相同的电话号码，您期望什么结果？

【参考方案1】：

您可以通过遍历列表 patt 来获取号码，然后访问 patt 项目中的每个电话键，并使用 == 运算符与您要查找的电话号码进行比较。下面的函数可以完成这项工作。

def answer(search):
    for data in patt: # iterate over each item in patt
        if data["Phone"] == search: # compare the current value if it is the searched one
            return data["Name"] # return the name in the dictionary when the number is found
    return None # if it is not found, return None
print(answer('0718604545'))

【参考方案2】：

下面应该工作。对于每个项目，检查“电话”键是否具有匹配 x 的值。如果是，则返回 'name' 键的值。

x = input("Enter Phone")
for item in Patt:
  if item["Phone"] == x:
    print(item["Name"])

【参考方案3】：

在准备答案时，我没有注意到一个正确的答案已经发布并被接受。不过，在下面发布我的答案版本，增加了连续提问的功能和退出程序的能力。

Patt = [
    'Phone': "0718604545", 'Name': "Tom", 'Age': '2007',
    'Phone': "0718123567", 'Name': "Katy", 'Age': '1998',
    'Phone': "0718604578", 'Name': "BillyW", 'Age': '1970',
    'Phone': "0714565778", 'Name': "Sony", 'Age': '1973'
]

print(Patt)


def search_phone_records(user_phone):
    for record in Patt:  # Iterate over all the phone records in the dictionary
        if user_phone == record['Phone']:  # stop if found phone number in the dictionary
            return record['Name']  # Return user's name from the phone record
    return None


while True:
    user_input = input("Enter Phone or press 'x' to exit: ")

    if user_input in ('x', 'X'):
        print("Have a nice day!!! Thank you using our service!!!")
        break  # End the programme

    # Search for the phone number
    user_name = search_phone_records(user_input)

    #print("[0]".format(user_name))
    if type(user_name) == type(None):  # Phone number is not found
        print("Oops!!! Entered phone number (0) is not found in the dictionary!!!".format(user_input))
    else:  # Phone number is found
        print("Entered phone number (0) is found in the dictionary!!!".format(user_input))
        print("It is 1's phone number.".format(user_input, user_name))

使用字典理解的另一种解决方案：

Patt = [
    'Phone': "0718604545", 'Name': "Tom", 'Age': '2007',
    'Phone': "0718123567", 'Name': "Katy", 'Age': '1998',
    'Phone': "0718604578", 'Name': "BillyW", 'Age': '1970',
    'Phone': "0714565778", 'Name': "Sony", 'Age': '1973'
]

print(Patt)


def search_phone_records_using_dictionary_comprehension(user_phone):
    return 'Name': record['Name'] for record in Patt if user_phone == record['Phone']


while True:
    user_input = input("Enter Phone or press 'x' to exit: ")

    if user_input in ('x', 'X'):
        print("Have a nice day!!! Thank you using our service!!!")
        break  # End the programme

    result = search_phone_records_using_dictionary_comprehension(user_input)
    print("result = 0".format(result))

    if len(result) == 0:   # Phone number is not found
        print("Oops!!! Entered phone number (0) is not found in the dictionary!!!".format(user_input))
    else:  # Phone number is found
        print("Entered phone number (0) is found in the dictionary!!!".format(user_input))
        print("It is 1's phone number.".format(user_input, result['Name']))