使用值从字典中删除一个键

Question

我可以用键从字典中删除一个值吗?函数del_contact应该只使用联系人的名字来删除contact，但不幸的是，我有字典和值，但没有键。如何解决这个问题?

my_contacts = {
1: {
"Name": "Tom Jones",
"Number": "911",
"Birthday": "22.10.1995",
"Address": "212 street"
},
2: {
"Name": "Bob Marley",
"Number": "0800838383",
"Birthday": "22.10.1991",
"Address": "31 street"
}
}

def add_contact():
user_input = int(input("please enter how many contacts you wanna add: "))
index = len(my_contacts) + 1
for _ in range(user_input):
details = {}
name = input("Enter the name: ")
number = input("Enter the number: ")
birthday = input("Enter the birthday")
address = input("Enter the address")
details["Name"] = name
details["Number"] = number
details["Birthday"] = birthday
details["Address"] = address
my_contacts[index] = details
index += 1
print(my_contacts)

def del_contact():
user_input = input("Please enter the name of the contact you want to delete: ")
my_contacts.pop(user_input)

add_contact()
print(my_contacts)

问题是我的字典键是1或Name，我希望能够仅使用Name的值来删除联系人。

Answer 1

基本上你能做的就是迭代字典并只保存不包含用户名的键。

下面的代码可以达到这个目的:

my_contacts = {1: {"Name": "Tom Jones",
"Number": "911",
"Birthday": "22.10.1995",
"Address": "212 street"},
2: {"Name": "Bob Marley",
"Number": "0800838383",
"Birthday": "22.10.1991",
"Address": "31 street"}
}
user_input = "Tom Jones"
my_contacts = {key: value for key, value in my_contacts.items() if value["Name"] != user_input}

Answer 2

这里有两个步骤:

1. 查找与
匹配的键/值对
2. 删除找到的密钥

如果你只删除一个条目(即使有多个匹配项)，它们总是可以组合在一起:

def del_contact():
user_input = input("Please enter the name of the contact you want to delete: ")
for k, v in my_contacts.items():  # Need both key and value, we test key, delete by value
if v["Name"] == user_input:
del my_contacts[k]
break  # Deleted one item, stop now (we'd RuntimeError if iteration continued)
else:
# Optionally raise exception or print error indicating name wasn't found

如果您可能删除多个条目，并且必须在适当的位置操作，那么您将拆分步骤(因为在更改键集的同时继续迭代dict是非法的):

def del_contact():
user_input = input("Please enter the name of the contact you want to delete: ")
to_delete = [k for k, v in my_contacts.items() if v["Name"] == user_input]
if not to_delete:
# Optionally raise exception or print error indicating name wasn't found
for k in to_delete:
del my_contacts[k]

或者，如果您愿意替换原始的dict，而不是对其进行就地修改，则可以将多次删除的情况一行写成:

def del_contact():
global my_contacts  # Assignment requires explicitly declared use of global
user_input = input("Please enter the name of the contact you want to delete: ")
my_contacts = {k: v for k, v in my_contacts.items() if v["Name"] != user_input}  # Flip test, build new dict
# Detecting no matches is more annoying here (it's basically comparing
# length before and after), so I'm omitting it

Answer 3

我想这就是你需要的，希望这对你有帮助。

代码:

my_contacts = {1: {"Name": "Tom Jones",
"Number": "911",
"Birthday": "22.10.1995",
"Address": "212 street"},
2: {"Name": "Bob Marley",
"Number": "0800838383",
"Birthday": "22.10.1991",
"Address": "31 street"}
}
def add_contact():
user_input = int(input("please enter how many contacts you wanna add: "))
index = len(my_contacts) + 1
for _ in range(user_input):
details = {}
name = input("Enter the name: ")
number = input("Enter the number: ")
birthday = input("Enter the birthday")
address = input("Enter the address")
details["Name"] = name
details["Number"] = number
details["Birthday"] = birthday
details["Address"] = address
my_contacts[index] = details
index += 1
print(my_contacts)
add_contact()
print(my_contacts)
def del_contact():
user_input = input("Please enter the name of the contact you want to delete: ")
values = [key for key in my_contacts.keys() if user_input in my_contacts[key].values()]
for value in values:
my_contacts.pop(value)
del_contact()
print(my_contacts)

输入:

{1: {'Name': 'Tom Jones', 'Number': '911', 'Birthday': '22.10.1995', 'Address': '212 street'}, 2: {'Name': 'Bob Marley', 'Number': '0800838383', 'Birthday': '22.10.1991', 'Address': '31 street'}, 3: {'Name': 'myname', 'Number': '1233455', 'Birthday': '12-12-22', 'Address': 'blabla street'}}

Please enter the name of the contact you want to delete: Bob Marley

输出:

{1: {'Name': 'Tom Jones', 'Number': '911', 'Birthday': '22.10.1995', 'Address': '212 street'}, 3: {'Name': 'myname', 'Number': '1233455', 'Birthday': '12-12-22', 'Address': 'blabla street'}}

Answer 4

由于索引用作字典的键，因此range函数可以如下使用:

def del_contact():
user_input = input("Please enter the name of the contact you want to delete: ")
for i in range(1,len(my_contacts)+1):
if my_contacts[i]['Name'] == user_input:
del my_contacts[i]
break

如果可以有多个相同名称的联系人需要删除，则删除break语句。

但是正如在评论中提到的，没有必要使用索引作为字典的键。您可能会遇到的一个潜在错误是，如果删除了名称为Tom Jones的第一个条目，则dict的长度为1，只有一个键- 2。然后当您尝试添加更多联系人时，当您检查字典index = len(my_contacts) + 1的长度时，由于长度为1，索引将为2。因此，my_contacts[index] = details将使用密钥2或"Name": "Bob Marley"更新联系人，而不是添加新联系人。